4.21 Poisson Distribution

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

Again we have

$\begin{array}{l} {μ'}_{3} & = \sum_{x = 0}^{n} x^{3} p (x) \\ = \sum_{x = 0}^{n} [x (x - 1) (x - 2) + 3 x (x - 1) + x] p (x) \\ (since x^{3} = x (x - 1) (x - 2) + 3 x (x - 1) + x) \\ = \sum_{x = 0}^{n} [x (x - 1) (x - 2) + 3 \sum_{x = 0}^{n} x (x - 1) + p (x) + \sum_{x = 0}^{n} x P (x) \\ = n (n - 1) (n - 2) p^{3} {(q + p)}^{n - 3} + 3 n (n - 1) p^{2} {(q + p)}^{n - 2} + n p {(q + p)}^{n - 1} \\ = n (n - 1) (n - 2) p^{3} + 3 n (n - 1) p^{2} + n p \end{array}$ $\begin{array}{l} {μ'}_{3} & = \sum_{x = 0}^{n} x^{3} p (x) \\ = \sum_{x = 0}^{n} [x (x - 1) (x - 2) + 3 x (x - 1) + x] p (x) \\ (since x^{3} = x (x - 1) (x - 2) + 3 x (x - 1) + x) \\ = \sum_{x = 0}^{n} [x (x - 1) (x - 2) + 3 \sum_{x = 0}^{n} x (x - 1) + p (x) + \sum_{x = 0}^{n} x P (x) \\ = n (n - 1) (n - 2) p^{3} {(q + p)}^{n - 3} + 3 n (n - 1) p^{2} {(q + p)}^{n - 2} + n p {(q + p)}^{n - 1} \\ = n (n - 1) (n - 2) p^{3} + 3 n (n - 1) p^{2} + n p \end{array}$

Since $μ_{3} = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 μ_{1}^{' 3}$ $μ_{3} = {μ'}_{3} - 3 {μ'}_{2} {μ'}_{1} + 2 μ_{1}^{' 3}$

We get

$\begin{array}{l} μ_{3} & = n (n - 1) (n - 2) p^{3} + 3 n (n - 1) p^{2} + n p - 3 (n^{2} p^{2} - n p^{2} + n p) (n p) + 2 n^{3} p^{3} \\ = n p q (q - p) \end{array}$ $\begin{array}{l} μ_{3} & = n (n - 1) (n - 2) p^{3} + 3 n (n - 1) p^{2} + n p - 3 (n^{2} p^{2} - n p^{2} + n p) (n p) + 2 n^{3} p^{3} \\ = n p q (q - p) \end{array}$

Similarly we get

${μ'}_{4} = n (n - 1) (n - 2) (n - 3) p^{4} + 6 n (n - 1) (n - 2) p^{3} + 7 n (n - 1) p^{2} + n p$ ${μ'}_{4} = n (n - 1) (n - 2) (n - 3) p^{4} + 6 n (n - 1) (n - 2) p^{3} + 7 n (n - 1) p^{2} + n p$

Since $μ_{4} = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 μ_{2} {μ'}^{2} - 3 {μ'}_{1}^{4}$ $μ_{4} = {μ'}_{4} - 4 {μ'}_{3} {μ'}_{1} + 6 μ_{2} {μ'}^{2} - 3 {μ'}_{1}^{4}$

We get

$μ_{4} = n p q [1 + 3 (n - 2) p q]$ $μ_{4} = n p q [1 + 3 (n - 2) p q]$

Hence we get

$\begin{array}{l} β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = {(n p q)}^{2} \frac{{(q - p)}^{2}}{{(n p q)}^{3}} = \frac{{(1 - 2 p)}^{2}}{n p q} \\ r_{1} = \frac{1 - 2 p}{\sqrt{n p q}} \\ β_{2} = \frac{μ_{4}}{μ_{2}^{2}} = 3 + \frac{1 - 6 p q}{n p q} and r_{2} = \frac{1 - 6 p q}{n p q} \end{array}$ $\begin{array}{l} β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = {(n p q)}^{2} \frac{{(q - p)}^{2}}{{(n p q)}^{3}} = \frac{{(1 - 2 p)}^{2}}{n p q} \\ r_{1} = \frac{1 - 2 p}{\sqrt{n p q}} \\ β_{2} = \frac{μ_{4}}{μ_{2}^{2}} = 3 + \frac{1 - 6 p q}{n p q} and r_{2} = \frac{1 - 6 p q}{n p q} \end{array}$

When $p < (1 / 2)$ $p < (1 / 2)$ , the distribution is positively skewed and when $p > (1 / 2)$ $p > (1 / 2)$ , the distribution is negatively skewed. When $p = (1 / 2)$ $p = (1 / 2)$ the distribution is symmetric.

Since coefficient of kurtosis is given by $r_{2} = β_{2} - 3 = (1 - 6 p q) / n p q$ $r_{2} = β_{2} - 3 = (1 - 6 p q) / n p q$

If 6pq<1, i.e., $p q < \frac{1}{6}$ $p q < \frac{1}{6}$ , r₂>0, the distribution is Leptokurtic.

If 6pq=1, i.e., $p q = \frac{1}{6}$ $p q = \frac{1}{6}$ , r₂=0, the distribution is Mesokurtic.

If 6pq>1, i.e., $p q > \frac{1}{6}$ $p q > \frac{1}{6}$ , r₂<0, the distribution is Platykurtic.

4.20.5 Moment Generating Function of a Binomial Distribution

From the definition we have $P (X = x) = {}^{n}C_{x} p^{x} q^{n - x}$ $P (X = x) = {}^{n}C_{x} p^{x} q^{n - x}$

Therefore

$\begin{array}{l} M (t) & = E [e^{t} X] \\ = \sum_{1}^{n} e^{t x} {}^{n}C_{x} p^{x} q^{n - x} \\ = \sum_{1}^{n} {}^{n}C_{x} {(p e^{t})}^{x} q^{n - x} \\ = {(q + p e^{t})}^{n} \end{array}$ $\begin{array}{l} M (t) & = E [e^{t} X] \\ = \sum_{1}^{n} e^{t x} {}^{n}C_{x} p^{x} q^{n - x} \\ = \sum_{1}^{n} {}^{n}C_{x} {(p e^{t})}^{x} q^{n - x} \\ = {(q + p e^{t})}^{n} \end{array}$ (4.13)

(4.13)

$\begin{array}{l} M (t) & = q + p {(1 + t + \frac{t^{2}}{2!} + \frac{t^{3}}{3! + \dots})}^{n} \\ = {(q + p + p t + \frac{p t^{2}}{2!} + \frac{p t^{3}}{3! + \dots})}^{n} \\ = {(1 + p t + \frac{p t^{2}}{2!} + \frac{p t^{3}}{3! + \dots})}^{n} \end{array}$ $\begin{array}{l} M (t) & = q + p {(1 + t + \frac{t^{2}}{2!} + \frac{t^{3}}{3! + \dots})}^{n} \\ = {(q + p + p t + \frac{p t^{2}}{2!} + \frac{p t^{3}}{3! + \dots})}^{n} \\ = {(1 + p t + \frac{p t^{2}}{2!} + \frac{p t^{3}}{3! + \dots})}^{n} \end{array}$ (4.14)

(4.14)

is the mgf for the Binomial distribution.

