Exercise 7.4 - Chapter 7 Probability Distributions 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $7.4$
Choose the correct answer:
Question 1.
Normal distribution was invented by
(a) Laplace
(b) De-Moivre
(c) Gauss
(d) all the above
Answer:
(b) De-Moivre
Question 2.
If $\mathrm{X} \sim \mathrm{N}(9,81)$ the standard normal variate $\mathrm{Z}$ will be
(a) $\mathrm{Z}=\frac{x-81}{9}$
(b) $\mathrm{Z}=\frac{X-9}{81}$
(c) $Z=\frac{X-9}{9}$
(d) $Z=\frac{9-x}{9}$
Answer:
(c) $\mathrm{Z}=\frac{X-9}{9}$
Hint:
$
\begin{aligned}
& \mu=9, \sigma=9 \\
& Z=\frac{X-9}{9}
\end{aligned}
$
Question 3.
If $\mathrm{Z}$ is a standard normal variate, the proportion of items lying between $\mathrm{Z}=-0.5$ and $\mathrm{Z}=$ $-3.0$ is
(a) $0.4987$
(b) $0.1915$
(c) $0.3072$
(d) $0.3098$
Answer:
(c) $0.3072$
Hint:

$
\mathrm{P}(-0.5<\mathrm{Z}<-3)
$
By symmetry we want $\mathrm{P}(0.5<\mathrm{Z}<3)$
$
\begin{aligned}
& =\mathrm{P}(0<\mathrm{Z}<3)-\mathrm{P}(0<\mathrm{Z}<0.5) \\
& =0.49865-0.1915 \\
& =0.30715 \sim 0.3072
\end{aligned}
$
Question 4.
If $\mathrm{X} \sim \mathrm{N}\left(\mu, \sigma^2\right)$, the maximum probability at the point of inflexion of normal distribution is
(a) $\overline{\left(\frac{1}{\sqrt{2 \pi}}\right)} e^{\frac{1}{2}}$
(b) $\left(\frac{1}{\sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}$
(c) $\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}$
(d) $\left(\frac{1}{\sqrt{2 \pi}}\right)$
Answer:
(c) $\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}$
Hint:
p.d.f is $f(x)=\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)\left(\frac{x-\mu}{\sigma}\right)^2}$
The points of inflexion of the curve are $x=\mu \pm \sigma$. Using this,
$
f(x)=\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)\left(\frac{\mu+\sigma-\mu}{\sigma}\right)^2}=\left(\frac{1}{\sigma \sqrt{2 \pi}}\right) e^{\left(-\frac{1}{2}\right)}
$

Question 5.
In a parametric distribution the mean is equal to variance is
(a) binomial
(b) normal
(c) Poisson
(d) all of the above
Answer:
(c) Poisson
Question 6.
In turning out certain toys in a manufacturing company, the average number of defectives is $1 \%$. The probability that the sample of 100 toys there will be 3 defectives is
(a) $0.0613$
(b) $0.613$
(c) $0.00613$
(d) $0.3913$
Answer:
(a) $0.0613$
Hint:
$
\begin{aligned}
& n=100, p=1 \%=\frac{1}{100} \\
& \lambda=n p=(100)\left(\frac{1}{100}\right)=1 \\
& \mathrm{P}(\mathrm{X}=3)=\frac{e^{-\lambda} \lambda^x}{x !}=\frac{e^{-1}(1)^3}{3 !}=\frac{e^{-1}}{6}=0.0613
\end{aligned}
$
Question 7.
The parameters of the normal distribution $\mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{72 \pi}} \frac{e^{(-x-10)^2}}{72}-\infty (a) $(10,6)$
(b) $(10,36)$
(c) $(6,10)$
(d) $(36,10)$
Answer:
(b) $(10,36)$
Hint:

