Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Define possion distribution.
Solution:
Poisson distribution was derived in 1837 by a French Mathematician Simeon D. Poisson.
A random variable X is said to follow a Poisson distribution with parameter X if it assumes only non-negative values and its probability mass function is given by

Question 2.
Write any 2 examples for possion distribution
Solution:
Examples of Poisson distribution are given by

• The number of printing mistakes per page in a textbook.
• A number of lightning per second.
• The number of bacteria in one cubic centimetre.

Question 3.
Write the condition for which the Poisson distribution is limiting case of binomial distribution
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:
(i) n, the number of trials is indefinitely large i.e n → ∞
(ii) p, the constant probability of success in each trial is very small, i.e. p → 0.
(iii) np = λ is finite. Thus p = \(\frac { λ }{n}\) and q = 1 – (\(\frac { λ }{n}\))
where λ, is a positive real number.

Question 4.
Derive the mean and variance of the Poisson distribution.
Solution:
Derivation of Mean and variance of Poisson distribution

Variance (X) = E(X²) – E(X)²

Variance (X) = E(X²) – E(X)²
= λ² + λ – (λ)²
= λ

Question 5.
Mention the properties of Poisson distribution.
Solution:

1. Poisson distribution is the only distribution in which the mean and variance are equal.
2. The probability that an event occurs in a given time, distance, area, or volume is the same.

Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? Give e^(-2.8) = 0.06
Solution:
Since the mortality rate for a certain disease in 7 in loop
∴ p = \(\frac { 7 }{1000}\) and n = 400
The value of mean A = λp = 400 × \(\frac { 7 }{1000}\)
∴ λ = 2.8
Let x be a random variable following
distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
∴ the distribution is p(x = 2) = \(\frac { e^{-2-8}(2.8)^2 }{2!}\)
\(\frac { 0.06×7.84 }{2}\)
= 0.2352

Question 7.
Mention the properties of Poisson distribution.
Solution:
Given p = \(\frac{5}{100}\) = 0.05 and n = 120
⇒ λ = np = (0.05) (120) = 6
Thus X is a Poisson random variable with P (X = x) = \(\frac{e^{-6} 6^{x}}{x !}\)
We want P (no defective bulb) = P (X = 0)
= \(\frac{e^{-6} 6^{0}}{0 !}\)
= e^-6
= 0.0025 (Using exponent table)
Thus the probability that a sample of 120 bulbs will not contain any defective bulb is 0.0025.

Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poisson variate, with a mean of 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused
Solution:
In a poisson distribution n=2
mean λ = 1.5
x follows poison distribution
With in p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(neither car is used) = p(x = 0)
\(\frac { e^{-1.5}(1.5)^0 }{0!}\) = e^-1.5
= 0.2231

(ii) p(some demand is refused) = p(x > 2)

= 1 – 0.2231 [1 + 1.5 + 1.125]
= 1 – 0.2231 [3.625]
= 1-0.8087
= 0.1913

Question 9.
The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be (i) no phone at all (ii) exactly 3 calls (iii) atleast 5 calls
Solution:
The average number of phone cells per minute into the switch board of a company is λ = 2.5
x follows poisson distribution with

(iii) p(atleast 5 calls) = p(x ≥ 5)
= p(x = 5) + p(x = 6) + …………..
= 1 – p(x < 5)

Question 10.
The distribution of the number of road accidents pre-day in a city is Poisson with mean 4. Find the number of days out of 100 days when there will be (i) no accident (ii) atleast 2 accidents and (iii) at most 3 accidents.
Solution:
In a possion distribution
mean λ = 4
n = 100
x follows possion distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) = \(\frac { e^{-4}(4)^x }{x!}\)
(i) p(no accident) = p(x = 0)
= \(\frac { e^{-4}(4)^0 }{0!}\) = e^-4 = 0.0183
out of 100 days there will be no accident
= n × p(x = 0)
= 100 × 0.0183 = 1.83
= 2 days (approximately)

(ii) p(atleast 2 accidents)
= p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 4) + …………
= 1 – p(x < 2)
= 1 – [p(x = 0) + p(x = 1)]
= 1 – [\(\frac { e^{-4}(4)^0 }{0!}\) + \(\frac { e^{-4}(4)^1 }{1!}\)]
= 1 – e^-4 [l + 4]
= 1 – 0.0183(5) = 1 – 0.0915
= 0.9085
= Out of 100 days there will be atleast 2 accidents = n × p(x ≥ 2)
= 100 × 0.9085
= 90.85
= 91 days (approximately)

(iii) p(atmost 3 accident) = p(x ≤ 3)
= p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

out of 100 days there will be at most 3 acccident = n × p(x ≤ 3)
= 100 × 0.4331
= 43.31
= 43 days(approximately)

Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200/ calculate the probability that in a factory employing 300 workers there will be atleast two fatal accidents in year, (given e^-0.25 = 0.7788 Solution:
Let p be the probability of a fatal accident in a factory during the yeart
p = \(\frac { 1 }{1200}\) and n = 300 1200
λ = np = 300 × \(\frac { 1 }{1200}\) = \(\frac { 1 }{4}\)
λ = 0.25
x follows poison distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) + \(\frac { e^{-0.25}(0.25) }{x!}\)
p(atleasttwo fatal accidents) = p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 3) + p(x = 4) + ……….
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}
= 1 – 0.7788 [1 + 0.25]
= 1 – 0.7788(1.25)
= 1 – 0.9735 = 0.0265
∴ p(x ≥ 2) = 0.0265

Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute (i) No customer appears (ii) three or more customers appear.
Solution:
The average number of customers ,who appear in a counter of a certain bank per minute = 2
∴ λ = 2
x follows poisson distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
x follows poisson distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(no customber appears) = p(x = 0)
= \(\frac { e^{-2}(2)^0 }{0!}\) = e^-2
= 0.1353

(ii) p(three or more customers appears) = p(x ≥ 3)
= p(x = 3) + p{x = 4) + p(x = 5) + ……
= 1 – p(x < 3)
= 1 – {p(x = 0) + p(x = 1) + p(x = 2)}

= 1 – 2^-2 [1 + 2 + 2]
= 1 – 0.1353(5)
= 1 – 0.6765
= 0.3235

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1.
Define Biomial distribution.
Solution:
Binomial distribution was discovered by James Bernoulli (1654-1705) in the year 1700.
A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non-negative value and its probability mass function in given by

Question 2.
Define Bernoulli trials.
Solution:
A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q, is called a Bernoulli trial.
Example 1, Tossing of a coin (Head or Tail)
Example 2, Writing an exam (Pass or Fail)

Question 3.
Derive the mean and variance of binomial distribution
Solution:
Derivation of the Mean and Variance of Binomial distribution:
The mean of the binomial distribution

= np(q + p)n – 1 [since p + q = 1]
= np
E(X) = np
The mean of the binomial distribution is np.
Var(X) = E(X²) – E(X²)

= n(n – 1)p²(q + p)(n – 2) + np
n(n – 1 )p² + np
Variance = E(X²) – [E(X)]²
= n²p² – np² + np – n²p²
= np(1 – p) = npq
Hence, the mean of the BD is np and the Variance is npq.

Question 4.
Write down the condition for which the binomial distribution can be used.
Solution:
The binomial distribution can be used under the following conditions:

• The number of trials (or) observations ‘n’ is fixed (finite).
• Each observation is independent of the other.
• In every trial, there are only two possible outcomes – success or failure.
• The probability of success ‘p’ is the same for each outcome.

Question 5.
Mention the properties of the binomial distribution.
Solution:
Properties of Binomial distribution
1. Binomial distribution is symmetrical if p = q = 0.5, It is skew-symmetric if p ≠ q. It is positively skewed if p < 0.5 and it is negatively skewed it p > 0.5.
2. For Binomial distribution, variance is less than mean
Variance npq = (np) q < np

Question 6.
If 5% of the items produced turn out to be defective, then find out the probability that out of 10 items selected at random there are
(i) exactly three defectives
(ii) atleast two defectives
(iii) exactly 4 defectives
(iv) find the mean and variance.
Solution:
Probability of getting a defective item

In binomial distribution
p(X = x) = nC_xp^xq^n-x

Let x = (19)²
log x = 7 log(19)⁷
= 7 log 19
= 7 × 1.2788
= 8.9516
x = Antilog 8.956
= 8.945 × 10⁸
ut y = (20)¹⁰
log y = (10) log 20
= 10 × 1.3010
= 13.010
y = Anti log 13.010
= 1.023 × 10³

(ii) p(atleast two defectives)
= p(x ≥ 2)
= p(x = 2) = p(x = 3) + ………….. + p(x = 10)
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}

Let x = (19)⁹
log x = 9 log 19
= 9 × 1.2788
= 11.5092
x = Antilog 11.5092
= 3.229 × 1011

(iii) p(extactly 4 defectives) = p(X = 4)

Let x = (19)⁶
log x = 6 log 19
= 6 × 1.2788
log x = 7.66728
Antilog (7.66728)
x = 4.648 × 10⁷

(iv) mean E(x) = np
= 10 × \(\frac { 1 }{20}\) = \(\frac { 1 }{2}\) = 0.5
Varaince = npq
= 10 × \(\frac { 1 }{20}\) = \(\frac { 19 }{20}\) = \(\frac { 19 }{40}\) = 0.475

Question 7.
In a particular university 40% of the students are having news paper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that
(i) none of those selected have news paper reading habit
(ii) all those selected have news paper reading habit
(iii) atleast two third have news paper reading habit.
Solution:
let p to the probability of having newspaper reading habit
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1\(\frac { 2 }{5}\) = \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\) and n = 9
In the binomial distribution p(x = 4) = nc_x p^xq^n-r
The binomial distribution P(x) = 9c_x (\(\frac { 2 }{5}\))^x (\(\frac { 3 }{5}\))^9-x

(i) p(none of those selected have newspaper reading

(ii) P(all those selected have newspaper reading habit)

(iii) p(at least two third have newspaper reading habit)

Question 8.
In a family of 3 children, what is the probability that there will be exactly 2 girls?
Solution:
let p be the probability of getting a girl child= 1/2 q = 1 – p ⇒ = 1 – 1/2
∴ q = \(\frac { 1 }{2}\) and n = 3
In a binomial distibution
p(x = x) nc_x p^xq^n-x
p(exactly 2 girls) p(x = 2)

