Exercise 7.2 - Chapter 7 Probability Distributions 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $7.2$
Question 1.
Define Poisson distribution.
Solution:
Poisson distribution is a discrete frequency distribution which gives the probability of a number of independent events occurring in a fixed time. It is useful for characterizing events with very low probabilities of occurrence within some definite time or space.
Question 2.
Write any 2 examples for Poisson 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 conditions for which the Poisson distribution is a limiting case of the binomial distribution.
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:
- the number of trials ' $n$ ' is indefinitely large i.e, $\rightarrow \infty$
- the probability of success ' $\mathrm{p}$ ' in each trial is very small, i.e, $\mathrm{p} \rightarrow 0$
- $\mathrm{np}=\lambda$ is finite. Thus $\mathrm{p}=\frac{\lambda}{n}$ and $\mathrm{q}=1-\frac{\lambda}{n}, \lambda>0$
Question 4.
Derive the mean and variance of the Poisson distribution.
Solution:
Let $\mathrm{X}$ be a Poisson random variable with parameter $\lambda$. The p.m.f is given by
$
\begin{aligned}
\mathrm{P}(x, \lambda)=\mathrm{P}(\mathrm{X} & =x)=\frac{e^{-\lambda} \lambda^x}{x !} \\
\text { Mean } \mathrm{E}(\mathrm{X}) & =\sum_{x=0}^{\infty} x p(x, \lambda) \\
& =\sum_{x=0}^{\infty} x \frac{e^{-\lambda} \lambda^x}{x !}
\end{aligned}
$

$
\begin{aligned}
& =\lambda e^{-\lambda}\left\{\sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1) !}\right\} \\
& =\lambda e^{-\lambda}\left\{1+\frac{\lambda}{1 !}+\frac{\lambda^2}{2 !}+\frac{\lambda^3}{3 !}+\ldots\right\} \\
& =\lambda e^{-\lambda}\left(e^\lambda\right)=\lambda
\end{aligned}
$
$
\begin{aligned}
& \text { Variance }(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& \mathrm{E}\left(\mathrm{X}^2\right)=\sum_{x=0}^{\infty} x^2 p(x, \lambda) \\
& =\sum_{x=0}^{\infty} x^2 \frac{e^{-\lambda} \lambda^x}{x !} \\
& =\sum_{x=0}^{\infty}[x(x-1)+x] \frac{e^{-\lambda} \lambda^x}{x !} \\
& =\sum_{x=0}^{\infty} x(x-1) \frac{e^{-\lambda} \lambda^x}{x !}+\sum_{x=0}^{\infty} x \frac{e^{-\lambda} \lambda^x}{x !} \\
& =\sum_{x=2}^{\infty} \frac{e^{-\lambda} \lambda^x}{(x-2) !}+\mathrm{E}(\mathrm{X}) \\
& =e^{-\lambda} \lambda^2 \sum_{x=2}^{\infty} \frac{\lambda^{x-2}}{(x-2) !}+\lambda \\
& =e^{-\lambda} \lambda^2\left[1+\frac{\lambda}{1 !}+\frac{\lambda^2}{2 !}+\ldots . .\right]+\lambda \\
& =e^{-\lambda} \lambda^2\left(e^\lambda\right)+\lambda \\
& =\lambda^2+\lambda \\
& \operatorname{Var}(\mathrm{X})=\lambda^2+\lambda-\lambda^2=\lambda \\
&
\end{aligned}
$
Thus the mean and variance of Poisson distribution are both equal to $\lambda$.
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 ? [Given $\mathrm{e}^{-2.8}=0.06$ ]
Solution:
Let $\mathrm{X}$ denote the number of deaths due to the disease
$\mathrm{P}($ death $)=\frac{7}{1000}=0.007 \Rightarrow \mathrm{p}=0.007$ and $\mathrm{n}=400$
The value of mean $\lambda=n p=(0.007)(400)=2.8$
Hence $\mathrm{X}$ follows a Poisson distribution with
$
\begin{aligned}
& \begin{aligned}
& \mathrm{P}(\mathrm{X}=x)=\frac{e^{-2.8}(2.8)^x}{x !} \\
& \text { We want } \mathrm{P}(\mathrm{X}=2)=\frac{e^{-2.8}(2.8)^2}{2 !} \\
&=\frac{(0.06)(2.8)^2}{2}=0.2352
\end{aligned}
\end{aligned}
$
So the probability of just 2 deaths on account of this disease in a group of 400 is $0.2352$.
Question 7.
It is given that $5 \%$ of the electric bulbs manufactured by a company are defective. Using Poisson distribution find the probability that a sample of 120 bulbs will contain no defective bulb.
Solution:
$
\begin{aligned}
& \text { Given } \mathrm{p}=\frac{5}{100}=0.05 \text { and } \mathrm{n}=120 \\
& \Rightarrow \lambda=\mathrm{n} \mathrm{p}=(0.05)(120)=6
\end{aligned}
$
Thus $\mathrm{X}$ is a Poisson random variable with $\mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{e^{-6} 6^x}{x !}$
We want $P$ (no defective bulb) $=\mathrm{P}(\mathrm{X}=0)$
$
\begin{aligned}
& =\frac{e^{-6} 6^0}{0 !} \\
& =\mathrm{e}^{-6}
\end{aligned}
$
$=0.0025$ (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 mean 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused.
Solution:
Let $\mathrm{X}$ be the Poisson variable denoting the demand for the cars.