4.20.6 Characteristics of a Binomial Distribution

Binomial distribution is a discrete probability distribution. It is applied when the number of repeated trials of any experiment is finite and fixed. Each trial is a Bernoulli’s trial. The trials are all independent. The probability of success of any trial is the same. It is denoted by p. If n denotes the number of trials, n and p are called the parameters of the binomial distribution. The mean of a Binomial distribution is np and the variance is npq. If p=0.5 the binomial distribution is symmetric. If p is very small and less than 0.5, then the Binomial distribution is skewed to the right. If p>0.5, the distribution is skewed to the left.

4.20.6.1 Solved Examples

Example 4.38: A die is thrown 4 times. Getting a number greater than 2 is a success. Find the probability of getting (1) exactly one success; (2) less than 3 success?

Solution: We have n=4

P=probability of getting a number greater than 2= $\frac{4}{6} = \frac{2}{3}$ $\frac{4}{6} = \frac{2}{3}$

$∴ q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$ $∴ q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$

1. Probability of getting one success=P(X=1)

$\begin{array}{l} = {}^{4}C_{1} {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{4 - 1} \\ = 4 \cdot \frac{2}{3} \cdot \frac{1}{27} = \frac{8}{81} \end{array}$ $\begin{array}{l} = {}^{4}C_{1} {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{4 - 1} \\ = 4 \cdot \frac{2}{3} \cdot \frac{1}{27} = \frac{8}{81} \end{array}$

2. Probability of getting less than three successes=P(X<3)

$\begin{array}{l} = P (X = 0) + P (X = 1) + P (X = 2) \\ = {}^{4}C_{0} {(\frac{2}{3})}^{0} {(\frac{1}{3})}^{4} + {}^{4}C_{1} {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{4 - 1} + {}^{4}C_{2} {(\frac{2}{3})}^{2} {(\frac{1}{3})}^{4 - 2} \\ = \frac{1}{81} + 4 \cdot \frac{2}{3} \cdot \frac{1}{27} + 6 \cdot \frac{4}{9} \cdot \frac{1}{9} \\ = \frac{1 + 8 + 24}{81} = \frac{33}{81} \end{array}$ $\begin{array}{l} = P (X = 0) + P (X = 1) + P (X = 2) \\ = {}^{4}C_{0} {(\frac{2}{3})}^{0} {(\frac{1}{3})}^{4} + {}^{4}C_{1} {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{4 - 1} + {}^{4}C_{2} {(\frac{2}{3})}^{2} {(\frac{1}{3})}^{4 - 2} \\ = \frac{1}{81} + 4 \cdot \frac{2}{3} \cdot \frac{1}{27} + 6 \cdot \frac{4}{9} \cdot \frac{1}{9} \\ = \frac{1 + 8 + 24}{81} = \frac{33}{81} \end{array}$

Example 4.39: If the chance that any one of five telephone lines is busy at any instant 0.01, what is probability that all the lines are busy? What is the probability that more than three lines are busy?

Solution: It is given that n=5, p=0.01

$q = 1 - p = 1 - 0.01 = 0.99$ $q = 1 - p = 1 - 0.01 = 0.99$

Therefore the probability that all the lines are busy=P(X=5)

$\begin{array}{l} = {}^{5}C_{5} {(0.01)}^{5} {(0.99)}^{0} \\ = {(0.01)}^{5} \end{array}$ $\begin{array}{l} = {}^{5}C_{5} {(0.01)}^{5} {(0.99)}^{0} \\ = {(0.01)}^{5} \end{array}$

Probability that more than two lines are busy=P(X>3)=P(X=4)+P(X=5)

$\begin{array}{l} = {}^{5}C_{4} {(0.01)}^{4} {(0.99)}^{1} + {}^{5}C_{5} {(0.01)}^{5} {(0.99)}^{0} \\ = 5 {(0.01)}^{4} (0.99) + {(0.01)}^{5} \\ = {(0.01)}^{4} (5 \times 0.99 + 0.01) \\ = {(0.01)}^{4} (4.95 + 0.01) \\ = {(0.01)}^{4} (4.96) \end{array}$ $\begin{array}{l} = {}^{5}C_{4} {(0.01)}^{4} {(0.99)}^{1} + {}^{5}C_{5} {(0.01)}^{5} {(0.99)}^{0} \\ = 5 {(0.01)}^{4} (0.99) + {(0.01)}^{5} \\ = {(0.01)}^{4} (5 \times 0.99 + 0.01) \\ = {(0.01)}^{4} (4.95 + 0.01) \\ = {(0.01)}^{4} (4.96) \end{array}$

Example 4.40: A box contains 100 tickets each bearing one of the numbers from 1 to 100. If five tickets are drawn successively with replacement from the box, find the probability that all the tickets bear number divisible by 10?

Solution: The numbers that are divisible by 10 are 10, 20, 30, …, 100.

There are 10 numbers (between 1 and 100 (inclusive)), which are divisible by 10

We have, n=10, $p = \frac{1}{100} = \frac{1}{10}$ $p = \frac{1}{100} = \frac{1}{10}$ , $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$ $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$

Probability that all the tickets drawn bear the number divisible by 10=P(X=5)

$= {}^{5}C_{5} {(\frac{1}{10})}^{5} {(\frac{9}{10})}^{0} = \frac{1}{10^{5}}$ $= {}^{5}C_{5} {(\frac{1}{10})}^{5} {(\frac{9}{10})}^{0} = \frac{1}{10^{5}}$

Example 4.41: A die is thrown three times. Getting a “3” or a “6” is considered to be success. Find the probability of getting at least two success?

Solution: Probability of getting a 3 or 6= $p = \frac{2}{6} = \frac{1}{3}$ $p = \frac{2}{6} = \frac{1}{3}$

Probability of getting at least two successes=P(X≥2)=P(X=2)+P(X=3)

$\begin{array}{l} = {}^{3}C_{2} {(\frac{1}{3})}^{2} {(\frac{2}{3})}^{1} = {}^{3}C_{3} {(\frac{1}{3})}^{3} {(\frac{2}{3})}^{0} \\ = 3 \cdot \frac{1}{9} \cdot \frac{2}{3} + \frac{1}{27} = \frac{7}{27} \end{array}$ $\begin{array}{l} = {}^{3}C_{2} {(\frac{1}{3})}^{2} {(\frac{2}{3})}^{1} = {}^{3}C_{3} {(\frac{1}{3})}^{3} {(\frac{2}{3})}^{0} \\ = 3 \cdot \frac{1}{9} \cdot \frac{2}{3} + \frac{1}{27} = \frac{7}{27} \end{array}$

Example 4.42: If 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random (a) 1 (b) 0, will be defective.

Solution:

Probability that the bolt is defective= $p = \frac{20}{100} = 0.2$ $p = \frac{20}{100} = 0.2$

$q = 1 - p = 1 - 0.2 = 0.8$ $q = 1 - p = 1 - 0.2 = 0.8$

We have n=4

(a) $\begin{array}{l} P (X = 1) & = {}^{4}C_{1} {(0.2)}^{1} {(0.8)}^{3} \\ = 4 (0.2) {(0.8)}^{3} = 0.4096 \end{array}$ $\begin{array}{l} P (X = 1) & = {}^{4}C_{1} {(0.2)}^{1} {(0.8)}^{3} \\ = 4 (0.2) {(0.8)}^{3} = 0.4096 \end{array}$

(b) $P (X = 0) = {}^{4}C_{0} {(0.2)}^{0} {(0.8)}^{4} = 0.4096$ $P (X = 0) = {}^{4}C_{0} {(0.2)}^{0} {(0.8)}^{4} = 0.4096$

Example 4.43: Out of 1000 families of 3 children each, how many families would you expect to have two boys and one girl assuming that boys and girls are equally likely?