Question 9.
Defects in yarn manufactured by a local mill can be approximated by a distribution with a mean of 1.2 defects for every 6 metres of length. If lengths of 6 metres are to be inspected, find the probability of less than 2 defects.
Solution:
Given n = 6
mean np = 1.2
⇒ 6p = 1.2
p = \(\frac { 1.2 }{6}\) p = 0.2 (or) p = \(\frac { 1 }{5}\)
q = 1 – p = 1 – \(\frac { 1 }{5}\)
∴ q = \(\frac { 4 }{5}\)
The binomial distribution
p(x) = 6C_x (0.2)^x(0.8)^6-x
p(x < 2) = p(x = 0) + p(x = 1)

Question 10.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
Solution:
Let X be the random variable denoting the number of defective bolts.
The probability of defective bolts p = \(\frac{18}{100}\) = 0.18 ⇒ q = 0.82.
Also n = 4
The p.m.f is P (X = x ) = \(^{4} \mathrm{C}_{x}(0.18)^{x}(0.82)^{4-x}\)
(i) P (exactly one defective) = P(X = 1)
= \(^{4} \mathrm{C}_{1}(0.18)^{1}(0.82)^{3}\)
= 4 (0.18) (0.82)³
= 0.3969
(ii) P (no defective) = P(X = 0)
= \(^{4} \mathrm{C}_{0}(0.18)^{0}(0.82)^{4}\)
= (0.82)⁴
= 0.45212
(iii) P (atmost 2 defective) = P(X ≤ 2)
= P(X = 2) + P(X = 1) + P(X = 0)
= \(^{4} \mathrm{C}_{2}\) (0.18)² (0.82)² + 0.3969 + 0.45212
= 0.1307 + 0.3969 + 0.45212
= 0.97972

Question 11.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
(i) exactly one will be defective
(ii) none will be defective
(iii) atmost 2 will be defective
Solution:
p = 0.09 = \(\frac { 9 }{100}\)
q = 1 – p ⇒ q – 1 – \(\frac { 9 }{100}\)
q = \(\frac { 100-9 }{100}\)
q = \(\frac { 91 }{100}\)
In a binomial distribution p(x = x) = nc_xp^xq^n-x
p(atleast one success) = p(x ≥ 1)

Taking log on both sides log ≤ (0.66) ≤ n log(0.91)
∴ n ≥ log \(\frac { log(0.66) }{log(0.91)}\)
n ≥ 5
∴ 5 or more trials are needed

Question 12.
Among 28 professors of a certain department, 18 drive foreign cars and 10 drive local made cars. If 5 of these professors are selected at random, what is the probability that atleast 3 of them drive foreign cars?
Solution:
Let p be the probability of foreign cars driven by the professors
p = \(\frac { 18 }{28}\) = \(\frac { 9 }{14}\)
Let q be the probability of local made cars driv¬en by the professors
q = \(\frac { 10 }{28}\) = \(\frac { 5 }{14}\)
and n = 5
The binomial distribution p(x = x) = nc_xp^xq^n-x

Question 13.
Out of 750 families with 4 children each, how many families would be expected to have (i) atleast one boy (ii) atmost 2 girls (iii) and children of both sexes? Assume equal probabilities for boys and girls?
Solution:
Assume equal probabilities for boys girls let p be the probability of having a boy Let x be the random variable for getting either a boy or a girl
∴ p = \(\frac { 1 }{2}\) and q \(\frac { 1 }{2}\) and n = 4
In binomial distribution p(x = 4) = nc_xp^xq^n-x
Here the binomial distribution is p(X= x)

= 1 – 0.0625
= 0.9375
for 750 families (p ≥ 1) = 750 × 0.9375
= 703.125
= 703(approximately)

(ii) p(almost 2 girls) = p(x ≤ 2)
= p(x = 0) = p(x = 1) + p(x = 2)

For 750 families p(x ≤ 2) = 0.6875 × 750
= 515.625
= 516 (approximately)

(iii) p(children of both sexes) = p(x = 1) + p(x = 2)p(x = 3)

= 0.875 x 750
For 750 families p(x = 2) = 0.875 × 750
= 656.25
= 656 (approximately)

Question 14.
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers,
Solution:
Given n = 5
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1 – \(\frac { 2 }{5}\) \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\)
The binomial distributionp (X = x) = 15c_x (\(\frac { 2 }{5}\))^x (\(\frac { 3 }{5}\))^15-x
(i) p(probability that 3 will have a laptop) = p(x = 3)

(ii) p(12 of the traels will not have a laptop)
= 1 – p(x = 12)
= 1 – 15c₁₂ (\(\frac { 2 }{5}\))¹² (\(\frac { 3 }{5}\))^15-12

(iii) p(at least three of the travelers have a laptop)
= p(x ≥ 3)
p(x ≥ 3) = 1 – p (x < 3)
= 1 – [p(x = 0) + p(x = 1) + p(x = 2)]
p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)

Question 15.
A pair 5f dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes.
Solution:
In a throw of a pair dice the doublets are (1, 1),(1, 2) (3, 3),(4, 4),(5, 5),(6, 6)
probability of getting a doublet p = \(\frac { 6 }{36}\) = \(\frac { 1 }{6}\)
⇒ q = 1 – q = \(\frac { 5 }{6}\) and n = 4 is given
The probability of success = \(\left[\begin{array}{l}
4 \\
x
\end{array}\right]\) [latex]\frac { 1 }{6}[/latex]^x [latex]\frac { 5 }{6}[/latex]^4-x
Therefore the probability of 2 success are

Question 16.
The mean of a binomial distribution is 5 and the standard deviation is 2. Determine the distribution.
Solution:
In a binomial distribution
mean np = 5 → (1)
Standard deviation \(\sqrt { npq}\) = 2
squaring on body sides
npq = 4 → (2)

n = 5 × 5 ⇒ n = 25
∴ the binomial distribution is
P(X = x) = nc_xp^xq^n-x
(i.e) p(X = x) = \(\left[\begin{array}{l}
25 \\
x
\end{array}\right]\) (\(\frac { 1 }{5}\))^x (\(\frac { 4 }{5}\))^(25-x)

Question 17.
Determine the binomial distribution for which the mean is 4 and variance 3. Also find P(X = 15)
Solution:
In a binomial distribution
mean np = 4 → (1)
variance npq = 3 → (2)

n = 4 × 4 ⇒ n = 16
The binomial distribution is p(X = x)nc_xp^xq^n-x

Question 18.
Assume that a drug causes a serious side effect at a rate of three patients per one hundred. What is the probability that atleast one person will have side effects in a random sample of ten patients taking the drug?
Solution:
Here n = 10

= 0.5344

Question 19.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease.
Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
n = 5
Let probability of recovery p = 0.73
q = 1 – p = 1 – 0.73
∴ q = 0.27
The binomial distribution is
p (x = x) = nC_xp^xq^n-x
p(x = x) = 5C_x(0.73)^x(0.27)^5-x
p(atleast 3 of the 5 mice recover) = p(x ≥ 3)
= p(x = 3) + p(x = 4) + p(x = 5)
= 5C₃ (0.73)³(0.27)^5-3 + 5C₄ (0.73)⁴ (0.27)^5-4 + 5C₅ (0.73)⁵ (0.27)^5-5
5C₂ (0.73)³ (0.27)² + 5c₁ (0.73)⁴ (0.27)¹ + 5c₀(0.73)⁵(0.27)°
[\(\frac { 5×4 }{1×2}\) × 0.389017 × 0.0729] + [5 × 0.28398241 × 0.27] + (1 × 0.2073071593 × 1)
= 0.283593393 + 0.3833762535 + 0.2073071593
= 0.2836 + 0.3834 + 0.2073
= 0.8743

Question 20.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease. Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
Success = 2 × fails
p = 2q ⇒ p = 2(1 – p)
p = 2 – 2p ⇒ p + 2p = 2
3p = 2 and p = 2/3
q = 1 – p = 1 – 2/3
q = 1/3 and n = 5
The binomial destribution is
p (X = x) = nC_xp^xq^n-x
= 5C(2/3)^x (1/3)
(i) p(three successes) = p(x = 3)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1

Question 1.
What is the population?
Solution:
A population is a set of similar items or events which is of interest for some question or experiment. A population can be specific or vague. Examples of population defined vaguely include the number of newborn babies in Tamil Nadu, the total number of tech startups in India, the average height of all exam candidates, mean weight of taxpayers in Chennai etc. Examples of population defined specifically include a number of fans produced in a particular factory, the number of students in a class, the number of boys and girls in a tuition centre etc.

Question 2.
What is the sample?
Solution:
A selection of a group of observation/individuals/population in such a way that is represents the population is called a sample.

Question 3.
What is a statistic?
Solution:
A statistic is used to estimate the value of a population parameter. For instance, we selected a random sample of 100 students from a school with 1000 students. The average height of the sampled students would be an example of a statistic. Examples, sample variance, sample quartiles, sample percentiles, sample moments, etc.

Question 4.
Define parameter.
Solution:
Parameter: The statistical constants of the population like mean (µ), variance (σ²) are referred to as population parameters.

Question 5.
What is the sampling distribution of a statistic?
Solution:
Sampling distribution of a statistic is the frequency distribution which is formed with various of a statistic computed from different samples of the same size drawn from the same population.

Question 6.
What is a standard error?
Solution:
The standard error (S.E) of a statistic is the standard deviation of its sampling distribution. If the parameter or the statistic is the mean, it is called the standard error of the mean (SEM). The standard error provides a rough estimate of the interval in which the population parameters are likely to fall.

Question 7.
Explain in detail about simple random sampling with a suitable example.
Solution:
In this technique, the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. Simple random sampling may be done, with or without replacement of samples selected. In a simple random sampling with replacement, there is a possibility of selecting the same sample any number of times.

So, simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units, the probability of drawing any unit at the first draw is \(\frac { 1 }{N}\), the probability of drawing any unit in the second draw from among the available (N – 1) units is \(\frac { 1 }{(N-1)}\), and so on.

For example, if we want to select 10 students, out of 100 students, then we must write the names/roll number of all the 100 students on slips of the same size and mix them, then we make a blindfold selection of 10 students. This method is called unrestricted random sampling because units are selected from the population without any restriction. This method is mostly used in lottery draws. If the population or universe is infinite, this method is inapplicable.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
In stratified random sampling, first divide the population into subpopulations, which are called strata. Then,the samples are selected from each of the strata through random techniques. The collection of all the samples from all strata gives the stratified random samples.