It is given that mean is $1.5 \Rightarrow \lambda=1.5$
(i) $\mathrm{P}$ (Neither car is used $)=\mathrm{P}(\mathrm{X}=0)=\frac{e^{-1.5}(1.5)^0}{0 !}=e^{-1.5}=0.2231$
(ii) Some demand is refused when demand is more than 2 since the firm has only 2 cars. So we want $P(X>2)$
Now $P(X>2)=1-P(X \leq 2)$
$
\begin{aligned}
& =1-[\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=0)] \\
& =1-\left[\frac{e^{-1.5}(1.5)^2}{2 !}+\frac{e^{-1.5}(1.5)^1}{1 !}+\frac{e^{-1.5}(1.5)^0}{0 !}\right] \\
& =1-e^{-1.5}\left[\frac{2.25}{2}+1.5+1\right] \\
& =1-e^{-1.5}(3.625) \\
& =1-(0.2231)(3.625) \\
& =1-0.8087 \\
& =0.1913
\end{aligned}
$
Question 9.
The average number of phone calls per minute into the switchboard of a company between $10.00$ am and $2.30 \mathrm{pm}$ is $2.5$. Find the probability that during one particular minute there will be
(i) no phone at all
(ii) exactly 3 calls
(iii) at least 5 calls.
Solution:
Let $\mathrm{X}$ be the Poisson variable denoting the number of phone calls per minute.
Given that mean $=\lambda=2.5$. The p.m. $\mathrm{f}$ is $p(\mathrm{X}=x)=\frac{e^{-2.5}(2.5)^x}{x !}$
(i) $\mathrm{P}$ (no phone call $)=\mathrm{P}(\mathrm{X}=0)$
$
=\frac{e^{-2.5}(2.5)^0}{0 !}
$

$
\begin{aligned}
&=e^{-2.5}=0.08208 \text { (use tables) } \\
& \text { (ii } \mathrm{P}(\text { exactly } 3 \text { calls) }=\mathrm{P}(\mathrm{X}=3) \\
&=\frac{e^{-2.5}(2.5)^3}{3 !}=\frac{(0.08208)(15.625)}{6} \\
&=0.2138 \\
&\text { (iii }) \mathrm{P}(\text { atleast } 5 \text { calls })=\mathrm{P}(\mathrm{X} \geq 5) \\
&=1-\mathrm{P}(\mathrm{X}<5) \\
&=1-[\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=0)] \\
& \mathrm{P}(\mathrm{X}=4)=\frac{e^{-2.5}(2.5)^4}{4 !}=\frac{(0.08208)(39.0625)}{24} \\
&=0.1336 \\
& \mathrm{P}(\mathrm{X}=2)=\frac{e^{-2.5}(2.5)^2}{2 !}=\frac{(0.08208)(6.25)}{2} \\
&=0.2565 \\
& \mathrm{P}(\mathrm{X}=1)=\frac{e^{-2.5}(2.5)}{1 !}=(0.08208)(2.5)=0.2052
\end{aligned}
$
Using the above values and $\mathrm{P}(\mathrm{X}=0)$ and $\mathrm{P}(\mathrm{X}=3)$ from the previous subdivisions in (A) we get, $\mathrm{P}(\mathrm{X} \geq 5)=1-[0.1336+0.2138+0.2565+0.2052+0.08208]$ $=1-0.89118$
$
=0.10882
$
Question 10.
The distribution of the number of road accidents per 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) at least 2 accidents and
(iii) at most 3 accidents.
Solution:
Let $\mathrm{X}$ be the Poisson variable denoting the number of accidents per day.
Given that mean is 4 (i.e, $) \lambda=4$. The p.m.f is given by $P(X=x)=\frac{e^{-4} 4^x}{x !}$
(i) $\mathrm{P}$ (no accident $)=\mathrm{P}(\mathrm{X}=0)=\mathrm{e}^{-4}=0.0183$

