Exercise 7.3 - Chapter 7 Probability Distributions 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $7.3$
Question 1.
Define Normal distribution.
Solution:
A random variable $\mathrm{X}$ is said to follow a normal distribution with parameters $\mu$ (mean) and $\sigma^2$ (variance) if its probability density function is given by
$
f(x: \mu, \sigma)=\frac{1}{\sigma \sqrt{2 \pi}} \exp \left\{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right\} \begin{gathered}
-\infty<x<\infty, \\
-\infty<\mu<\infty, \\
\sigma>0
\end{gathered}
$
Question 2.
Define Standard normal variate.
Solution:
A random variable $\mathrm{Z}=\frac{X-\mu}{\sigma}$ is called a standard normal variate with mean 0 and standard deviation 1 (i.e.) $\mathrm{Z} \sim \mathrm{N}(0,1)$. Its probability density function is given by $\varphi(z)=\frac{1}{\sqrt{2 \pi}} e^{\frac{-z^2}{2}} \quad-\infty<z<\infty$
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:
- $\mathrm{n}$, the number of trials is infinitely large, i.e. $\mathrm{n} \rightarrow \infty$
- neither $\mathrm{p}$ (or $\mathrm{q}$ ) is very small.
Question 4.
Write down any five chief characteristics of Normal probability curve.
Solution:
Chief Characteristics of the Normal Probability Curve are as follows:

- The curve is bell-shaped and symmetrical about the line $\mathrm{x}=\mu$.
- Mean, median and mode of the distribution coincide.
- The total area under the normal curve is equal to unity.
- For a given $\mu$ and $\sigma$, there is only one normal distribution.
- The Points of inflexion are given by $x=\mu \pm \sigma$
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 $\mathrm{X}$ be the numbers of hours for which the bulbs are in use. It is given that $\mathrm{X}$ is normally distributed with mean 2040 hours and a standard deviation of 60 hours, (i.e) $\mathrm{X}$ $\sim \mathrm{N}\left(2040,60^2\right)$
(i) $\mathrm{P}(\mathrm{X}>2150)$
We change to the standard normal variate.

The total area to the right of $Z=0$ is $0.5$.
The area between $\mathrm{Z}=0$ and $1.833$ is $0.4664$ (from tables)
So $P(Z>1.833)=0.5-0.4664=0.0336$
The number of bulbs likely to bum for more than 2150 hours is $2000 \times 0.0336=67.2 \sim$ 67

(ii) We want $\mathrm{P}(\mathrm{X}<1950)$
$
\begin{aligned}
& =\mathrm{P}\left(\frac{\mathrm{X}-\mu}{\sigma}<\frac{1950-2040}{60}\right) \\
& =\mathrm{P}(\mathrm{Z}<-1.5)
\end{aligned}
$
The area between $\mathrm{Z}=-1.5$ and $\mathrm{Z}=0$ is same as area between $\mathrm{Z}=0$ and $\mathrm{Z}=1.5$. From the tables, area between $Z=0$ and $Z=1.5$ is $0.4332$
$
\mathrm{P}(\mathrm{Z}<-1.5)=0.5-0.4332=0.0668
$
Hence the number of bulbs likely to bum for less than 1950 hours is $2000 \times 0.0668=$ $133.6 \sim 134$

(iii) We want $\mathrm{P}(1920<\mathrm{X}<2100)$
When $\mathrm{X}=1920$,
$
Z=\frac{X-\mu}{\sigma}=\frac{1920-2040}{60}=-2
$
When $\mathrm{X}=2100$,
$
\mathrm{Z}=\frac{2100-2040}{60}=\frac{60}{60}=1
$

$
\begin{aligned}
& \text { So } \mathrm{P}(1920<\mathrm{X}<2100)=\mathrm{P}(-2<\mathrm{Z}<1) \\
& =\mathrm{P}(-2<\mathrm{Z}<0)+\mathrm{P}(0<\mathrm{Z}<1) \\
& =\mathrm{P}(0<\mathrm{Z}<2)+\mathrm{P}(0<\mathrm{Z}<1) \\
& =0.4772+0.3413 \\
& =0.8185
\end{aligned}
$
Hence the number of bulbs likely to bum for more than 1920 hours but less than 2100 hours is $2000 \times 0.8185=1637$.