Solution: Probability of having a boy= $p = \frac{1}{2}$ $p = \frac{1}{2}$

We have

$q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}, n = 3, N = 1000$ $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}, n = 3, N = 1000$

Expected number of families having two boys and one girl= $1000 \cdot {}^{3}C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{1}$ $1000 \cdot {}^{3}C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{1}$

$= 1000 \cdot 3 \cdot \frac{1}{4} \cdot \frac{1}{2} = 375$ $= 1000 \cdot 3 \cdot \frac{1}{4} \cdot \frac{1}{2} = 375$

Example 4.44: The average percentage of failures in certain examination is 40. What is the probability that out of a group of six candidates, at least four pass in the examination?

Solution: Let p be the probability that a candidate passes the examination.

$∴ p = \frac{60}{100} (given)$ $∴ p = \frac{60}{100} (given)$

We have p=0.6, q=0.4, n=6

$\begin{array}{l} P (X \geq 4) & = Probability that at least four candidates pass the examination \\ = P (X = 4) + P (X = 5) + P (X = 6) \\ = {}^{6}C_{4} {(0.6)}^{4} {(0.4)}^{2} + {}^{6}C_{5} {(0.6)}^{5} (0.4) + {}^{6}C_{6} {(0.6)}^{6} {(0.4)}^{0} \\ = (15) {(0.6)}^{4} {(0.4)}^{2} + (6) {(0.6)}^{5} (0.4) + {(0.6)}^{6} \\ = 0.5443 \end{array}$ $\begin{array}{l} P (X \geq 4) & = Probability that at least four candidates pass the examination \\ = P (X = 4) + P (X = 5) + P (X = 6) \\ = {}^{6}C_{4} {(0.6)}^{4} {(0.4)}^{2} + {}^{6}C_{5} {(0.6)}^{5} (0.4) + {}^{6}C_{6} {(0.6)}^{6} {(0.4)}^{0} \\ = (15) {(0.6)}^{4} {(0.4)}^{2} + (6) {(0.6)}^{5} (0.4) + {(0.6)}^{6} \\ = 0.5443 \end{array}$

Example 4.45: X follows binomial distribution such that 4P(x=4)=P(x=2)

If n=6, find p the probability of success?

Solution: We have

$4 P (x = 4) = P (x = 2) (g i v e n)$ $4 P (x = 4) = P (x = 2) (g i v e n)$

$4 \cdot {}^{6}C_{4} \cdot p^{4} q^{2} = {}^{6}C_{2} \cdot p^{2} q^{4}$ $4 \cdot {}^{6}C_{4} \cdot p^{4} q^{2} = {}^{6}C_{2} \cdot p^{2} q^{4}$

$4 p^{2} = q^{2}$ $4 p^{2} = q^{2}$

$4 p^{2} = {(1 - p)}^{2} = 1 + p^{2} - 2 p$ $4 p^{2} = {(1 - p)}^{2} = 1 + p^{2} - 2 p$

$3 p^{2} = 2 p - 1 = 0$ $3 p^{2} = 2 p - 1 = 0$

$\begin{array}{l} p & = \frac{- 2 \pm \sqrt{4 + 12}}{(2) (3)} = \frac{- 2 \pm 4}{6} = \frac{- 6}{6} or \frac{2}{6} \\ = - 1 or \frac{1}{3} \end{array}$ $\begin{array}{l} p & = \frac{- 2 \pm \sqrt{4 + 12}}{(2) (3)} = \frac{- 2 \pm 4}{6} = \frac{- 6}{6} or \frac{2}{6} \\ = - 1 or \frac{1}{3} \end{array}$

p cannot be negative, therefore we consider $p = 1 / 3$ $p = 1 / 3$

Thus $p = 1 / 3$ $p = 1 / 3$ (since p=−1 is admissible)

Example 4.46: Find the maximum n such that the probability of getting no head in tossing a coin n times is greater than 0.1?

Solution: Let p denote the probability of getting a head. Then

$p = \frac{1}{2}, q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$ $p = \frac{1}{2}, q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$

Let n be the number of trials, such that P(X=0)>0.1

i.e.,

${}^{n}C_{0} p^{0} q^{n - 0} > 0.1$ ${}^{n}C_{0} p^{0} q^{n - 0} > 0.1$

${(\frac{1}{2})}^{n} > 0.1$ ${(\frac{1}{2})}^{n} > 0.1$

When

n=1, we get $\frac{1}{2} > 0.5$ $\frac{1}{2} > 0.5$

n=2, we get ${(\frac{1}{2})}^{2} = 0.25 > 0.1$ ${(\frac{1}{2})}^{2} = 0.25 > 0.1$

n=3, we get ${(\frac{1}{2})}^{3} = 0.125 > 0.1$ ${(\frac{1}{2})}^{3} = 0.125 > 0.1$

n=4, we get ${(\frac{1}{2})}^{4} = 0.0625 < 0.1$ ${(\frac{1}{2})}^{4} = 0.0625 < 0.1$

and the maximum value for such that P(X=0)>.1 holds is 3. Hence the maximum n such that P(X=0)>0.1 is n=4.

Example 4.47: If the sum of the mean and variance of a binomial distribution of 5 trials is $9 / 5$ $9 / 5$ , find the binomial distribution?

Solution: Let n be the number of trials, p be the probability of success, and q be the probability of failure.

Then we have

$n p + n p q = \frac{9}{5} (given), n = 5$ $n p + n p q = \frac{9}{5} (given), n = 5$

$5 p + 5 p (1 - p) = \frac{9}{5}$ $5 p + 5 p (1 - p) = \frac{9}{5}$

$25 p + 25 p - 25 p^{2} = 9$ $25 p + 25 p - 25 p^{2} = 9$

$25 p^{2} - 50 p + 9 = 0$ $25 p^{2} - 50 p + 9 = 0$

$p = \frac{50 \pm \sqrt{2500 - 900}}{(2) (25)} = \frac{50 \pm 40}{50} = \frac{9}{5} or \frac{1}{5}$ $p = \frac{50 \pm \sqrt{2500 - 900}}{(2) (25)} = \frac{50 \pm 40}{50} = \frac{9}{5} or \frac{1}{5}$

$p = (9 / 5) > 1$ $p = (9 / 5) > 1$ is not admissible. Therefore $p = (1 / 5)$ $p = (1 / 5)$ .

The required binomial distribution is $P (X = x) = {}^{5}C_{x} {(\frac{1}{5})}^{x} {(\frac{4}{5})}^{5 - x}, 0 \leq x \leq 5$ $P (X = x) = {}^{5}C_{x} {(\frac{1}{5})}^{x} {(\frac{4}{5})}^{5 - x}, 0 \leq x \leq 5$ .

Example 4.48: If the probability of a defective bolt is $1 / 10$ $1 / 10$ .