For 100 days we have $100 \times 0.0183=1.83 \sim 2$
Hence out of 100 days there will be no accident for 2 days.
(ii) $P$ (atleast 2 accidents) $=P(X \geq 2)$
$=1-\mathrm{P}(\mathrm{X}<2)$
$=1-[\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=0)]$
$=1-\left[\mathrm{e}^{-4}(4)+\mathrm{e}^{-4}\right]$
$=1-(0.0183)(5)$
$=1-0.0915$
$=0.9085$
For 100 days we have $100 \times 0.9085 \sim 91$
Hence out of 100 days there will be at least 2 accidents for 91 days.
(iii) $P$ (atmost 3 accidents) $=P(X \leq 3)$
$=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)$
$=e^{-4}\left[1+\frac{4}{1}+\frac{16}{2}+\frac{64}{6}\right]$
$=(0.0183)[23.6667]$
$=0.4331$
For 100 days we have $100 \times 0.4331 \sim 43$
Hence out of 100 days, there will be atmost 3 accidents for 43 days.
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 at least two fatal accidents in a year, (given $\mathrm{e}^{-}$ $0.25=0.7788)$
Solution:
Let $\mathrm{X}$ denote the number of accidents.
Given that probability of accidents ' $\mathrm{p}$ ' is $1 / 1200$ and $\mathrm{n}=300$
$\Rightarrow \lambda=n p=300\left(\frac{1}{1200}\right)=1 / 4=0.25$
So $\mathrm{X}$ is poisson variable with mean $0.25$
The p.m.f is $\mathrm{P}(\mathrm{X}=x)=\frac{e^{-0.25}(0.25)^x}{x !}$
We want $\quad P(X \geq 2)=1-P(X<2)$
$=1-[P(X=0)+P(X=1)]$

$
\begin{aligned}
& =1-\left[\mathrm{e}^{-0.25}+\mathrm{e}^{-0.25}(0.25)\right] \\
& =1-\mathrm{e}^{-0.25}(1.25) \\
& =1-(0.7788)(1.25)
\end{aligned}
$
$
=0.0265
$
Thus the probability that there will be atleast two fatal accidents in a year is $0.0265$.
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:
Let $\mathrm{X}$ denote the number of customers.
Given $\lambda=2$
$
\begin{aligned}
& \text { (i) } \mathrm{P} \text { (no customer) }=\mathrm{P}(\mathrm{X}=0) \\
& =\frac{e^{-2}(2)^0}{0 !} \\
& =\mathrm{e}^{-2} \\
& =0.1353
\end{aligned}
$
(ii) $\mathrm{P}$ ( 3 or more customers $)=\mathrm{P}(\mathrm{X} \geq 3)$
$
\begin{aligned}
& =1-\mathrm{P}(\mathrm{X}<3) \\
& =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)] \\
& =1-\left[e^{-2}+e^{-2}(2)+\frac{e^{-2}(2)^2}{2 !}\right] \\
& =1-e^{-2}[1+2+2] \\
& =1-(0.1353)(5)=0.3235
\end{aligned}
$
Thus during a given minute, the probability that three or more customers appear is $0.3235$.