When the population is heterogeneous or different segments or groups with respect to the variable or characteristic understudy, then stratified Random sampling methods are studied. First, the population is divided into a homogeneous number of sun-groups of strata before the sample is drawn. A sample from each stratum at random. The following steps are involved in selecting a random sample in a stratified random sampling method.

(a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.

(b) After the population is stratified, a sample of a specified size is drawn at random from each stratum using the Lottery Method or table of random number method.

Question 9.
Explain in detail systematic random sampling with example.
Solution:
In systematic sampling, randomly select the first sample from the first k units. Then every kth member, starting with the first selected sample, is included in the sample.

Systematic sampling is a commonly used technique if the complete and up-to-date list of the sampling units is available. We can arrange the items in numerical, alphabetical, selecting the first at random, the rest being automatically selected according to some pre-determined pattern. A systematic is formed by selecting every item from the population, where k refers to the sample interval. The sampling interval can be determined by divided the size of the population by the size of the sample to be chosen.

That is K = \(\frac { N }{n}\), where k is an integer.
k = sampling interval, size of the population, sample size
Procedure for selection of samples by systematic sampling method

(i) If we want to select a sample of 10 students from a class of 100 students,the sampling interval is Calculated as k = \(\frac { N }{n}\) = \(\frac { 100 }{10}\) = 10
Thus sampling interval = 10 denotes that for every 10 samples one sample

(ii) The first sample is selected from the first 10(sampling interval) samples through selection procedures.

(iii) If the selected first random sample is 5, then the rest of the samples are automatically selected by incriminating the value of the sampling interval 9k = 10. i.e, 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.

Question 10.
Explain in detail about sampling error.
Solution:
Sampling Errors: Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling errors are inherent in the method of sampling. They may arise accidentally without any bias or prejudice.

Sampling Errors arise primarily due to the following reasons:

• Faulty selection of the sample instead of the correct sample by defective sampling technique.
• The investigator substitutes a convenient sample if the original sample is not available while investigating.
• In area surveys, while dealing with borderlines it depends upon the investigator whether to include them in the sample or not. This is known as the Faulty demarcation of sampling units.

Question 11.
Explain in detail the non-sampling error.
Solution:
Non-sampling Errors:
The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating, or using measuring instruments (tape, scale) are called Non-sampling errors. It may arise in the following ways:

• Due to negligence and carelessness on the part of either investigator or respondents
• Due to a lack of trained and qualified investigators.
• Due to the framing of the wrong questionnaire.
• Due to applying the wrong statistical measure.
• Due to incomplete investigation and sample survey.

Question 12.
State any two merits of simple random sampling.
Solution:
Merits:

• In simple random sampling personal bias is completely eliminated.
• This method is economical as it saves time, money and labour.

Question 13.
State any three merits of stratified random sampling.
Solution:
Merits:
(a) A random stratified sample is superior to a sample random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
(b) A stratified random sample can be kept small in size without losing its accuracy
(c) it is easy to administer, if the population under study is sub divided
(d) It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had the geographical areas.

Question 14.
State any two demerits of systematic random sampling.
Solution:
Demerits:
1. Systematic samples are not random samples.
2. If N is not multiple of n-then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

Question 15.
State any two merits for systematic random sampling.
Solution:
Merits of systematic sampling are given below:

• This method distributes the sample more evenly over the entire listed population.
• The time and work are reduced much.

Question 16.
Using the following Tippet’s random number table

Draw a sample of 10 three-digit numbers which are even numbers.
Solution:
There are many ways to select 10 random samples from the given Tippets random number table since the population size is three-digit numbers, Here the door numbers must be even (ie) the unit digit must be even. Here we consider the column-wise selection of random numbers starting from the first column.

So the first sample is 416 and the other 9 samples are 056, 664, 952, 748, 524, 914, 154, 340, and 140.
Tippets random number Table

Question 17.
A wholesaler in apples claims that only 4% of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning good apples.
Solution:
sample size = 600; Number of success = 600 – 36 = 564
sample proportion p = \(\frac { 564 }{600}\) = 0.94
600
population proportion (p) = probability of getting good apple = 96%
= \(\frac { 96 }{100}\) {∵ 4% of the apples 100 are defective}
P = 0.96
Q = 1 – p = 1 – 0.96
Q = 0.04
The S.E for a sample proporation is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.96)(0.04) }{600}}\)
\(\sqrt{\frac { 0.0384 }{600}}\) = \(\sqrt{0.000064}\)
∴ S.E = 0.008
Hence the standard error foe sample proportion is S.E = 0.008

Question 18.
A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Tamil Nadu State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. Calculate standard error of mean.
Solution:
Given n = 1000; \(\bar{x}\) = 119 lbs (pounds)
s = 30 lbs is known in this problem.
since σ is unknown, so we consider \(\bar{σ}\) = s and µ = 120 lbs
S.E = \(\frac { \bar{σ} }{√n}\) = \(\frac { s }{√n}\) = \(\frac { 30 }{\sqrt{1000}}\)
= \(\frac { 30 }{31.623}\) = 0.9487
Therefore the standard error for the average weight of large group of students of 120 lbs is 0.9487

Question 19.
A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. Calculate the suitable standard error that this sample is taken from a population with standard deviation 3?
Solution:
Sample size n = 60
Sample S.D S = 2.5
population S.D a = 3
The standard error for sample S.D is given by
\(\sqrt{\frac { σ^2 }{2n}}\) = \(\sqrt{\frac { (3)^2 }{2(60)}}\) = \(\frac { 3 }{\sqrt{120}}\)
= \(\frac { 3 }{10.954}\) = 0.27387
= 0.2739
Thus standard error for sample S.D = 0.2739

Question 20.
In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village?
Solution:
sample size n = 400
case (i):
sample proporation of vegetarian p = \(\frac { 3 }{10.954}\) = \(\frac { 230 }{400}\)
p = 0.575
q = 1 – p
= 1 – 0.575
q = 0.425
Sample error S.E= \(\sqrt{\frac { pq }{n}}\)
= \(\sqrt{\frac { 0.575×0.425 }{400}}\) = \(\sqrt{\frac { 0.223125 }{400}}\)
\(\sqrt{0.0005578125}\)
S.E = 0.2361

Case(ii):
sample size n = 400
since both vegetarian and non- vegetarian foods are equally popular in that village
sample proparation of vegetarian p = \(\frac { 1 }{2}\) = 0.5
q = 1 -p ⇒ q = 1 – 0.5
q = 0.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Miscellaneous Problems

Question 1.
A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain
(a) no more than 2 rejects?
(b) at least 2 rejects?
Solution:
In a binomial distribution
n = 10; p = \(\frac { 12 }{100}\) = \(\frac { 3 }{25}\); q = 1 – p = 1 – \(\frac { 3 }{25}\); q = \(\frac { 22 }{25}\)
p(X = x) = nc_xp^xq^n-x
(a)p(no more than 2 rejects)
p(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

(b) p (at least 2 rejects) = p (x ≥ 2)

Question 2.
Hospital records show that of patients suffering from a certain disease 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?
Solution:
Let x be the random variable of patience suffering from a certain disease
In a binomial distribution
p (X = x) = nc_xp^xq^n-x

Question 3.
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.
Solution:
In a poisson distribution
mean(λ) = \(\frac { 3 }{20}\) = 0.15
p(X = x) = \(\frac { e^{-λ}λ^x }{x!}\)
p(not be more than one failure) = p(x ≤ 1)
p(x = 0) + p(x = 1)

= 0.86074 × (1.15)
= 0.98981

Question 4.
Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
1. Find the probability that none passes in a given minute.
2. What is the expected number passing in two minutes?
Solution:
In a poisson distribution
Average per hour = 300 vehicles
mean per minute = \(\frac { 300 }{60}\) = 5
∴ λ = 5

= e^-5(12.5)
= 0.0067379 × 12.5
= 0.08422375
= 0.08422375 × 10²

Question 5.
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Raghul wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Raghul takes the test and scores 585. Will he be admitted to this university?
Solution:
Let x denotes the scores of a national test mean
µ = 500 and standard deviation σ = 100
standard normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-5000 }{100}\)
when x = 585
z = \(\frac { 585-500 }{100}\) = \(\frac { 85 }{100}\) = 0.85
p(x ≤ 585) = p(z ≤ 0.85)
p(z ≤ 0.85) = p(-∞ < z < 0) + p(0 < z < 0.85)
= 0.5 + 0.3023
= 0.8023
for n = 100;
p(z ≤ 0.85) = 100 × 0.8023
= 80.23
∴ Raehul scores 80.23%

We can determine the scores of 70% of the students as follows:
from the table for the area 0.35
We get z₁ = -1.4(as z₁ lies to left of z = 0)
similarly z₂ = 1.4

Now z₁ = \(\frac { x_1-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)
-1.4 × 100 = x₁ – 500 ⇒ x₁ 500 – 140
x₁ = 360
Again z₂ = \(\frac { x_2-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)
1.4 × 100 = x₂ – 500 ⇒ x₂ = 140 + 500
= x₂ = 640
Hence 70% of students score between 360 and 640
But Raghul scored 585. His score is not better than the score of 70% of the students.
∴ He will not be admitted to the university.