(1) Find the mean; (2) Variance; (3) moment of coefficient skewness; and (4) Kurtosis for the distribution of defective bolts is a total of 400?

Solution: p=Probability of a defective bolt= $\frac{1}{10}$ $\frac{1}{10}$ (given)

We have n=400, $p = \frac{1}{10}$ $p = \frac{1}{10}$ , $q = \frac{9}{10}$ $q = \frac{9}{10}$

Therefore

1. $Mean = E (X) = n p = 400 \cdot \frac{1}{10} = 40$ $Mean = E (X) = n p = 400 \cdot \frac{1}{10} = 40$

2. $Variance = n p q = 400 \cdot \frac{1}{10} \cdot \frac{9}{10} = 36$ $Variance = n p q = 400 \cdot \frac{1}{10} \cdot \frac{9}{10} = 36$

3. $\begin{array}{l} Moment of coefficient skewness r_{1} & = \frac{Q - 6}{\sqrt{n p q}} = \frac{\frac{9}{10} - \frac{1}{10}}{\sqrt{36}} = \frac{\frac{8}{10}}{6} \\ = \frac{8}{60} = \frac{2}{15} = 0.1333 \end{array}$ $\begin{array}{l} Moment of coefficient skewness r_{1} & = \frac{Q - 6}{\sqrt{n p q}} = \frac{\frac{9}{10} - \frac{1}{10}}{\sqrt{36}} = \frac{\frac{8}{10}}{6} \\ = \frac{8}{60} = \frac{2}{15} = 0.1333 \end{array}$

4. $\begin{array}{l} Coefficient of Kurtosis r_{2} & = \frac{1 - 6 p q}{n p q} = \frac{1 - 6 \cdot \frac{1}{10} \cdot \frac{9}{10}}{36} = \frac{1 - 0.54}{36} \\ = \frac{0.46}{36} = 0.01277 \end{array}$ $\begin{array}{l} Coefficient of Kurtosis r_{2} & = \frac{1 - 6 p q}{n p q} = \frac{1 - 6 \cdot \frac{1}{10} \cdot \frac{9}{10}}{36} = \frac{1 - 0.54}{36} \\ = \frac{0.46}{36} = 0.01277 \end{array}$

Example 4.49: Using recurrence formula of the binomial distribution, compute P(x successes) for x=1, 2, 3, 4, 5 given n=5 and $p = \frac{1}{6}$ $p = \frac{1}{6}$ .

Solution: We have $p = \frac{1}{6}$ $p = \frac{1}{6}$ , $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$ $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$ , n=5

Therefore

$\begin{array}{l} P (X = x) = {}^{n}C_{x} p^{x} q^{n - x}; 0 \leq x \leq n = {}^{5}C_{x} {(\frac{1}{6})}^{x} {(\frac{5}{6})}^{5 - x}; 0 \leq x \leq 5 \end{array}$ $\begin{array}{l} P (X = x) = {}^{n}C_{x} p^{x} q^{n - x}; 0 \leq x \leq n = {}^{5}C_{x} {(\frac{1}{6})}^{x} {(\frac{5}{6})}^{5 - x}; 0 \leq x \leq 5 \end{array}$

The recurrence formula of binomial distribution is $P (x + 1) = \frac{n - x}{x + 1} \cdot \frac{p}{q} \cdot P (x), 0 \leq x \leq n - 1$ $P (x + 1) = \frac{n - x}{x + 1} \cdot \frac{p}{q} \cdot P (x), 0 \leq x \leq n - 1$

We have

$P (X = 0) = P (0) = {}^{5}C_{0} {(\frac{1}{6})}^{0} {(\frac{5}{6})}^{5 - 0} = \frac{3125}{7776}$ $P (X = 0) = P (0) = {}^{5}C_{0} {(\frac{1}{6})}^{0} {(\frac{5}{6})}^{5 - 0} = \frac{3125}{7776}$

Using the recurrence formula we get

$\begin{array}{l} P (1) = \frac{5 - 0}{0 + 1} \cdot \frac{(\frac{1}{6})}{(\frac{5}{6})} P (0) = 5 \cdot \frac{1}{5} P (0) = \frac{3125}{7776} \\ P (2) = \frac{5 - 1}{1 + 1} \cdot \frac{(1)}{(5)} P (1) = \frac{2}{5} \frac{3125}{7776} = \frac{1250}{7776} \\ P (3) = \frac{5 - 2}{2 + 1} \cdot \frac{(1)}{(5)} P (2) = \frac{3}{3} \cdot \frac{1}{5} \cdot \frac{1250}{7776} = \frac{250}{7776} \\ P (4) = \frac{5 - 3}{3 + 1} \cdot \frac{(1)}{(5)} P (3) = \frac{2}{4} \cdot \frac{1}{5} \cdot \frac{250}{7776} = \frac{25}{7776} \\ P (5) = \frac{5 - 4}{4 + 1} \cdot \frac{(1)}{(5)} P (4) = \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{25}{7776} = \frac{1}{7776} \end{array}$ $\begin{array}{l} P (1) = \frac{5 - 0}{0 + 1} \cdot \frac{(\frac{1}{6})}{(\frac{5}{6})} P (0) = 5 \cdot \frac{1}{5} P (0) = \frac{3125}{7776} \\ P (2) = \frac{5 - 1}{1 + 1} \cdot \frac{(1)}{(5)} P (1) = \frac{2}{5} \frac{3125}{7776} = \frac{1250}{7776} \\ P (3) = \frac{5 - 2}{2 + 1} \cdot \frac{(1)}{(5)} P (2) = \frac{3}{3} \cdot \frac{1}{5} \cdot \frac{1250}{7776} = \frac{250}{7776} \\ P (4) = \frac{5 - 3}{3 + 1} \cdot \frac{(1)}{(5)} P (3) = \frac{2}{4} \cdot \frac{1}{5} \cdot \frac{250}{7776} = \frac{25}{7776} \\ P (5) = \frac{5 - 4}{4 + 1} \cdot \frac{(1)}{(5)} P (4) = \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{25}{7776} = \frac{1}{7776} \end{array}$

The required probabilities are $\frac{3125}{7776}$ $\frac{3125}{7776}$ , $\frac{1250}{7776}$ $\frac{1250}{7776}$ , $\frac{250}{7776}$ $\frac{250}{7776}$ , $\frac{25}{7776}$ $\frac{25}{7776}$ , and $\frac{1}{7776}$ $\frac{1}{7776}$ .

Example 4.50: In 256 sets of 12 tosses of a coin, in how many cases may one expect 8 heads and 4 tails?

Solution: Probability of getting a head in a single toss= $p = \frac{1}{2}$ $p = \frac{1}{2}$

We have n=12, $q = \frac{1}{2}$ $q = \frac{1}{2}$ , x=number of heads.