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:
Let $\mathrm{X}$ be the normal random variable denoting the number of items in the distribution. It is given that $30 \%$ of items are under 50 $\Rightarrow \mathrm{P}(\mathrm{X}<50)=30 \%=0.3$.
Also, $10 \%$ are over 86 $\Rightarrow \mathrm{P}(\mathrm{X}>86)=10 \%=0.1$.
We have to find $\mu$ and $\sigma$.
Representing the given data diagrammatically,
Where $\mathrm{Z}_1=\frac{50-\mu}{\sigma}$ and $\mathrm{Z}_2=\frac{86-\mu}{\sigma}$

From the diagram, $\mathrm{P}\left(-\mathrm{Z}_1<\mathrm{Z}<0\right)=0.2$
By symmetry $P\left(0<Z<Z_1\right)=0.2$
$\mathrm{Z}_1=0.525$ (from the normal table)
Hence $-0.525=\frac{50-\mu}{\sigma}$
(i.e.) $50-\mu=-0.525 \sigma$
Again $\mathrm{P}\left(0<\mathrm{Z}<\mathrm{Z}_2\right)=0.4$
$\mathrm{Z}_2=1.28$
Hence $\frac{86-\mu}{\sigma}=1.28$

(or) $86-\mu=1.28 \sigma$
Solving (1) and (2)
$
\begin{aligned}
& 50-\mu=-0.525 \sigma \\
& 86-\mu=1.28 \sigma
\end{aligned}
$
Subtracting,
$
\begin{aligned}
& 36=1.28 \sigma+0.525 \sigma \\
& 36=1.805 \sigma \\
& \sigma=19.94
\end{aligned}
$
Using this in (2),
$
\begin{aligned}
& 86-\mu=1.28(19.94) \\
& 86-\mu=25.52 \\
& \mu=60.48
\end{aligned}
$
Hence the mean of the distribution is $60.48$ and standard deviation is $19.94$.
Question 7.
$\mathrm{X}$ is normally distributed with mean 12 and SD 4. Find $\mathrm{P}(\mathrm{X} \leq 20)$ and $\mathrm{P}(0 \leq \mathrm{X} \leq 12)$.
Solution:
Given $X \sim N(12,42)$, (i.e) mean $(\mu)=12$ and s.d $(\sigma)=4$
$
\begin{aligned}
& \mathrm{P}(\mathrm{X} \leq 20) \\
& =\mathrm{P}\left(\frac{\mathrm{X}-\mu}{\sigma} \leq \frac{20-12}{4}\right) \\
& =\mathrm{P}(\mathrm{X} \leq 2)
\end{aligned}
$
Now $\mathrm{P}(\mathrm{Z} \leq 2)=\mathrm{P}(\mathrm{Z} \leq 0)+\mathrm{P}(0 \leq \mathrm{Z} \leq 2)$
$
=0.5+0.4772
$
$
=0.9772
$

$
\begin{aligned}
& \mathrm{P}(0 \leq \mathrm{X} \leq 12) \\
& =\mathrm{P}\left(\frac{0-12}{4} \leq \frac{X-\mu}{\sigma} \leq \frac{12-12}{4}\right) \\
& =\mathrm{P}(-3 \leq \mathrm{Z} \leq 0) \\
& =\mathrm{P}(0 \leq \mathrm{Z} \leq 3) \\
& \mathrm{P}(0 \leq \mathrm{Z} \leq 3)=0.49865 \text { (from normal tables) }
\end{aligned}
$

Question 8.
If the heights of 500 students are normally distributed with mean $68.0$ inches and standard deviation of $3.0$ inches, how many students have height
(a) greater than 72 inches
(b) less than or equal to 64 inches
(c) between 65 and 71 inches.
Solution:
Given $\mathrm{X}$ is the normal random variables denoting the height of the students with mean $\mu$ $=68$ and s.d $(\sigma)=3$.
(a) To find $\mathrm{P}(\mathrm{X}>72)$
$
\begin{aligned}
& =\mathrm{P}\left(\mathrm{Z}>\frac{72-68}{3}\right) \\
& =\mathrm{P}\left(\mathrm{Z}>\frac{4}{3}\right) \\
& =\mathrm{P}(\mathrm{Z}>1.33)
\end{aligned}
$
Now $\mathrm{P}(\mathrm{Z}>1.33)=0.5-\mathrm{P}(0<\mathrm{Z}<1.33)$
$
\begin{aligned}
& =0.5-0.4082 \\
& =0.0918
\end{aligned}
$
Hence number of students whose height is greater than 72 inches is $500 \times 0.0918=45.9$ $\sim 46$