Question 6.
The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time.
(i) less than 19.5 hours?
(ii) between 20 and 22 hours?
Solution:
Let x denotes the time taken to assemable cars mean µ = 20 hours and S.D σ = 2 hours
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-20 }{2}\)
(i) p(less than 19.5 hours) = p(x < 19.5)
when x = 19.5
z = \(\frac { 19.5 }{2}\) = \(\frac { -0.5 }{2}\) = 0.25
p(x < 19.5) = p(z < – 0.25)
= p(-∞ < z < 0) – p(-0.25 < z < 0)
= 0.5 – p(0 < z < 0.25)
= 0.5 – 0.0987
= 0.4013

(ii) p(between 20 and 22 hours) = p(20 < x < 22)

when x = 20
z = \(\frac { 20-20 }{2}\) = \(\frac { 0 }{2}\) = 0
when x = 22;
z = \(\frac { 22-20 }{2}\) = \(\frac { 2 }{2}\) = 1
p(20 < x < 22) = p(0 < z < 1)
= 0.3413

Question 7.
The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000.
(a) What percent of people earn less than $40,000?
(b) What percent of people earn between $45,000 and $65,000?
(c) What percent of people earn more than $75,000
Solution:
Let x denotes the annual salaries of employees in a large company
mean µ = 50,000 and S.D σ = 20,000
Standard normal variate z = \(\frac { x-µ }{σ}\)
(a) p(people earn less than $40,000) = p(x < 40,000)
when x = 40,000
z = \(\frac { 40,000-50,000 }{20,000}\) = \(\frac { 10,000 }{20,000}\)
z = -0.5
p(x < 40,000) = p(z < -0.5)
= p(-∞ < z < 0) – p(-0.5 < z < 0)
= 0.5 -p(-0.5 < z <0)
= 0.5 – p(0 < z < 0.5) (due to symmetry)
= 0.5 – 0.01915
= 0.3085
= p(x < 40,000) in percentage = 0.3085 × 100 = 30.85

(b) p(people ear between $45,000 and $65,000)
p(45000 < x < 65000)
When x = 45,000;
z = \(\frac { 45,000-50,000 }{20,000}\) = \(\frac { -5000 }{20,000}\) = \(\frac { -1 }{4}\)
z = -0.25
when x = 65,000;
z = \(\frac { 65,000-50,000 }{20,000}\) = \(\frac { 15000 }{20,000}\) = \(\frac { 3 }{4}\)
z = 0.75
p(45000 < x < 65000) = p(-0.25 < z < 0.75)
= p(-0.25 < z < 0) + p(0 < z < 0.75)
= p(0 < z < 0.25) + p(0 < z < 0.75)
= p(0 < z < 0.25) + p(0 < z < 0.75)
= 0.0987 + 0.2734 = 0.3721
p(45000 < x < 65000) in percentage = 0.3721 × 100
= 37.21

p(people earn more than$75,000) = p(x > 70000)
when x = 75,000;
z = \(\frac { 75,000-50,000 }{20,000}\) = \(\frac { 25000 }{20,000}\) = \(\frac { 5 }{4}\)
z = 1.25
p(x > 75,000) = p(x > 1.25)
= p(0 < z < ∞) – p(0 < z < 1.25) = 0.5 – 0.3944 = 0.1056 p(x > 750,000)in percent = 01056 × 100
= 10.56

Question 8.
X is a normally normally distributed variable with mean µ = 30 and standard deviation σ = 4. Find
(a) P(x < 40) (b) P(x > 21)
(c) P(30 < x < 35)
Solution:
x is a normally distributed variable with mean µ = 30 and standard deviation σ = 4

Then the normal variate z = \(\frac { x-µ }{σ}\) = \(\frac { x-30 }{4}\)

(a) p(x < 40) = ?
when x = 40;
z = \(\frac { 40-30 }{4}\) = \(\frac { 10 }{4}\) = 2.5
p(x < 40) = p(z < 2.5)
= p(-∞ < z < 0) + p(0 < z < 2.5) = 0.5 + 0.4938 = 0.9938 (b) p(x > 21) = ?
when x = 21;
z = \(\frac { 40-30 }{4}\) = \(\frac { 10 }{4}\) = -2.25
p(x > 21) = p(z > -2.25)
= p(-2.25 < z < 0) + p(0 < z < ∞)
= p(0 < z < 2.25) + 0.5
= 0.4878 + 0.5
= 0.9878

(c) p(30 < x < 35) = ?
when x = 30;
z = \(\frac { 30-30 }{4}\) = \(\frac { 0 }{4}\) = 0
when x = 35;
z = \(\frac { 35-30 }{4}\) = \(\frac { 5 }{4}\) = 1.25
p(30 < x < 35) = p(0 < z < 1.25)
= 0.3944

Question 9.
The birth weight of babies is Normally distributed with mean 3,500 g and standard deviation 500 g. What is the probability that a baby is born that weighs less than 3,100 g?
Solution:
Let x be a normally distributed variable with mean 3,500 g and standard deviation 500 g

Here µ = 3500 and σ = 500
The standard normal variate z = \(\frac { x-µ }{σ}\)
p(weight less than variate 3100 g) = p(x < 3100)
when x = 3100;
z = \(\frac { 3100-3500 }{500}\) = \(\frac { -400 }{500}\) = \(\frac { -4 }{5}\)
z = -0.8
∴ p(z < 3100) = p(z < -0.8)
= p(-∞ < z < 0) – p(-0.8 < z < 0)
= 0.5 – p(0 < z < 0.8)
= 0.5 – 0.2881
= 0.2119

Question 10.
People’s monthly electric bills in chennai are normally distributed with a mean of Rs 225 and a standard deviation of Rs 55. Those people spend a lot of time online. In a group of 500 customers, how many would we expect to have a bill that is Rs 100 or less?
Solution:
Let X be a normally distributed variable with a mean of Rs 225 and a standard deviation of Rs 55

Given mean µ = 225 and s.d σ = 55
Now the probability that the bill will be ₹100 or less is P (X ≤ 100)
= P(Z ≤ \(\frac{100-225}{55}\))
= P(Z ≤ -2.27)
= 0.5 – P(-2.27 < Z < 0)
= 0.5 – P(0 < Z < 2.27)
= 0.5 – 0.4884
= 0.0116
Thus, in a group of 500 customers, we expect to have 500 × 0.0116 = 5.8 ~ 6 customers whose electric bills will be ₹ 100 or less.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3

Question 1.
Define normal distribution.
Solution:
A random variable X is said to follow a normal distribution with parameters mean µ and varaince σ², if its probability density function is given by

Question 2.
Define standard normal variate.
Solution:
A random variable Z = (X – µ)/σ follows the standard normal distribution. Z is called the standard normal variate with mean 0 and standard deviation 1 i.e Z – N (0, 1). Its Probability density function is given by:
φ(z) = \(\frac { 1 }{\sqrt {2π}}\) e^-x²/2 -∞ < z < ∞

Question 3.
Write down the conditions in which the normal distribution is a limiting case of the binomial distribution.
Solution:
The Normal distribution is a limiting case of Binomial distribution under the following conditions:

• n, the number of trials is infinitely large, i.e. n → ∞
• neither p (or q ) is very small.

Question 4. m
Write down any five characteristics of the normal probability curve.
Solution:
Chief Characteristics or Properties of Normal Probability distribution and Normal probability Curve.
The normal probability curve with mean µ and standard deviation σ has the following properties:
(i) the curve is bell-shaped and symmetrical about the line x = u.
(ii) Mean, median, and mode of the distribution coincide.
(iii) x-axis is an asymptote to the curve, (tails of the curve never touches the horizontal (x) axis)
(iv) No portion of the curve lies below the x-axis as f(x) being the probability function can never be negative.
(v) The points of inflexion of the curve are x = µ ± σ

Question 5.
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and a standard deviation of 60 hours. Estimate the number of bulbs likely to burn for
(i) more than 2,150 hours
(ii) less than 1,950 hours
(iii) more 1,920 hours but less than 2,100 hours.
Solution:
Let x denote the burning of the bulb follows normal distribution with mean 2,040 and standard deviation 60 hours.
Here m = 2040; σ = 60 and N = 2000
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-2040 }{60}\)
(i) p(morethan 2,150 hours)
p(x > 2150)
when x = 2150
z = \(\frac { 2150-2040 }{60}\) = \(\frac { 110 }{60}\)= 1.833
p(x > 2150) = p(z > 1.833)
= p(0 < z < ∞) – p(0 < z < 1.833)
= 0.5 – 0.4664
= 0.0336
∴ Number of bulbs whose burning time is more than 2150 hours=0.0336 × 2000
= 67.2 = 67(approximately)

(ii) p(less than 1950 hours)
p(x < 1950)
when x = 1950
z = \(\frac { 1950-3040 }{60}\) = \(\frac { -90 }{60}\)= -1.5
p(x < 1950) = p(z < -1.5) = p(z > 1.5)
= 0.5 – 0.4332
= 0.068

Numbers of bulbs whose burning time is less than
1950 = 0.0668 × 2000 = 133.6
= 134 (approximately)

(iii) p(more 1,920 hours but less than 2,100 hours)
= p(1920 < x < 2100)
when x = 1950
z = \(\frac { 1920-2040 }{60}\) = \(\frac { -120 }{60}\)= -2
when x = 2100
z = \(\frac { 2100-2040 }{60}\) = \(\frac { -60 }{60}\)= -2
∴ p(1920 < x < 2040) = p(-2 < z < 1)
= p(0 < z < 2) + p(0 < z < 1)
= 0.4772 + 0.3413
= 0.8185

∴ Number of bulbs whose burning time more than 1920 hours but less than 2100 hours) = 0.8185 × 2000
= 1637

Question 6.
In a distribution, 30% of the items are under 50 and 10% are over 86. Find the mean and standard deviation of the distribution.
Solution:

z = \(\frac { x-µ }{σ}\)
Given that
p(x < 50) = 0.3 p(x > 86) = 0.1
p(z < -c) = 0.3
p(-c < z < 0) = 0.5 – 0.3
p(-c < z < 0) = 0.2 {from the table}
p(0 < z < c) = 0.2
c = 0.53
then -c = -0.53
∴ \(\frac { 50-µ }{σ}\) = -0.53
50 – µ = -0.53σ
µ – 0.53σ = 50 → 1
p(x < 50) = 0.1
p(0 < z < ∞) = -p(0 < z < c₁) = 0.1
p(0 < z < ∞) = p(0 < z < c₁) + 0.1
0.5 = p(0 < z < c₁) + 0.1
p(0 < z < c₁) = 0.5 – 0.1
p = (0 < z < c₁) = 0.4
c₁ = 1.29
∴ \(\frac { 86-µ }{σ}\) = 1.29
86 – µ = 1.29 σ
µ + 1.29σ = 86 → 2
solving eqn 1 & 2
eqn 2 ⇒ m + 1.29σ = 86
eqn 1 ⇒ m + 0.53σ = 50
– + – ………….
………… 1.82 ………… σ = 36 ………….
σ = \(\frac { 36 }{1.82}\) ∴ = 19.78
Substitute σ = 19.78 in eqn 1
µ – 0.53(19.78) = 50
µ -10.48 = 50
µ = 50 + 10.48
µ = 60.48
Mean = 60.48 and standard deviation = 19.78