Comparing $\mathrm{f}(\mathrm{x})$ with p.d.f of normal distribution, $\mu=10$,
$
\begin{aligned}
& \sigma \sqrt{2 \pi}=\sqrt{72 \pi}=\sqrt{36 \times 2 \pi}=6 \sqrt{2 \pi} \\
& \sigma=6
\end{aligned}
$
Question 8.
A manufacturer produces switches and experiences that 2 per cent switches are defective. The probability that in a box of 50 switches, there are at most two defective is:
(a) $2.5 \mathrm{e}^{-1}$
(b) $\mathrm{e}^{-1}$
(c) $2 \mathrm{e}^{-1}$
(d) none of the above
Answer:
(a) $2.5 \mathrm{e}^{-1}$
Hint:
$
n=50, p=\frac{2}{100}=0.02, \quad \lambda=n p=1
$
We want $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$
$
\begin{aligned}
& =e^{-1}\left[1+\frac{1}{11}+\frac{1^2}{2 !}\right]=e^{-1}\left[\frac{5}{2}\right] \\
& =2.5 e^{-1}
\end{aligned}
$
Question 9.
An experiment succeeds twice as often as it fails. The chance that in the next six trials, there shall be at least four successes is
(a) $\frac{240}{729}$
(b) $\frac{489}{729}$
(c) $\frac{496}{729}$
(d) $\frac{251}{729}$
Answer:

(c) $\frac{496}{729}$
Hint:
Let $\mathrm{X}$ be the binomial random variable. Given $\mathrm{p}=2 \mathrm{q}$
From $\mathrm{p}+\mathrm{q}=1$, we get $\mathrm{p}=\frac{2}{3}, \mathrm{q}=\frac{1}{3}$
We want $\mathrm{P}(\mathrm{X} \geq 4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)$
$
\begin{aligned}
& ={ }^6 C_4\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^2+{ }^6 C_5\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)^1+{ }^6 C_6\left(\frac{2}{3}\right)^6\left(\frac{1}{3}\right)^0 \\
& =15\left(\frac{16}{729}\right)+6\left(\frac{32}{729}\right)+\left(\frac{64}{729}\right)=\frac{496}{729}
\end{aligned}
$
Question 10.
If for a binomial distribution $\mathrm{b}(\mathrm{n}, \mathrm{p})$ mean $=4$ and variance $=\frac{4}{3}$, the probability, $\mathrm{P}(\mathrm{X} \geq$
5) is equal to
(a) $(2 / 3)^6$
(b) $(2 / 3)^5(1 / 3)$
(c) $(1 / 3)^6$
(d) $4(2 / 3)^6$
Answer:
(d) $4(2 / 3)^6$
Hint:
$
\begin{aligned}
n p=4, n p q & =\frac{4}{3} \Rightarrow q=\frac{1}{3} \text { and } p=\frac{2}{3} \\
n & =\frac{4}{p}=\frac{4}{\frac{2}{3}}=6 \\
\mathrm{P}(\mathrm{X} \geq 5) & =\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)
\end{aligned}
$

$
\begin{aligned}
& ={ }^6 \mathrm{C}_5\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)+{ }^6 \mathrm{C}_6\left(\frac{2}{3}\right)^6 \\
& =6\left(\frac{2^5}{3^6}\right)+\frac{2^6}{3^6}=3\left(\frac{2^6}{3^6}\right)+\frac{2^6}{3^6}=4\left(\frac{2}{3}\right)^6
\end{aligned}
$
Question 11.
The average percentage of failure in a certain examination is 40 . The probability that out of a group of 6 candidates atleast 4 passed in the examination are
(a) $0.5443$
(b) $0.4543$
(c) $0.5543$
(d) $0.4573$
Answer:
(a) $0.5443$
Hint:
The percentage of success $\mathrm{p}=0.6 \Rightarrow \mathrm{q}=0.4$
$
\begin{aligned}
& P(X \geq 4)=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)+(0.6)^6 \\
& =15(0.6)^4(0.16)+6(0.6)^5(0.4)+(0.6)^6 \\
& =0.31104+0.186624+0.046656 \\
& =0.54432
\end{aligned}
$

