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_{x}p^{x}q^{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_{x}p^{x}q^{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 z_{1} = -1.4(as z_{1} lies to left of z = 0)

similarly z_{2} = 1.4

Now z_{1} = \(\frac { x_1-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)

-1.4 × 100 = x_{1} – 500 ⇒ x_{1} 500 – 140

x_{1} = 360

Again z_{2} = \(\frac { x_2-500 }{100}\) ⇒ -1.4 = \(\frac { x_1-500 }{100}\)

1.4 × 100 = x_{2} – 500 ⇒ x_{2} = 140 + 500

= x_{2} = 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.