Question 7.
X is normally distributed with mean 12 and sd 4. Find P (x ≤) 20 and P (0 ≤ x ≤ 12)
Solution:
x is normally distribution with mean 12 and sd 4
∴ µ = 12 and σ = 4
Standard normal variable
z = \(\frac { x-µ }{σ}\) = \(\frac { x-12 }{4}\)
(i) p(x ≤ 20)

when x = 20
z = \(\frac { 20-12 }{4}\) = \(\frac { 8 }{4}\) = 2
v(x ≤ 20) = \(\frac { 8 }{4}\) = 2
p(x ≤ 20) = p(z ≤ 2)
= 0.5 + p(0 < z < 2)
= 0.5 + 0.4772
= 0.9772

(ii) p(0 < x < 12 )

when x = 0
z = \(\frac { 0-12 }{4}\) = \(\frac { -12 }{4}\) = -3
when x = 12
z = \(\frac { 12-12 }{4}\) = \(\frac { 0 }{4}\) = 0
p(0 ≤ x ≤ 12) = p(-3 ≤ z ≤ 0)
= p(0 ≤ z ≤ 3)
= 0.4987

Question 8.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height
(a) greater than 2 inches
(b) less than or equal to 64 inches
(c) between 65 and 71 inches.
Solution:
let x denote the height of a student N = 500; m = 68.0 inches and σ = 3.0 inches the standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-68 }{3}\)
a(greater than 72 inches)
p = p(x > 72)
when x = 72
z = \(\frac { 72-68 }{3}\) = \(\frac { 4 }{3}\) = 1.33
p(x > 72) = p(z > 1.33)
= 0.5 – 0.4082
= 0.0918

Number of students whose height are greater than 72 inches
= 0.0918 × 500
= 45.9
= 46 (approximately)

(b) p(less than or equal to 64 inches)
p(x ≤ 64)
when x = 64
z = \(\frac { 64-68 }{3}\) = \(\frac { -4 }{3}\) = -1.33
p(x ≤ 64) = p(z ≤ -1.33)
p(z ≥ -1.33)
= 0.5 – 0.4082
= 0.0918

∴ Number of heights whose ate less than or equal to 64 inches = 0.0918 × 500
= 45.9
= 46 (approximately)

(c) p(between 65 and 71 inches)
p(65 ≤ x ≤ 71)
when x = 65

z = \(\frac { 65-68 }{3}\) = \(\frac { -3 }{3}\) = -1
when x = 71
z = \(\frac { 71-68 }{3}\) = \(\frac { 3 }{3}\) = 1
p(65 ≤ x ≤ 71) = p(-1 < z < 1)
= p(-1 < z < 0) + p(0 < z < 1)
= p(0 < z < 1) + p(0 < z < 1)
= 2 × [p(0 < z < 1)]
= 2 × 0.3413
= 0.6826
∴ Number of students whose height between 65 and 7 inches = 0.6826 × 500
= 341.3
= 342 (approximately)

Question 9.
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second. Find the probability that it will take less than 16.35 seconds to develop prints.
Solution:
let x be the random variable have long the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second

µ = 16,28 and σ = 0.12
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-16.28 }{0.12}\) = 1
p(less than 16.35 seconds) = p(x < 16.35)
when x = 16.35
z = \(\frac { 16.35-16.28 }{0.12}\) = \(\frac { 0.07 }{0.12}\) = 0.583
p(x< 16.35) = p(z < 0.583)
= p(—∞ < z < 0) + p(0 < z < 0.583)
= 0.5 + 0.2190
= 0.7190

Question 10.
If the heights of 500 students are normally distributed with mean of 68.0 inches and standard deviation of 3.0 inches, how many students have a height (a) greater than 2 inches (b) less than or equal to 64 inches (c) between 65 and 71 inches.
Solution:
Let x be a normal variate with mean of 400 labour days and a standard deviation of 100 labour days
m = 400 and σ = 100
The construction work should be completed within 450 days.
The standard normal variate
\(\frac { x-µ }{6}\) = \(\frac { x-400 }{100}\)
personality for 1 labour day = Rs 10,000
If personality amount is = 2,00,000 than No of excess

days = \(\frac { 200000 }{10000}\) = 20
∴ x = 450 + 20 = 470
when x = 470
z = \(\frac { 470-400 }{100}\) = \(\frac { 70 }{100}\) = 0.7
= p(x ≥ 470) = p(z ≥ 0.7)
= 0.5 – 0.2580
= 0.2420

(ii) p(at most 500 days) = p(x ≤ 500 )
when x = 500

z = \(\frac { 500-400 }{100}\) = \(\frac { 100 }{100}\) = 1
p(x ≤ 500) = p(z ≤ 1)
= p(∞ < z < 0) -r- p(0 < z < 1)
= 0.5 + 0.3415
= 0.8413

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.
Find the expected value for the random variable of an unbiased die
Solution:
When a un based die is thrown , any one of the number 1, 2, 3, 4, 5, 6, can turn up that x denote the random variable taking the values from 1 to 6

Question 2.
Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table

Solution:
Expected value of X, E(X) = \(\sum_{x} x P_{x}(x)\)
E(X) = (0 × 0.2) + (1 × 0.1) + (2 × 0.4) + (3 × 0.3)
= 0 + 0. 1 + 0.8 + 0.9
= 1.8

Question 3.
The following table is describing about the probability mass function of the random variable X

Find the standard deviation of x.
Solution:
Let x be the random variable taking the values 3, 4, 5
E(x) = Σpixi
= (0.1 × 3) + (0.1 × 4) + (0.2 × 5)
0.3 + 0.4 + 1.0
E(x) = 1.7
E(x²) = Σpixi²
= (0.1 × 3²)+ (0.1 × 4²) + (0.2 × 5²)
= (0.1 × 9) + (0.1 × 16) + (0.2 × 25)
E(x²) = 7.5
Var(x) = E(x²) – (E(x)]²
= 7.5 – (1.7)²
= 7.5 – 2.89
Var(x) = 4.61
Standard deviation(S.D) = \(\sqrt { var (x)}\)
= \(\sqrt { 4.61}\)
σ = 2.15

Question 4.
Let X be a continuous random variable with probability density function
f_x (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected value of X.
Solution:
Let x be a continuous random variable. In the probability density function, Expected

Question 5.
Let X be a continuous random variable with probability density function
f_x (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the mean and variance of X.
Solution:
Let x be a continuous random variable. In the probability density function,

Question 6.
In investment, a man can make a profit I of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38. Find the expected gain.
Solution:

X	5000	-8000
P(x = x)	0.62	0.38

Let x be the random variable of getting gain in an Investment
E(x) be the random variable of getting gain in an Investment
E(x) = ΣPixi
= (0.62 × 5000) + [0.38 × (-8000)]
= 3100 – 3040
E(x) = 60
∴ Expected gain = ₹ 60

Question 7.
What are the properties of mathematical expectation?
Solution:
The properties of Mathematical expectation are as follows:

1. E(a) = a, where ‘a’ is a constant
2. Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)
3. Multiplication theorem: E(XY) = E(X) E(Y)
4. E(aX) = aE(X), where ‘a’ is a constant
5. For constants a and b, E(aX + b) = a E(X) + b

Question 8.
What do you understand by mathematical expectation?
Solution:
The average value of a random Phenomenon is termed as mathematical expectation or expected value.
The expected value is a weighted average of the values of a random variable may assume

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
Let X be a random variable. Let E(X) denote the expectation of X.
Then the variance is defined in terms of the mathematical expectation as follows.
(a) X is discrete r.v with p.m.f p(x). Then Var(X) = \(\sum[x-\mathrm{E}(\mathrm{X})]^{2} p(x)\)
(b) X is continuous r.v with p.d.f f_x(x). Then Var(X) = \(\int_{-\infty}^{\infty}[X-E(X)]^{2} f_{X}(x) d x\)

Question 10.
Define mathematical expectation in tears of a discrete random variable?
Solution:
Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by
E(X) = \(\sum_{ x }\) x p(x) ……(1)

Question 11.
State the definition of mathematical expectation using continuous random variables.
Solution:
Let X be a continuous random variable with probability density function f(x). Then the expected value of X is
\(\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x\)
If the integral exists, E(X) is the mean of the values, otherwise, we say that the mean does not exist.

Question 12.
In a business venture, a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What are his expected, variance and standard deviation of profit?
Solution:
Let X be the random variable of getting profit in a business

X	2000	-1000
P(x = x)	0.4	0.6

E(x) = Σ_xxp_x(x)
= (0.4 × 2000) +[0.6 × (-1000)]
= 800 – 600
E(X) = 200
∴ Expected value of profit = ₹ 200
E(X²) = Σx² P_x(x)
= [(2000)² × 0.4] + [(-1000)² × 0.6]
= (4000000 × 0.4) + (1000000 × 0.6)
E(X²) = 2200000
Var(X) = E(X²) – [E(X)]²
= 22000000 – (200)²
= 2200000 – 40000
Var(X) = 21,60,000
Variance of his profit = ₹ 21,60,000
Standard deviation(S.D) = \(\sqrt { var (x)}\)
σ = \(\sqrt { 2160000}\)
σ = 1469.69
Standard deviation of his profit is ₹ 1,469.69

Question 13.
The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f.

Find the expected number of miles (in thousands) a tire would last until it reaches the critical tread wear point.
Solution:
We know that,

Question 14.
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.
Solution:
Let X denote the amount the person receives in a game
Then X takes values 4,-2 and
So P(X = 4) = P (of getting a head)
= \(\frac { 1 }{2}\)
P(X = – 2) = P (of getting a tail)
= \(\frac { 1 }{2}\)
Hence the Probability distribution is

X	4	-2
P(x = x)	1/2	1/2

E(x²) = 10
Var(x) = E(x²) – E(x²)
= 10 – (1)²
Var(x) = 9
∴ His expected gain = ₹ 1
His variance of gain = ₹ 9

Question 15.
Let X be a random variable and Y = 2X + 1. What is the variance of Y if the variance of X is 5?
Solution:
Given X is a random variable and Y = 2X + 1 and Var(X ) = 5
Var (Y) = Var (2X + 1) = (2)² = 4
Var X = 4(5) = 20

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Miscellaneous Problems

Question 1.
Using three yearly moving averages, Determine the trend values from the following data.

Solution:

Question 2.
From the following data, calculate the trend values using fourly moving averages.