Probability of getting 8 heads and 4 tails= $P (X = 8) = {}^{12}C_{8} p^{8} q^{4}$ $P (X = 8) = {}^{12}C_{8} p^{8} q^{4}$

$= {}^{12}C_{8} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{4}$ $= {}^{12}C_{8} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{4}$

Expected value of getting 8 heads and 4 tails in 256 tosses is

$\begin{array}{l} N . P (X = 8) & = 256 {(\frac{1}{2})}^{12} \\ = 31 \end{array}$ $\begin{array}{l} N . P (X = 8) & = 256 {(\frac{1}{2})}^{12} \\ = 31 \end{array}$

Example 4.51: Fit a binomial distribution and calculate the expected frequencies

x	0	1	2	3	4	5
f	10	20	30	15	15	10

Solution: Mean of the given distribution is

$μ = E (X) = \frac{(0) (10) + (1) (20) + (3) (15) + (4) (15) + (5) (10)}{10 + 20 + 30 + 15 + 15 + 10} = \frac{235}{100} = 2.35$ $μ = E (X) = \frac{(0) (10) + (1) (20) + (3) (15) + (4) (15) + (5) (10)}{10 + 20 + 30 + 15 + 15 + 10} = \frac{235}{100} = 2.35$

We have n=5

Since µ=np=2.35, we get 5p=2.35

or $p = \frac{2.35}{5} = 0.47$ $p = \frac{2.35}{5} = 0.47$

Hence

$\begin{array}{l} q = 1 - p & = 1 - 0.47 = 0.53 \\ P (X = 0) & = {}^{n}C_{0} p^{0} q^{n - 0} \\ = {}^{5}C_{0} {(0.47)}^{0} {(0.53)}^{5 - 0} \\ = {(0.53)}^{5} \\ = 0.0418 \\ N = \sum_{i = 1}^{5} f_{i} & = 10 + 20 + 30 + 15 + 15 + 10 = 100 \end{array}$ $\begin{array}{l} q = 1 - p & = 1 - 0.47 = 0.53 \\ P (X = 0) & = {}^{n}C_{0} p^{0} q^{n - 0} \\ = {}^{5}C_{0} {(0.47)}^{0} {(0.53)}^{5 - 0} \\ = {(0.53)}^{5} \\ = 0.0418 \\ N = \sum_{i = 1}^{5} f_{i} & = 10 + 20 + 30 + 15 + 15 + 10 = 100 \end{array}$

Therefore expected frequencies are

$\begin{array}{l} N . P (X = 0) & = 100 \times (0.0418) = 4.18 ≅ 4 \\ N . P (X = 1) & = 100 \times {}^{5}C_{1} (0.47) {(0.53)}^{4} \\ = 100 \times 5 (0.47) {(0.53)}^{4} = 18.53 ≅ 19 \\ N . P (X = 2) & = 100 \times {}^{5}C_{2} {(0.47)}^{2} {(0.53)}^{3} \\ = 32.87 ≅ 33 \\ N . P (X = 3) & = 100 \times {}^{5}C_{3} {(0.47)}^{3} {(0.53)}^{2} \\ = 29.15 ≅ 29 \\ N . P (X = 4) & = 100 \times {}^{5}C_{4} {(0.47)}^{4} (0.53) \\ = 12.93 ≅ 13 \\ N . P (X = 5) & = 100 \times {}^{5}C_{5} {(0.47)}^{5} (0.53) \\ = 2.29 ≅ 2 \end{array}$ $\begin{array}{l} N . P (X = 0) & = 100 \times (0.0418) = 4.18 ≅ 4 \\ N . P (X = 1) & = 100 \times {}^{5}C_{1} (0.47) {(0.53)}^{4} \\ = 100 \times 5 (0.47) {(0.53)}^{4} = 18.53 ≅ 19 \\ N . P (X = 2) & = 100 \times {}^{5}C_{2} {(0.47)}^{2} {(0.53)}^{3} \\ = 32.87 ≅ 33 \\ N . P (X = 3) & = 100 \times {}^{5}C_{3} {(0.47)}^{3} {(0.53)}^{2} \\ = 29.15 ≅ 29 \\ N . P (X = 4) & = 100 \times {}^{5}C_{4} {(0.47)}^{4} (0.53) \\ = 12.93 ≅ 13 \\ N . P (X = 5) & = 100 \times {}^{5}C_{5} {(0.47)}^{5} (0.53) \\ = 2.29 ≅ 2 \end{array}$

The expected frequencies are 4, 19, 33, 29, 13, and 2.

The required Binomial distribution is

X	0	1	2	3	4	5
N·P(X=x)	4	19	33	29	13	2

That is, the required Binomial distribution of fit is N (p+q)ⁿ=100(0.53+0.47)⁵

Example 4.52: Five dice were thrown 96 times and the number of times an odd number actually turned one in the experiment is given below. Calculate the expected frequencies?

Number of dice showing 1 or 3 or 5	0	1	2	3	4	5
Observed frequency	1	10	24	35	18	8

Solution: Probability of getting an odd number when a dice is thrown= $\frac{3}{6} = \frac{1}{2}$ $\frac{3}{6} = \frac{1}{2}$

∴ $q = \frac{1}{2}$ $q = \frac{1}{2}$

We have

$\begin{array}{l} N = 1 + 10 + 24 + 35 + 18 + 8 = 96 \\ n = 5 \end{array}$ $\begin{array}{l} N = 1 + 10 + 24 + 35 + 18 + 8 = 96 \\ n = 5 \end{array}$

The expected (theoretical) frequencies are

$\begin{array}{l} N \cdot P (X = 0) & = 96 \times {}^{5}C_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{5 - 0} = 3 \\ N \cdot P (X = 1) & = 96 \times {}^{5}C_{1} {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{5 - 1} = 15 \\ N \cdot P (X = 2) & = 96 \times {}^{5}C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{5 - 2} = 30 \\ N \cdot P (X = 3) & = 96 \times {}^{5}C_{3} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{5 - 3} = 30 \\ N \cdot P (X = 4) & = 96 \times {}^{5}C_{4} {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{5 - 4} = 15 \\ N \cdot P (X = 5) & = 96 \times {}^{5}C_{5} {(\frac{1}{2})}^{5} {(\frac{1}{2})}^{0} = 3 \end{array}$ $\begin{array}{l} N \cdot P (X = 0) & = 96 \times {}^{5}C_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{5 - 0} = 3 \\ N \cdot P (X = 1) & = 96 \times {}^{5}C_{1} {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{5 - 1} = 15 \\ N \cdot P (X = 2) & = 96 \times {}^{5}C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{5 - 2} = 30 \\ N \cdot P (X = 3) & = 96 \times {}^{5}C_{3} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{5 - 3} = 30 \\ N \cdot P (X = 4) & = 96 \times {}^{5}C_{4} {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{5 - 4} = 15 \\ N \cdot P (X = 5) & = 96 \times {}^{5}C_{5} {(\frac{1}{2})}^{5} {(\frac{1}{2})}^{0} = 3 \end{array}$

The binomial distribution of fit is $N {(q + p)}^{n} = 96 {(\frac{1}{2} + \frac{1}{2})}^{5}$ $N {(q + p)}^{n} = 96 {(\frac{1}{2} + \frac{1}{2})}^{5}$ .

Example 4.53: Show that for the binomial distribution

$μ_{r + 1} = p q (n μ_{r + 1} + \frac{d μ_{r}}{d p})$ $μ_{r + 1} = p q (n μ_{r + 1} + \frac{d μ_{r}}{d p})$

where μ_r is the rth moment about the mean. Hence obtain μ₂, μ_3, and μ₄.

Solution: We know that mean of the binomial distribution is μ=np.

