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:

- Poisson distribution is the only distribution in which the mean and variance are equal.
- 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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