Solution:

Question 3.
Fit a straight line trend by the method of least squares to the following data.

Solution:

Therefore, the required equation of the straight line trend is given by
y = a + bx
y = 55.9875 + 0.830 x
⇒ y = 55.9875 + 0.83 (\(\frac { x-1983.5 }{0.5}\))
The trend values can be obtained by
When x = 1980
y = 55.9875 + 0.83 (\(\frac { 1980-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-7)
= 55.9875 – 5.81
= 50.1775
When x = 1981
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-5)
= 55.9875 – 4.15
= 51.8375
When x = 1982
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-3)
= 55.9875 – 2.49
= 53.4975
When x = 1983
y = 55.9875 + 0.83 (\(\frac { 1983-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-1)
= 55.9875 – 0.83
= 55.1575
When x = 1984
y = 55.9875 + 0.83 (\(\frac { 1984-1983.5 }{0.5}\))
55.9875 + 0.83 (1)
= 56.8175
when x = 1985
y = 55.9875 + 0.83 (\(\frac { 1985-1983.5 }{0.5}\))
= 55.9875 + 0.83 (3)
= 55.9875 + 2.49
= 58.4775
when x = 1986
y = 55.9875 + 0.83 (\(\frac { 1986-1983.5 }{0.5}\))
= 55.9875 + 0.83 (5)
= 55.9875 + 4.15
= 60.1375
when x = 1987
y = 55.9875 + 0.83 (\(\frac { 1987-1983.5 }{0.5}\))
= 55.9875 + 0.83 (7)
= 55.9875 + 5.81
= 61.7975

Question 4.
Fit a straight line trend by the method of least squares to the following data.

Solution:

Lasperyre’s price Index number


Hence Fisher’s Ideal Index satisfies Time reversal test

Question 5.
Using the following data, construct Fisher’s Ideal Index Number and Show that it satisfies Factor Reversal Test and Time Reversal Test?

Solution:

Factor reversal test

Hence Fisher’s Ideal Index satisfies Factor reversal test.

Question 6.
Compute the consumer price index for 2015 on the basis of 2014 from the following data.

Solution:

Question 7.
An Enquiry was made into the budgets of the middle class families in a city gave the following information.

What changes in the cost of living have taken place in the middle class families of a city?
Solution:

Conclusion:
The cost of living has increased up to 26.10% in 2011 as compared to 2010.

Question 8.
From the following data, calculate the control limits for the mean and range chart.

Solution:

UCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 51 + 0.577(6.5)
= 51 + 3.7505
= 54.7505
= 54.75
CL = \(\bar { \bar x}\) = 51
UCL = \(\bar { \bar x}\) – A₂\(\bar { R}\)
= 51 – 0.577(6.5)
= 51 – 3.7505
= 47.2495
= 47.25
The control limits for Range chart is
UCL = D₄\(\bar { R}\)
= 2.114(6.5)
= 13.741
CL = \(\bar { R}\) = 6.5
LCL = D₃\(\bar { R}\) = 0(6.5) = 0

Question 9.
The following data gives the average life(in hours) and range of 12 samples of 5 lamps each. The data are

Construct control charts for mean and range Comment on the control limits.
Solution:

UCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.6675
= 1614.1675
= 1614.17
CL = \(\bar { \bar x}\) = 1367.5
LCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 1367.5 – 0.577(427.5)
= 1367.5 – 246.6675
= 1120.8325
= 1120.83
The control limits for Range chart is
UCL = D₄\(\bar { R}\)
= 2.115(427.5)
= 904.1625
= 904.16
CL = \(\bar { R}\) = 427.5
LCL = D₃\(\bar { R}\)
= 0(427.5)
= 0

Question 10.
The following are the sample means and I ranges for 10 samples, each of size 5. Calculate ; the control limits for the mean chart and range chart and state whether the process is in control or not.

Solution:

UCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 4.982 + 0.577(0.36)
= 4.982 + 0.20772
= 5.18972
= 5.19
CL = \(\bar { \bar x}\) = 4.982
LCL = \(\bar { \bar x}\) + A₂\(\bar { R}\)
= 4.982 – 0.577(0.36)
= 4.982 – 0.20772
= 4.77428
= 4.774
The control limits for range chart is
UCL = D₂\(\bar { R}\) = 2.115(3.6)
= 7.614
CL = \(\bar { R}\) = 3.6
LCL = D₃\(\bar { R}\)
= 0(0.36) = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 1.
Construct cumulative distribution function for the given probability distribution.

Solution:
F(0) = P(x ≤ 0) = p(0) = 0.3
F(1) = P(x ≤ 1) = p(0) + p(1)
= 0.3 + 0.2 = 0.5
F(2) = P(x ≤ 2) = P(0) + P(1) + P(2)
= 0.3 + 0.2 + 0.4 = 0.9
F(3) = P(x ≤ 3) = P(0) + P(2) + P(3) + P(4)
= 0.3 + 0.2 + 0.4 + 0.1 = 1

Question 2.
Let X be a discrete random variable with the following p.m.f

Find and plot the c.d.f. of X.
Solution:
F(3) = P(x ≤ 3) = P(3) = 0.3
F(5) = P(x ≤ 5) = P(x = 3) + (x = 5)
= 0.3 + 0.2 = 0.5
F(8) = P(x ≤ 8) = P(3) + P(5) + P(8)
= 0.3 + 0.2 + 0.3
= 0.8
F(10) = P(x ≤ 10)
= P(3) + P(5) + P(8) + P(10)
= 0.3 + 0.2 + 0.3 + 0.2
= 1

Question 3.
The discrete random variable X has the following probability function.

where k is a constant. Show that k = \(\frac { 1 }{18}\)
Solution:
From the data
P(x = 2) = kx = 2k
P(x = 4) = kx = 4k
P(x = 6) = kx = 6k
P(x = 8) = k(x – 2)
= k(8 – 2)
= 6k
Since P(X = x) is a probability mass function
\(\sum_{x=2}^{8}\) P(X = x) = 1
\(\sum_{i=2}^{∞}\) P(xi) = 1
(i.e) P(x = 2) + P(x = 4) + P(x = 6) + P(x = 8) = 1
2k + 4k + 6k + 6k = 1
18k = 1
∴ k = \(\frac { 1 }{18}\)

Question 4.
The discrete random variable X has the probability function

Solution:
\(\sum_{i=1}^{\infty} p\left(x_{i}\right)=1\)
gives k + 2k + 3k + 4k = 1
⇒ 10k = 1
⇒ k = \(\frac{1}{10}\) = 0.1

Question 5.
Two coins are tossed simultaneously. Getting a head is termed a success. Find the t probability distribution of the number of successes.
Solution:
Let X is the random variable which counts the Number of Heads when the coins are tossed the outcomes are stated below

Question 6.
The discrete random variable X has the probability function.

(i) Find k
(ii) Evaluate p( x < 6), p(x ≥ 6)and p(0 < x < 5) (iii) If P(X ≤ x) > 1, 2 then find the minimum value of x.
Solution:
(i) Since the condition of probability mass function
\(\sum_{i=1}^{\infty} p\left(x_{i}\right)=1\)
\(\sum_{i=0}^{7} p\left(x_{i}\right)=1\)
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k – 1 = 0
⇒ (10k – 1) (k + 1) = 0
⇒ k = \(\frac{1}{10}\) and k = -1
Since p(x) cannot be negative, k = -1 is not applicable. Hence k = \(\frac{1}{10}\)

(ii) P(x < 6)
P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)
= 0 + k + 2k +2k + 3k + k²
= 8k + k²

P(x ≥ 6) = P(x = 6) + p(x = 7)
= 2k² + 7k² + k
= 9k² + k

P(0 < x < 5)
= P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)
= k + 2k + 2k + 3k
= 8k = 8(\(\frac { 1 }{10}\))
∴ P(0 < x < 5) = \(\frac { 8 }{10}\)

(iii) We want the minimum value of x for which P(X ≥ x) > \(\frac { 1 }{2}\)
Now P(X ≤ 0) = 0 < \(\frac { 1 }{2}\)
P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k = k
\(\frac { 1 }{10}\) < \(\frac { 1 }{2}\)
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k = 3k
\(\frac { 3 }{10}\) \(\frac { 1 }{2}\)
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)
= 0 + k + 2k + 2k = 5k
= \(\frac { 5 }{10}\) = \(\frac { 1 }{2}\)
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)
= 0 + k + 2k + 2k + 3k = 8k
= \(\frac { 8 }{10}\) > \(\frac { 1 }{2}\)
This Shows that the minimum value of X for which P(X ≤ x) > \(\frac { 1 }{2}\) is 4

Question 7.
The distribution of a continuous random variable X in range (-3, 3) is given by p.d.f.

Verify that the area under the curve is unity.
Solution:
Since(-3, 3) in the range of given p.d.f
Area under the curve A = \(\int_{-∞}^{∞}\) f(x)dx = \(\int_{-3}^{3}\) f(x)dx

Hence the Area under the curve is unity

Question 8.
A continuous random variable X has the following distribution function:

Find (i) k and (ii) the probability density function.
Solution:
We have \(\frac{d}{d x}\)F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1
f(x) = 0 for x ≤ 1
Again F(x) = 1 for x > 3
f(x) = \(\frac{d}{d x}\) (1) = 0 for x > 3
In 1 < x ≤ 3, F(x) = k(x – 1)4
f(x) = \(\frac{d}{d x}\) (k(x – 1)4) = 4k(x – 1)3
(i) We know that \(\int_{-\infty}^{\infty} f(x) d x\) = 1
This gives \(\int_{1}^{3} 4 k(x-1)^{3} d x=1\)
\(\left[k(x-1)^{4}\right]_{1}^{3}=1\)
k[16 – 0] = 1
k = \(\frac{1}{16}\)
(ii) The probability density function is
f(x) = \(\frac{1}{4}(x-1)^{3}\), 1 < x ≤ 3
= 0, otherwise

Question 9.
The length of time (in minutes) that a certain person speaks on the telephone is found to be a random phenomenon, with a probability function specified by the probability density function f(x) as

(a) Find the value of A that makes f (x) a p.d.f.
(b) What is the probability that the number of minutes that person will talk over the phone is
(i) more than 10 minutes
(ii) less than 5 minutes, and
(iii) between 5 and 10 minutes.
Solution:
(a) Since f(x) is a probability density Function

b (i) more than 10 minutes

Question 10.
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function

(a) Is the distribution function continuous? If so, give its probability density function?
(b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes, and (iii) between 1 and 3 minutes?
Solution:
(i) Yes, the distribution function is continuous on [0, 4]
The probability density function

(b) The probability that a person will have to wait
(i) more than 3 minutes

(iii) between 1 and 3 minutes

Question 11.
Define random variable.
Solution:
A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as X, Y and Z etc. If X and Y are r.v’s then X + Y is also an r.v.