Therefore

$\begin{array}{l} μ_{r} & = \sum_{x = 0}^{n} {(x - μ)}^{r} p (x) \\ = \sum_{x = 0}^{n} {(x - n p)}^{r} p (x) \\ = \sum_{x = 0}^{n} {(x - n p)}^{r} \cdot {}^{n}C_{x} p^{x} q^{n - x} \end{array}$ $\begin{array}{l} μ_{r} & = \sum_{x = 0}^{n} {(x - μ)}^{r} p (x) \\ = \sum_{x = 0}^{n} {(x - n p)}^{r} p (x) \\ = \sum_{x = 0}^{n} {(x - n p)}^{r} \cdot {}^{n}C_{x} p^{x} q^{n - x} \end{array}$

Differentiating with respect to p we get

$\begin{array}{l} \frac{d μ_{r}}{d p} & = \sum_{x = 0}^{n} - r_{n} {(x - n p)}^{r - 1} \cdot {}^{n}C_{x} p^{x} q^{n - x} + \sum_{x = 0}^{n} [{(x - n p)}^{r} [{}^{n}C_{x} p^{x} q^{n - x} + {}^{n}C_{x} p^{x} (n - x) q^{n - x - 1}]] \\ = - n r \sum_{x = 0}^{n} {(x - n p)}^{r - 1} p^{x} q^{n - x} + \frac{1}{p q} [\sum_{x = 1}^{n} {}^{n}C_{x} p^{x} q^{n - x} {(x - n p)}^{r} (x q - n p + x p)] \\ = - n r μ_{r - 1} + \frac{1}{p q} [\sum_{x = 0}^{n} {}^{n}C_{x} p^{x} q^{n - x} {(x - n p)}^{r + 1}] \\ or & = \frac{d μ_{r}}{d p} + - n r μ_{r - 1} = \frac{1}{p q} [\sum_{x = 0}^{n} {}^{n}C_{x} p^{x} q^{n - x} {(x - n p)}^{r + 1}] \\ or & p q [\frac{d μ_{r}}{d p} + n r μ_{r - 1}] = μ_{r + 1} \end{array}$ $\begin{array}{l} \frac{d μ_{r}}{d p} & = \sum_{x = 0}^{n} - r_{n} {(x - n p)}^{r - 1} \cdot {}^{n}C_{x} p^{x} q^{n - x} + \sum_{x = 0}^{n} [{(x - n p)}^{r} [{}^{n}C_{x} p^{x} q^{n - x} + {}^{n}C_{x} p^{x} (n - x) q^{n - x - 1}]] \\ = - n r \sum_{x = 0}^{n} {(x - n p)}^{r - 1} p^{x} q^{n - x} + \frac{1}{p q} [\sum_{x = 1}^{n} {}^{n}C_{x} p^{x} q^{n - x} {(x - n p)}^{r} (x q - n p + x p)] \\ = - n r μ_{r - 1} + \frac{1}{p q} [\sum_{x = 0}^{n} {}^{n}C_{x} p^{x} q^{n - x} {(x - n p)}^{r + 1}] \\ or & = \frac{d μ_{r}}{d p} + - n r μ_{r - 1} = \frac{1}{p q} [\sum_{x = 0}^{n} {}^{n}C_{x} p^{x} q^{n - x} {(x - n p)}^{r + 1}] \\ or & p q [\frac{d μ_{r}}{d p} + n r μ_{r - 1}] = μ_{r + 1} \end{array}$ (4.15)

(4.15)

Hence proved.

Putting r=1, 2, 3 successively we get

$\begin{array}{l} μ_{2} & = p q [\frac{d μ_{1}}{d p} + n μ_{0}] = p q (0 + n) = n p q \\ μ_{3} & = p q [\frac{d μ_{2}}{d p} + n μ_{1}] = p (n q - n p + 0) = n p q (q - p) \\ and μ_{4} & = p q [\frac{d μ_{3}}{d p} + 3 μ_{2}] = p q (n (6 - p^{2} - 6 p + 1) + 3 n (n p) (1 - p) \\ = n p q [1 - 6 p (1 - p) + 3 n p q] \\ = n p q [1 - 6 p q] + 3 n^{2} p^{2} q^{2}] \end{array}$ $\begin{array}{l} μ_{2} & = p q [\frac{d μ_{1}}{d p} + n μ_{0}] = p q (0 + n) = n p q \\ μ_{3} & = p q [\frac{d μ_{2}}{d p} + n μ_{1}] = p (n q - n p + 0) = n p q (q - p) \\ and μ_{4} & = p q [\frac{d μ_{3}}{d p} + 3 μ_{2}] = p q (n (6 - p^{2} - 6 p + 1) + 3 n (n p) (1 - p) \\ = n p q [1 - 6 p (1 - p) + 3 n p q] \\ = n p q [1 - 6 p q] + 3 n^{2} p^{2} q^{2}] \end{array}$

Exercise 4.7

1. Eight coins are tossed simultaneously, find the probability of getting at least 6 heads?

Ans: $\frac{37}{256}$ $\frac{37}{256}$

2. Assuming that it is true that 2 out of 10 industrial accidents are due to fatigue, find that exactly 2 of 8 industrial accidents will be due to fatigue?

Ans: 0.29

3. The average percentage of failures in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 pass in the examination?

Ans: 0.5443

4. X is a binomial random variable. If 4 P(X=4)=P(X=2) find p?

Ans: $\frac{1}{3}$ $\frac{1}{3}$

5. In a binomial distribution, the mean and SD are 8 and 2, respectively. Find n and p?

Ans: n=16, $p = 1$ $p = 1$

6. An oil exploration firm finds that 55 of the test wells it drills yield a deposit of natural gas. If it drills 6 wells, find the probability that at least 1 well will yield gas?

Ans: 1−(0.95)⁶

7. A fair coin is tossed 6 times. Find the probability of getting at least 7 heads?

Ans: $\frac{15}{64}$ $\frac{15}{64}$

8. Ten coins are thrown simultaneously. Find the probability of getting at least 7 heads?

Ans: 0.1719

9. In four throws with a pair of dice, what is the probability of throwing a doublet at least twice?

Ans: $\frac{171}{1296}$ $\frac{171}{1296}$

10. In a throw of 4 dice, find the probability that at least one die shows up 4?

Ans: $\frac{671}{1296}$ $\frac{671}{1296}$

11. The probability of a man hitting a target is $1 / 4$ $1 / 4$ . He fires 7 times. What is the probability of his hitting at least twice the target?

Ans: $\frac{4547}{8192}$ $\frac{4547}{8192}$

12. The items produced by a firm are supposed to be 5% defective. What is the probability that 8 items will contain less than 2 defective items?

Ans: ${(\frac{19}{20})}^{7} (\frac{27}{20})$ ${(\frac{19}{20})}^{7} (\frac{27}{20})$

13. Assuming that half of the population is vegetarian so that the chance of an individual being vegetarian is $1 / 2$ $1 / 2$ and assuming that 100 investigators can take samples of 10 individuals to see whether they are vegetarian, how many investigators would you expect to report that three people or less were vegetarian?

Ans: 17

14. A box contains 100 tickets each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear number divisible by 10?

Ans: $\frac{1}{10^{5}}$ $\frac{1}{10^{5}}$

15. If the probability of success is $1 / 20$ $1 / 20$ , how many trials are necessary in order that the probability of at least one success is just greater than $1 / 2$ $1 / 2$ ?