(b) To find $\mathrm{P}(\mathrm{X} \leq 64)$
$
\begin{aligned}
& =\mathrm{P}\left(\mathrm{Z} \leq \frac{64-68}{3}\right) \\
& =\mathrm{P}(\mathrm{Z} \leq-1.33)
\end{aligned}
$
By Symmetry,
$\mathrm{P}(\mathrm{Z} \leq-1.33)=\mathrm{P}(\mathrm{Z} \geq 1.33)=0.0918$ (see before section)
Hence number of students whose heights are less than or equal to 64 inches is $0.0918 \times$ $500=46$

(c) To find $P(65<X<71)$
$
\begin{aligned}
& =\mathrm{P}\left(\frac{65-68}{3}<\mathrm{Z}<\frac{71-68}{3}\right) \\
& =\mathrm{P}(-1<\mathrm{Z}<1) \\
& =\mathrm{P}(-1<\mathrm{Z}<0)+\mathrm{P}(0<\mathrm{Z}<1) \\
& =2 \mathrm{P}(0<\mathrm{Z}<1)\{\text { By symmetry }\} \\
& =2(0.3413) \\
& =0.6826
\end{aligned}
$
Hence number of students with height between 65 and 71 inches is $(500)(0.6826)=$ $341.3 \sim 342$.

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 $\mathrm{X}$ be the normal random variable denoting the developing time of the prints.
Given that mean $\mu=16.28$ and s.d $\sigma=0.12$
To find $\mathrm{P}(\mathrm{X}<16.35)$
When $\mathrm{X}=16.35, \mathrm{Z}=\frac{\mathrm{X}-\mu}{\sigma}=\frac{16.35-16.28}{0.12}$
$\mathrm{Z}=0.58$
So $\mathrm{P}(\mathrm{X}<16.35)=\mathrm{P}(\mathrm{Z}<0.58)$
Now $\mathrm{P}(\mathrm{Z}<0.58)=0.5+\mathrm{P}(0<\mathrm{Z}<0.58)$
$=0.5+0.2190$
$=0.7190$
Thus the probability that it will take less than $16.35$ seconds to develop prints is $0.719$.

Question 10.
Time taken by a construction company to construct a flyover is a normal variate with mean 400 labour days and a standard deviation of 100 labour days. If the company promises to construct the flyover in 450 days or less and agree to pay a penalty of $₹$ 10,000 for each labour day spent in excess of 450 . What is the probability that
(i) the company pays a penalty of at least ₹ $2,00,000$ ?
(ii) the company takes at most 500 days to complete the flyover?
Solution:
Let $\mathrm{X}$ be the normal variate denoting the number of labour days.
Given mean $\mu=400$ and s.d $\sigma=100$
(i) The company pays penalty of ₹ $2,00,000$ at the rate of ₹ 10,000 per each extra labour day.
No. of extra days $=\frac{2,00,000}{10,000}=20$
So the probability that company pays a penalty of atleast ₹ $2,00,000$ is probability that labour days should be atleast $450+20=470$ days, (i.e.) $P(X \geq 470)$
Now $\mathrm{P}(\mathrm{X} \geq 470)=\mathrm{P}\left(\mathrm{Z} \geq \frac{470-400}{100}\right)=\mathrm{P}(\mathrm{Z} \geq 0.7)$
$
\mathrm{P}(\mathrm{Z} \geq 0.7)=0.5-\mathrm{P}(0 \leq \mathrm{Z} \leq 0.7)=0.5-0.258=0.242
$

(ii) $\mathrm{P}(\mathrm{X} \leq 500)=\mathrm{P}\left(\mathrm{Z} \leq \frac{500-40}{100}\right)=\mathrm{P}(\mathrm{Z} \leq 1)$
$
\mathrm{P}(\mathrm{Z} \leq 1)=0.5+\mathrm{P}(0 \leq \mathrm{Z} \leq 1)=0.5+0.3413=0.8413
$
Thus the probability that the company takes at most 500 days to complete the flyover is $0.8413$.