Question 12.
Explain What are the types of random variables.
Solution:
Random variables are classified into two types namely discrete and continuous random variables these are important for practical applications in the field of Mathematics and Statistics.

Question 13.
Define discrete Random Variable
Solution:
A variable which can assume a finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.
Examples:

1. Marks obtained in an exam.
2. The number of chocolates in a box.
3. The number of phone calls during a day.
4. The number of TV sets sold during a month by a dealer.

Question 14.
What do you understand by continuous random variables?
Solution:
A random variable X which can take on any value (integral as well as fraction) in the interval is called a continuous random variable.

Question 15.
Describe what is meant by a random variable
Solution:
A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as X, Y, and Z, etc. If X and Y are r.v’s then X + Y is also an r.v.

Question 16.
Distinguish between a discrete and continuous random variable
Solution:

Question 17.
Explain the distribution function of a random variable.
Solution:
The discrete cumulative distribution function or distribution function of a real-valued discrete random variable X takes the countable number of points x₁, x₂, …. with corresponding probabilities p(x₁), p(x₂),… and then the cumulative distribution function is defined by
F_x(x) = P(X ≤ x), for all x ∈ R
i.e. F_x (x) = \(\sum_{x \leq x}\) p(x_i)

Question 18.
Explain the terms (i) probability Mass function (ii) probability density function and (iii) probability
Solution:
(i) If X is a discrete random variable with distinct values x₁, x₂, …. x_n, …, then the function, denoted by P_x(x) and defined by

This is defined to be the probability mass function or discrete probability function of X.

(ii) The probability that a random variable X takes a value in the interval [t₁, t₂] (open or closed) is given by the integral of a function called the probability density function f_x(x):
P(t₁ ≤ X ≤ t₂) = \(\int_{t_{1}}^{t_{2}}\) f_x(x)dx.

(iii)The probability distribution of a random variable X is defined only when we have the various values oft the various values of the random variable e.g. x₁, x₂ …… x_n togather with respective probabilities p₁, p₂, p₃ …… p₄ satisfying
\(\sum_{i=1}^{n}\) Pi = 1

Question 19.
What are the properties of
(i) discrete random variable and
(ii) Continuous random variable
Solution:
Discrete Random Variable:

• A variable which can take only certain values.
• The value of the variables can increase incomplete numbers
• Binomial, Poisson, Hypergeometric probability distributions come under this category
• Example: Number of students who opt for commerce in class 11, say 30, 35, 40, 45, and 50.

Continuous random variable:

• A variable which can take any value in a particular limit.
• Its value increases infractions but not in jumps.
• Normal, student’s t and chi-square distribution come under this category.
• Example: Height, Weight, and age of family members: 50.5 kg, 30 kg, 42.8 kg, and 18.6 kg.

Question 20.
State the properties of the distribution function.
Solution:

• Property 1: The distribution function F is increasing, (i.e) if x ≤ y, then F(x) ≤ F(y)
• Property 2: F(x) is continuous from right, (i.e) for each x ∈ R, F (x⁺) = F (x)
• Property 3: F (∞) = 1
• Property 4: F (-∞) = 0
• Property 5: F'(x) = f(x)
• Property 6: P(a ≤ X ≤ b) = F(b) – F(a)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.4

Question 1.
A time series is a set of data recorded
(a) Periodically
(b) Weekly
(c) successive points of time
(d) all the above
Solution:
(d) all the above

Question 2.
A time series consists of
(a) Five components
(b) Four components
(c) Three components
(d) Two components
Solution:
(b) Four components

Question 3.
The components of a time series which is attached to short term fluctuation is
(a) Secular trend
(b) Seasonal variations
(c) Cyclic variation
(d) Irregular variation
Solution:
(d) Irregular variation

Question 4.
Factors responsible for seasonal variations are
(a) Weather
(b) Festivals
(c) Social customs
(d) All the above
Solution:
(d) All the above

Question 5.
The additive model of the time series with the components T, S, C and I is
(a) y = T + S + C × I
(b) y = T + S × C × I
(c) y = T + S + C + I
(d) y = T + S × C + I
Solution:
(c) y = T + S + C + I

Question 6.
Least square method of fitting a trend is
(a) Most exact
(b) Least exact
(c) Full of subjectivity
(d) Mathematically unsolved
Solution:
(a) Most exact

Question 7.
The value of ‘b’ in the trend line y = a + bx is
(a) Always positive
(b) Always negative
(c) Either positive or negative
(d) Zero
Solution:
(c) Either positive or negative

Question 8.
The component of a time series attached to long term variation is trended as
(a) Cyclic variation
(b) Secular variations
(c) Irregular variation
(d) Seasonal variations
Solution:
(b) Secular variations

Question 9.
The seasonal variation means the variations occurring with in
(a) A number of years
(b) within a year
(c) within a month
(d) within a week
Solution:
(b) within a year

Question 10.
Another name of consumer’s price index number is:
(a) Whole-sale price index number
(b) Cost of living index
(c) Sensitive
(d) Composite
Solution:
(b) Cost of living index

Question 11.
Cost of living at two different cities can be compared with the help of
(a) Consumer price index
(b) Value index
(c) Volume index
(d) Un-weighted index
Solution:
(a) Consumer price index

Question 12.
Laspeyre’s index = 110, Paasche’s index = 108, then Fisher’s Ideal index is equal to:
(a) 110
(b)108
(c) 100
(d) 109
Solution:
(d) 109
Hint:
p²₀₁ = 110; p^p₀₁ = 108
Fisher’s Ideal Index = \(\sqrt { 110×108 }\) = \(\sqrt { 11880 }\) = 108.99 = 109

Question 13.
Most commonly used index number is:
(a) Volume index number
(b) Value index number
(c) Price index number
(d) Simple index number
Solution:
(c) Price index number

Question 14.
Consumer price index are obtained by:
(a) Paasche’s formula
(b) Fisher’s ideal formula
(c) Marshall Edgeworth formula
(d) Family budget method formula
Solution:
(d) Family budget method formula

Question 15.
Which of the following Index number satisfy the time reversal test?
(a) Laspeyre’s Index number
(b) Paasche’s Index number
(c) Fisher Index number
(d) All of them.
Solution:
(c) Fisher Index number

Question 16.
While computing a weighted index, the current period quantities are used in the:
(a) Laspeyre’s method
(b) Paasche’s method
(c) Marshall Edgeworth method
(d) Fisher’s ideal method
Solution:
(b) Paasche’s method

Question 17.
The quantities that can be numerically measured can be plotted on a
(a) p – chart
(b) c – chart
(c) x bar chart
(d) np – chart
Solution:
(c) x bar chart

Question 18.
How many causes of variation will affect the quality of a product?
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c) 2

Question 19.
Variations due to natural disorder is known as
(a) random cause
(b) non-random cause
(c) human cause
(d) all of them
Solution:
(a) random cause

Question 20.
The assignable causes can occur due to
(a) poor raw materials
(b) unskilled labour
(c) faulty machines
(d) all of them
Solution:
(d) all of them

Question 21.
A typical control charts consists of
(a) CL, UCL
(b) CL, LCL
(c) CL, LCL, UCL
(d) UCL, LCL
Solution:
(c) CL, LCL, UCL

Question 22.
\(\bar { x}\) chart is a
(a) attribute control chart
(b) variable control chart
(c) neither Attribute nor variable control chart
(d) both Attribute and variable control chart
Solution:
(b) variable control chart

Question 23.
R is calculated using
(a) x_max – x_min
(b) x_min – x_max
(c) \(\bar { x}\)_max – \(\bar { x}\)_min
(d) \(\bar {\bar x}\)_max – \(\bar {\bar x}\)_min
Solution:
(a) x_max – x_min

Question 24.
The upper control limit for x chart is given by
(a) \(\bar { x}\) + A₂\(\bar { R}\)
(b) \(\bar {\bar x}\) + A₂R
(c) \(\bar {\bar x}\) + A₂\(\bar { R}\)
(d) \(\bar { x}\) + A₂\(\bar {\bar R}\)
Solution:
(c) \(\bar {\bar x}\) + A₂\(\bar { R}\)

Question 25.
The LCL for R chart is given by
(a) D₂\(\bar { x}\)
(b) D₂\(\bar {\bar R}\)
(c) D₃\(\bar {\bar R}\)
(d) D₃\(\bar { x}\)
Solution:
(d) D₃\(\bar { x}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.2 Textbook Questions and Answers, Notes. BANDHANBNK Pivot Point Calculator

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2

Question 1.
Find the slope of the tangent to the following curves at the respective given points.
(i) y = x⁴ + 2x² – x at x = 1
(ii) x = a cos³ t, y = b sin³ t at t = \(\frac { π }{ 2 }\)
Solution:
(i) y = x⁴ + 2x² – x
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = 4x³ + 4x – 1
Slope of the tangent (\(\frac { dy }{ dx }\))_(x=1)
= 4(1)³ + 4(1) – 1
= 4 + 4 – 1 = 7
(ii) x = a cos³ t, y = b sin³ t
Differenriating w.r.t. ‘t’
\(\frac { dx }{ dt }\) = – 3a cos² t sin t
\(\frac { dy }{ dt }\) = 3b sin² t sin t

Question 2.
Find the point on the curve y = x² – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.
Solution:
y = x² – 5x + 4
Differentiating w.r.t. ‘x’
Slope of the tangent \(\frac { dy }{ dx }\) = 2x – 5
Given line 3x + y = 7
Slope of the line = –\(\frac { 3 }{ 1 }\) = -3
Since the tangent is parallel to the line, their slopes are equal.
∴ \(\frac { dy }{ dx }\) = -3
⇒ 2x – 5 = -3
2x = 2
x = 1
When x = 1, y = (1)² – 5 (1) + 4 = 0
∴ Point on the curve is (1, 0).