Ans: 14

16. If the probability that a man aged 60 will live to be 70 of 0.65. What is the probability that out of 10 men now 60, at least 7 will live to be 70?

Ans: 0.5139

17. The mean and variance of a binomial variable X with parameters n and p are 16 and 8. Find P(X≥1) and P(X>2)?

Ans: $1 - \frac{1}{2^{32}}$ $1 - \frac{1}{2^{32}}$ and 0.999

18. In 256 sets of 12 tosses of a coin, in how many tosses one can expect 8 heads and 4 tails?

Ans: 31

19. If 3 of 20 tyres are defective and 4 of them are randomly chosen for inspection, what is the probability that only one of the defective tyre will be included?

Ans: 0.3685

20. The mean and variance of a binomial distribution are 4 and $4 / 3$ $4 / 3$ , respectively. Find P(X≥1)?

Ans: 0.998

21. Assume that 50% of all engineering students are good in mathematics, determine the probability that among 18 engineering students (1) exactly 10; (2) at least 10; (3) at most 8; (4) at least 2 and at most 9, are good in mathematics.

Ans: (1) 0.9982; (2) ${(\frac{1}{2})}^{18} ({}^{18}C_{10} + {}^{18}C_{11} + \dots + {}^{18}C_{18})$ ${(\frac{1}{2})}^{18} ({}^{18}C_{10} + {}^{18}C_{11} + \dots + {}^{18}C_{18})$ ; (3) ${(\frac{1}{2})}^{18} ({}^{18}C_{0} + {}^{18}C_{1} + \dots + {}^{18}C_{8})$ ${(\frac{1}{2})}^{18} ({}^{18}C_{0} + {}^{18}C_{1} + \dots + {}^{18}C_{8})$ ; (4) ${(\frac{1}{2})}^{18} ({}^{18}C_{2} + {}^{18}C_{3} + \dots + {}^{18}C_{9})$ ${(\frac{1}{2})}^{18} ({}^{18}C_{2} + {}^{18}C_{3} + \dots + {}^{18}C_{9})$

22. Six dice are thrown 729 times. How many times would you expect at least three dice to show a 5 or 6?

Ans: 233

23. Out of 800 families with 4 children each, how many would you expect to have 3 boys and 1 girl? Assume equal probabilities for boys and girls.

Ans: 200

24. The mean of a binomial distribution is 3 and variance is 3/16. Find (a) the value of n; (b) P(X≥7); (c) P(1≤X≤6)?

Ans: (a) n=12; (b) 0.1446; (c) 0.82

25. A pair of dice is thrown. Find the probability of getting a sum of 11 (a) once; (b) twice?

Ans: (a) 250; (b) 25

26. Fit a binomial distribution for the data.

X	0	1	2	3	4	5
f(x)	38	144	342	287	164	25

Ans: 33, 162, 316, 309, 151, 29

27. Seven coins are tossed at a time, 128 times. The number of heads observed at each throw is recorded and the results are given below:

No. of heads	0	1	2	3	4	5	6	7
Frequency	7	6	19	35	30	23	7	1

Fit a binomial distribution to the data assuming that the coins are (a) biased; (b) unbiased.

Ans: (a) 128[0.517+0.483]⁷; (b) 1, 7, 21, 35, 35,21, 7, 1

28. Fit a binomial distribution to the following frequency distribution:

x	0	1	2	3	4	5	6
f(x)	13	25	52	38	32	16	4

Ans: 6, 28, 56, 60, 36, 12, 2

29. The probability of a newly generated virus attacked to the computer system will corrupt 4 files out of 20 files, which are opened in an hour. If 20 files are opened in an hour, find the probability that

a. At least 10 files are corrupted

b. Exactly 3 files are corrupted

c. All the files are corrupted

d. All the files are safe.

Ans: (a) 0.0000045; (b) 0.001073; (c) $\frac{1}{512}$ $\frac{1}{512}$ ; (d) 0.9999

30. Fit a binomial distribution for the following data:

x	0	1	2	3	4	5
f(x)	2	14	20	34	22	8

Ans: 2, 10, 26, 34, 22, 6

31. Fit a binomial distribution whose mean is 9 and SD is $3 / 2$ $3 / 2$ .

Ans: ${}^{12}C_{x} {(\frac{3}{4})}^{x} {(\frac{1}{4})}^{12 - x} (0 \leq x \leq 12)$ ${}^{12}C_{x} {(\frac{3}{4})}^{x} {(\frac{1}{4})}^{12 - x} (0 \leq x \leq 12)$

32. The sum of mean and variance of a binomial distribution is 15 and the sum of their squares is 117. Find the distribution?

Ans: ${}^{27}C_{x} {(\frac{1}{3})}^{x} {(\frac{2}{3})}^{27 - x} (0 \leq x \leq 27)$ ${}^{27}C_{x} {(\frac{1}{3})}^{x} {(\frac{2}{3})}^{27 - x} (0 \leq x \leq 27)$

33. If on an average 1 vessel in every 10 is wrecked. Find the probability that out of 5 vessels expected to arrive, at least 4 will arrive safely?

Ans: 0.9185

34. The probability that a bulb produced by a factory will fuse after 160 days of use is 0.06. Find the probability that out of 5 such bulbs, at most 1 bulb will fuse after 160 days in use?

Ans: 0.9681

35. Find the expectation of the number of heads in 15 tosses of a coin?

Ans: 7.5

36. The incidence of an occupational disease in an industry is such that the workmen have a 20% chance of suffering from it. What is the probability that out of 6 workmen, 4 or more will contact the disease?

Ans: 0.016

37. The probability of a defective bolt is 0.2. Find (a) Mean, (b) SD of the distribution, (c) Coefficient of skewness β₁, and (d) coefficient of kurtosis β₂?

Ans: (a) 200; (b) 12.6; (c) β₁=0.225; ^(d) β₂=3.0025

38. Show that mean of a binomial distribution is always greater than its variance.

39. Determine the binomial distribution whose mean is 5 and SD is $\sqrt{2.5}$ $\sqrt{2.5}$ .

Ans: ${}^{10}C_{x} {(\frac{1}{2})}^{x} {(\frac{1}{2})}^{10 - x} (0 \leq x \leq 10)$ ${}^{10}C_{x} {(\frac{1}{2})}^{x} {(\frac{1}{2})}^{10 - x} (0 \leq x \leq 10)$

40. The mean of a binomial distribution is 20 and SD is 4. Find n, p, and q?

Ans: p=0.2, q=0.8, n=100

4.21 Poisson Distribution

A type of probability distribution useful in describing the number of events that will occur in a specific period of time or in a specific area or volume is the Poisson distribution. It is a probability distribution discovered by French mathematician Simon Denis Poisson (1781–1840) in the year 1837. A use for this distribution was not found until 1898, when an individual named Bortkiewicz was tasked by the Prussian army to investigate accidental deaths of soldiers attributed to being kicked by horses. The initial application of the Poisson distribution to number of deaths attributed to horse kicks in the Prussian army led to its use in analyzing the accidental deaths, and service requirements.