Question 3.
Find the points on curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y = 1729.
Solution:
y = x³ – 6x² + x+ 3
Differentiating w.r.t. ‘x’
Slope of the tangent \(\frac { dy }{ dx }\) = 3x² – 12x + 1
Slope of the normal = \(\frac { 1 }{ 3x^2 – 12x + 1 }\)
Given line is x + y = 1729
Slope of the line is – 1
Since the normal is parallel to the line, their slopes are equal.
\(\frac { 1 }{ 3x^2 – 12x + 1 }\) = -1
3x² – 12x + 1 = 1
3x² – 12x =0
3x(x – 4) = 0
x = 0, 4
When x = 0, y = (0)³ – 6(0)² + 0 + 3 = 3
When x = 4, y = (4)³ – 6(4)² + 4 + 3
= 64 – 96 + 4 + 3 = -25
∴ The points on the curve are (0, 3) and (4, -25).

Question 4.
Find the points on the curve y² – 4xy = x² + 5 for which the tangent is horizontal.
Solution:
y² – 4xy = x² + 5 ………… (1)
Differentiating w.r.t. ‘x’

When the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.
\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac { x+2y }{ y-2x }\) = 0
⇒ x + 2y = 0
x = -2y
Substituting in (1)
y² – 4 (-2y) y = (-2y)² + 5
y² + 8y² = 4y² + 5
5y² = 5 ⇒ y² = 1
y = ±1
When y = 1, x = -2
When y = – 1, x = 2
∴ The points on the curve are (- 2, 1) and (2, -1).

Question 5.
Find the tangent and normal to the following curves at the given points on the curve.
(i) y = x² – x⁴ at (1, 0)
(ii) y = x⁴ + 2e^x at (0, 2)
(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 3 }\)
Solution:
(i) y = x² – x⁴ at (1, 0)
Differentiating w.r.t. ‘x’
\(\frac { dx }{ dy }\) = 2x – 4x³
Slope of the tangent ‘m’ = (\(\frac { dx }{ dy }\))_{(1, 0)}
= 2 (1) – 4 (1)³ = -2
Slope of the normal –\(\frac { 1 }{ m }\) = \(\frac { -1 }{ -2 }\) = \(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m (x – x₁)
y – 0 = – 2 (x – 1)
y = -2x + 2
2x + y – 2 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
y – 0 = \(\frac { 1 }{ 2 }\)(x – 1)
2y = x- 1
x – 2y – 1 = 0

(ii) y = x⁴ + 2e^x at (0, 2)
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = 4x³ + 2e^x
Slope of the tangent ‘m’
(\(\frac { dy }{ dx }\))_{(0, 2)} = 4(0)³ + 2e⁰ = 2
Slope of the Normal –\(\frac { 1 }{ m }\) =-\(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – 2 = 2(x – 0)
⇒ y – 2 = 2x
⇒ 2x – y + 2 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\) (x – x₁)
y – 2 = –\(\frac { 1 }{ 2 }\)(x – 0)
2y – 4 = -x
x + 2y – 4 = 0

(iii) y = x sin x at (\(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))
Differentiating w.r.t. ‘x’
\(\frac { dy }{ dx }\) = x cos x + sin x
Slope of the tangent ‘m’ = (\(\frac { dy }{ dx }\))_{(π/2, π/2)}
= \(\frac { π }{ 2 }\) cos \(\frac { π }{ 2 }\) + sin \(\frac { π }{ 2 }\) = 1
Slope of the Normal –\(\frac { 1 }{ m }\) = -1
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – \(\frac { π }{ 2 }\) = 1 (x – \(\frac { π }{ 2 }\))
⇒ x – y = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
⇒ y – \(\frac { π }{ 2 }\) = -1(x – \(\frac { π }{ 2 }\))
⇒ y – \(\frac { π }{ 2 }\) = -x + \(\frac { π }{ 2 }\)
⇒ x + y – π = 0

(iv) x = cos t, y = 2 sin² t at t = \(\frac { π }{ 2 }\)
at t = \(\frac { π }{ 3 }\), x = cos \(\frac { π }{ 3 }\) = \(\frac { 1 }{ 2 }\)
at t = \(\frac { π }{ 3 }\), y = 2 sin² \(\frac { π }{ 3 }\) = 2(\(\frac { 3 }{ 4}\)) = \(\frac { 3 }{ 2 }\)
Point is (\(\frac { 1 }{ 2 }\), \(\frac { 3 }{ 2 }\))
Now x = cos t y = 2 sin² t
Differentiating w.r.t. ‘t’,
\(\frac { dx }{ dt }\) = -sin t; \(\frac { dy }{ dt }\) = 4 sin t cos t
Slope of the tangent

Slope of the Normal –\(\frac { 1 }{ m }\) = \(\frac { 1 }{ 2 }\)
Equation of tangent is
y – y₁ = m(x – x₁)
⇒ y – \(\frac { 3 }{ 2 }\) = -2(x – \(\frac { 1 }{ 2 }\))
⇒ 2y – 3 = – 4x + 2
⇒ 4x + 2y – 5 = 0
Equation of Normal is
y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
⇒ y – \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 2 }\)(x – \(\frac { 1 }{ 2 }\))
⇒ 2 (2y – 3) = 2x – 1
⇒ 4y – 6 = 2x – 1
⇒ 2x – 4y + 5 = 0

Question 6.
Find the equations of the tangents to the curve y = 1 + x³ for which the tangent is orthogonal with the line x + 12y = 12.
Solution:
Curve is y = 1 + x³
Differentiating w.r.t ‘x’,
Slope of the tangent ‘m’ = \(\frac { dy }{ dx }\) = 3x²
Given line is x + 12y = 12
Slope of the line is –\(\frac { 1 }{ 12 }\)
Since the tangent is orthogonal with the line, the slope of the tangent is 12.
∴ \(\frac { dy }{ dx }\) = 12
i.e 3x² = 12
x² = 4
x = ±2
When x = 2, y = 1 + 8 = 9 ⇒ point is (2, 9)
When x = -2, y = 1 – 8 = -7 ⇒ point is (-2, -7)
Equation of tangent with slope 12 and at the j point (2, 9) is
y – 9 = 12 (x – 2)
y – 9 = 12x – 24
12x – y – 15 = 0
Equation of tangent with slope 12 and at the point (-2, -7) is
y + 7 = 12 (x + 2)
y + 7 = 12x + 24
12x – y + 17 = 0

Question 7.
Find the equations of the tangents to the curve y = –\(\frac { x+1 }{ x-1 }\) which are parallel to the line x + 2y = 6.
Solution:

Given line is x + 2y = 6
Slope of the line = –\(\frac { 1 }{ 2 }\)
Since the tangent is parallel to the line, then the slope of the tangent is –\(\frac { 1 }{ 2 }\)
∴ \(\frac { dy }{ dx }\) = \(\frac { 2 }{ (x-1)^2 }\) = –\(\frac { 1 }{ 2 }\)
(x – 1)² = 4
x – 1 = ±2
x = -1, 3
When x = – 1, y = 0 ⇒ point is (-1, 0)
When x = 3, y = 2 ⇒ point is (3, 2)
Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (-1, 0) is
y – o = –\(\frac { 1 }{ 2 }\)(x + 1)
2y = -x – 1 ⇒ x + 2y + 1 = 0
Equation of tangent with slope –\(\frac { 1 }{ 2 }\) and at the point (3, 2) is 2
y – 2 = –\(\frac { 1 }{ 2 }\) (x – 3)
2y – 4 = -x + 3
x + 2y – 7 = 0.

Question 8.
Find the equation of tangent and normal to the curve given by x – 7 cos t andy = 2 sin t, t ∈ R at any point on the curve.
Solution:
x = 7 cos t and y = 2 sin t, t ∈ R
Differentiating w.r.t. ‘t’,
\(\frac { dx }{ dt }\) = -7 sin t and \(\frac { dy }{ dt }\) = 2 cos t
Slope of the tangent ‘m’
\(\frac { dy }{ dx }\) = \(\frac{\frac { dy }{ dt }}{\frac{ dx }{ dt }}\) = \(\frac { 2 cot t }{ -7 sin t }\)
Any point on the curve is (7 Cos t, 2 sin t)
Equation of tangent is y – y₁ = m (x – x₁)
y – 2 sint = –\(\frac { 2 cot t }{ 7 sin t }\) (x – 7 cos t)
7y sin t – 14 sin² t = -2x cos t + 14 cos² t
2x cos t + 7 y sin t – 14 (sin² t + cos² t) = 0
2x cos t + 7y sin t – 14 = 0
Now slope of normal is –\(\frac { 1 }{ 3 }\) = \(\frac { 7 sin t }{ 2 cos t }\)
Equation of normal is y – y₁ = –\(\frac { 1 }{ m }\)(x – x₁)
y – 2 sin t = \(\frac { 7 sin t }{ 2 cos t }\) (x – 7 cos t)
2y cos t – 4 sin t cos t = 7x sin t – 49 sin t cos t 7x sin t – 2y cos t – 45 sin t cos t = 0

Question 9.
Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0
Solution:
Given curves are xy = 2 ……… (1)
x² + 4y = 0 ………. (2)
Now solving (1) and (2)
Substituting (1) in (2)
⇒ x² + 4(2/x) = 0
x³ + 8 = 0
x³ = -8
x = -2
Substituting in (1) ⇒ y = \(\frac { 2 }{ -2 }\) = -1
∴ Point of intersection of (1) and (2) is (-2, -1)
xy = 2 ⇒ y = \(\frac { 2 }{ x }\) ……….. (1)
Differentiating w.r.t. ‘x’

The angle between the curves

Question 10.
Show that the two curves x² – y² = r² and xy = c² where c, r are constants, cut orthogonally.
Solution:
Given curves are x² – y² = r² ……….. (1)
xy = c² …….. (2)
Let (x₁, y₁) be the point of intersection of the given curves.
(1) ⇒ x² – y² = r²
Differentiating w.r.t ‘x’,
2x – 2y \(\frac { dx }{ dy }\) = 0
\(\frac { dx }{ dy }\) = \(\frac { x }{ y }\)
now (\(\frac { dx }{ dy }\))(_x1,y1) = m₁ = \(\frac { x_1 }{ y_1 }\)
(2) ⇒ xy = c²
Differentiating w.r.t ‘x’,

Hence, the given curves cut orthogonally.