The following are the examples of random variables for which the Poisson probability distribution provides a good model:

1. The number of car accidents in a year, on a road.

2. The number of earthquakes in a year.

3. The number of breakdowns of an electronic computer.

4. The number of printing mistakes at each page of the book.

5. The number of defective screws per box of 10 screws.

The distribution is useful in testing the randomness of a set of data and fitting of empirical data to a theoretical curve.

4.21.1 Conditions Under Which Poisson Distribution Is Used

The conditions for the applicability of the Poisson distribution are the same as those for the applicability of the Binomial distribution.

The additional requirement is that the probability of success is very small. The conditions are:

1. The random variable X is discrete.

2. The number of trials is indefinitely very large, i.e., n→∞.

3. The probability of success in a trail is very small, i.e., p is close to zero.

4. The product of n and p is constant.

A random variable that counts the number of successes in an experiment is called a Poisson variable.

4.21.2 Poisson Probability Function

In this section we show that Poisson distribution is the limiting form of a binomial distribution and define Poisson distribution.

Let an experiment satisfying the conditions for Poisson distribution be formed. Let n denote the number of trials, and p denote the probability of success. We assume that n is indefinitely very large and p is very small.

Using the definition of binomial distribution, we have,

$\begin{array}{l} P (X = x) & = Probability of x successes \\ = {}^{n}C_{x} p^{x} q^{n - x}, 0 \leq x \leq n, and q = 1 - p \end{array}$ $\begin{array}{l} P (X = x) & = Probability of x successes \\ = {}^{n}C_{x} p^{x} q^{n - x}, 0 \leq x \leq n, and q = 1 - p \end{array}$

Let np=λ, then $p = \frac{λ}{n}$ $p = \frac{λ}{n}$ and $q = 1 - p = 1 - \frac{λ}{n}$ $q = 1 - p = 1 - \frac{λ}{n}$

Therefore we get

$\begin{array}{l} P (X = x) & = {}^{n}C_{x} p^{x} q^{n - x} \\ = \frac{n!}{x! (n - x)!} {(\frac{λ}{n})}^{x} {(\frac{1 - λ}{n})}^{n - x} \\ = \frac{n (n - 1) \dots (n - x + 1) (n - x)!}{x! (n - x)!} {(\frac{λ}{n})}^{x} {(1 - \frac{λ}{n})}^{n - x} \\ = \frac{(\frac{n}{n}) (\frac{n - 1}{n}) \dots (\frac{n - (x - 1)}{n})}{x!} {(λ)}^{x} {(1 - \frac{λ}{n})}^{n} {(1 - \frac{λ}{n})}^{- x} \\ = \frac{λ^{x}}{x!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots (1 - \frac{x - 1}{n}) {(1 - \frac{λ}{n})}^{n} {(1 - \frac{λ}{n})}^{- x} \\ ∴ \begin{matrix} \lim \\ n \to \infty \end{matrix} P (X = x) & = \begin{matrix} \lim \\ n \to \infty \end{matrix} [\frac{λ^{x}}{x!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots (1 - \frac{x - 1}{n})] \\ [{[{(1 - \frac{λ}{n})}^{\frac{- n}{λ}}]}^{- λ} {(1 - \frac{λ}{n})}^{- x}] \\ = \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{λ^{x}}{x!} \begin{matrix} \lim \\ n \to \infty \end{matrix} (1 - \frac{1}{n}) (1 - \frac{2}{n}) \\ \dots (1 - \frac{x - 1}{n}) \begin{matrix} \lim \\ \frac{- n}{m} \to \infty \end{matrix} [{(1 - \frac{λ}{n})}^{\frac{- n}{λ}} \begin{matrix} \lim \\ n \to \infty \end{matrix} {(1 - \frac{λ}{n})}^{- x}] \\ = \frac{λ^{x}}{x!} (1 - 0) (1 - 0) \dots (1 - 0) \cdot e^{- x} \cdot (1 - 0) \\ = \frac{λ^{x} e^{- x}}{x!} \end{array}$ $\begin{array}{l} P (X = x) & = {}^{n}C_{x} p^{x} q^{n - x} \\ = \frac{n!}{x! (n - x)!} {(\frac{λ}{n})}^{x} {(\frac{1 - λ}{n})}^{n - x} \\ = \frac{n (n - 1) \dots (n - x + 1) (n - x)!}{x! (n - x)!} {(\frac{λ}{n})}^{x} {(1 - \frac{λ}{n})}^{n - x} \\ = \frac{(\frac{n}{n}) (\frac{n - 1}{n}) \dots (\frac{n - (x - 1)}{n})}{x!} {(λ)}^{x} {(1 - \frac{λ}{n})}^{n} {(1 - \frac{λ}{n})}^{- x} \\ = \frac{λ^{x}}{x!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots (1 - \frac{x - 1}{n}) {(1 - \frac{λ}{n})}^{n} {(1 - \frac{λ}{n})}^{- x} \\ ∴ \begin{matrix} \lim \\ n \to \infty \end{matrix} P (X = x) & = \begin{matrix} \lim \\ n \to \infty \end{matrix} [\frac{λ^{x}}{x!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots (1 - \frac{x - 1}{n})] \\ [{[{(1 - \frac{λ}{n})}^{\frac{- n}{λ}}]}^{- λ} {(1 - \frac{λ}{n})}^{- x}] \\ = \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{λ^{x}}{x!} \begin{matrix} \lim \\ n \to \infty \end{matrix} (1 - \frac{1}{n}) (1 - \frac{2}{n}) \\ \dots (1 - \frac{x - 1}{n}) \begin{matrix} \lim \\ \frac{- n}{m} \to \infty \end{matrix} [{(1 - \frac{λ}{n})}^{\frac{- n}{λ}} \begin{matrix} \lim \\ n \to \infty \end{matrix} {(1 - \frac{λ}{n})}^{- x}] \\ = \frac{λ^{x}}{x!} (1 - 0) (1 - 0) \dots (1 - 0) \cdot e^{- x} \cdot (1 - 0) \\ = \frac{λ^{x} e^{- x}}{x!} \end{array}$

Therefore when n is indefinitely very large.

$P (X = x) = \frac{λ^{x} e^{- x}}{x!}, x = 0, 1, 2, 3, \dots$ $P (X = x) = \frac{λ^{x} e^{- x}}{x!}, x = 0, 1, 2, 3, \dots$

which is called the Poisson probability function. The corresponding distribution is called Poisson distribution. λ is called the parameter of the distribution.

Definition 4.27

(Poisson distribution) A discrete random variable X taking the values 0, 1, 2, … is said to follow Poisson distribution with parameter λ, if its pmf is given by

$\begin{array}{l} P (X = x) & = \frac{λ^{x} e^{- x}}{x!}, & x = 0, 1, 2, 3, \dots \dots \\ = 0, & otherwise \end{array}$ $\begin{array}{l} P (X = x) & = \frac{λ^{x} e^{- x}}{x!}, & x = 0, 1, 2, 3, \dots \dots \\ = 0, & otherwise \end{array}$

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 4.21 Poisson Distribution

Create new playlist

Sign In

Sign Up

4.20.5 Moment Generating Function of a Binomial Distribution

4.20.6 Characteristics of a Binomial Distribution

4.20.6.1 Solved Examples

4.21 Poisson Distribution

4.21.1 Conditions Under Which Poisson Distribution Is Used

4.21.2 Poisson Probability Function

Table of Contents for
4.21 Poisson Distribution