Question 12.
Forty per cent of the passengers who fly on a certain route do not check in any luggage. The planes on this route seat 15 passengers. For a full flight, what is the mean of the number of passengers who do not check in any luggage?
(a) $6.00$
(b) $6.45$
(c) $7.20$
(d) $7.50$
Answer:
(a) $6.00$
Hint:
$
\mathrm{n}=15, \mathrm{p}=0.4 \Rightarrow \operatorname{mean}(\mathrm{np})=6
$
Question 13.
Which of the following statements is/are true regarding the normal distribution curve?
(a) it is a symmetrical and bell-shaped curve
(b) it is asymptotic in that each end approaches the horizontal axis but never reaches it
(c) its mean, median and mode are located at the same point
(d) all of the above statements are true
Answer:
(d) all of the above statements are true
Question 14.
Which of the following cannot generate a Poisson distribution?
(a) The number of telephone calls received in a ten-minute interval
(b) The number of customers arriving at a petrol station
(c) The number of bacteria found in a cubic foot of soil
(d) The number of misprints per page
Answer:
(b) The number of customers arriving at a petrol station
Question 15.
The random variable $\mathrm{X}$ is normally distributed with a mean of 70 and a standard deviation of 10 . What is the probability that $\mathrm{X}$ is between 72 and 84 ?
(a) $0.683$
(b) $0.954$
(c) $0.271$
(d) $0.340$

Answer:
(d) $0.340$
Hint:
$
\begin{aligned}
& \mu=70, \sigma=10 \\
& \mathrm{P}(72<\mathrm{X}<84)=\mathrm{P}\left(\frac{72-70}{10}<\mathrm{Z}<\frac{84-70}{10}\right) \\
& =\mathrm{P}(0.2<\mathrm{Z}<1.4) \\
& =\mathrm{P}(0<\mathrm{Z}<1.4)-\mathrm{P}(0<\mathrm{Z}<0.2) \\
& =0.4192-0.0793 \\
& =0.3399 \\
& =0.340
\end{aligned}
$
Question 16.
The starting annual salaries of newly qualified chartered accountants (CA's) in South Africa follow a normal distribution with a mean of ₹ 180,000 and a standard deviation of $₹ 10,000$. What is the probability that a randomly selected newly qualified CA will earn between ₹ 165,000 and ₹ 175,000 .
(a) $0.819$
(b) $0.242$
(c) $0.286$
(d) $0.533$
Answer:
(b) $0.242$
Hint:

$
\begin{aligned}
& \mu=180,000, \sigma=10,000 \\
& P(165,000<\mathrm{X}<175,000) \\
& =\mathrm{P}\left(\frac{165,000-180,000}{10,000}<\mathrm{Z}<\frac{175,000-180,000}{10,000}\right) \\
& =\mathrm{P}(-1.5<\mathrm{Z}<-0.5)
\end{aligned}
$
By symmetry of the normal curve,
$
\begin{aligned}
& =\mathrm{P}(0.5<\mathrm{Z}<1.5) \\
& =\mathrm{P}(0<\mathrm{Z}<1.5)-\mathrm{P}(0<\mathrm{Z}<0.5)
\end{aligned}
$
$
\begin{aligned}
& =0.4332-0.1915 \\
& =0.2417 \\
& =0.242
\end{aligned}
$
Question 17.
In a large statistics class the heights of the students are normally distributed with a mean of $172 \mathrm{~cm}$ and a variance of $25 \mathrm{~cm}$. What proportion of students are between $165 \mathrm{~cm}$ and $181 \mathrm{~cm}$ in height?
(a) $0.954$
(b) $0.601$
(c) $0.718$
(d) $0.883$
Answer:
(d) $0.883$
Hint:

$
\begin{aligned}
& \mu=172, \sigma^2=25 \Rightarrow \sigma=5 \\
& \mathrm{P}(165<\mathrm{X}<181) \\
& =\mathrm{P}\left(\frac{165-172}{5}<\mathrm{Z}<\frac{181-172}{5}\right) \\
& =\mathrm{P}(-1.4<\mathrm{Z}<1.8) \\
& =\mathrm{P}(-1.4<\mathrm{Z}<0)+\mathrm{P}(0<\mathrm{Z}<1.8) \\
& =\mathrm{P}(0<\mathrm{Z}<1.4)+\mathrm{P}(0<\mathrm{Z}<1.8) \\
& =0.4192+0.4641 \\
& =0.8833
\end{aligned}
$
Question 18.
A statistical analysis of long-distance telephone calls indicates that the length of these calls is normally distributed with a mean of 240 seconds and a standard deviation of 40 seconds. What proportion of calls lasts less than 180 seconds?
(a) $0.214$
(b) $0.094$
(c) $0.933$
(d) $0.067$
Answer:
(d) $0.067$
Hint:

$
\begin{aligned}
& \mu=240, \sigma=40 \\
& \mathrm{P}(\mathrm{X}<180) \\
& =\mathrm{P}\left(\mathrm{Z}<\frac{180-240}{40}\right) \\
& =\mathrm{P}(\mathrm{Z}<-1.5)=\mathrm{P}(\mathrm{Z}>1.5) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<1.5) \\
& =0.5-0.4332 \\
& =0.0668 \\
& =0.067
\end{aligned}
$
Question 19.
Cape town is estimated to have $21 \%$ of homes whose owners subscribe to the satellite service, DSTV. If a random sample of your home in taken, what is the probability that all four home subscribe to DSTV?
(a) $0.2100$
(b) $0.5000$
(c) $0.8791$
(d) $0.0019$
Answer:
(d) $0.0019$
Hint:
$
\begin{aligned}
& p=\frac{21}{100}=0.21, q=0.79, n=4 \\
& \mathrm{P}(\mathrm{X}=4)={ }^4 \mathrm{C}_4\left(\frac{21}{100}\right)^4(0.79)^0=(0.21)^4=0.0019
\end{aligned}
$

Question 20.
Using the standard normal table, the sum of the probabilities to the right of $\mathrm{z}=2.18$ and to the left of $\mathrm{z}=-1.75$ is:
(a) $0.4854$
(b) $0.4599$
(c) $0.0146$
(d) $0.0547$
Answer:
(d) $0.0547$
Hint:
$
\begin{aligned}
& \mathrm{P}(\mathrm{Z}>2.18)+\mathrm{P}(\mathrm{Z}<-1.75) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<2.18)+\mathrm{P}(\mathrm{Z}>1.75) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<2.18)+0.5-\mathrm{P}(0<\mathrm{Z}<1.75) \\
& =0.5-0.4854+0.5-0.4599 \\
& =0.0547
\end{aligned}
$
Question 21.
The time until first failure of a brand of inkjet printers is normally distributed with a mean of 1,500 hours and a standard deviation of 200 hours. What proportion of printers fails before 1000 hours?
(a) $0.0062$
(b) $0.0668$
(c) $0.8413$

(d) $0.0228$
Answer:
(a) $0.0062$
Hint:
$
\begin{aligned}
& \mu=1500, \sigma=200 \\
& \mathrm{P}(\mathrm{X}<1000) \\
& =\mathrm{P}\left(\mathrm{Z}<\frac{1000-1500}{200}\right) \\
& =\mathrm{P}(\mathrm{Z}<-2.5)=\mathrm{P}(\mathrm{Z}>2.5) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<2.5) \\
& =0.5-0.4938 \\
& =0.0062
\end{aligned}
$
Question 22.
The weights of newborn human babies are normally distributed with a mean of $3.2 \mathrm{~kg}$ and a standard deviation of $1.1 \mathrm{~kg}$. What is the probability that a randomly selected newborn baby weight less than $2.0 \mathrm{~kg}$ ?
(a) $0.138$
(b) $0.428$
(c) $0.766$
(d) $0.262$
Answer:
(a) $0.138$
Hint:
$
\begin{aligned}
& \mu=3.2, \sigma=1.1 \\
& P(X<2) \\
& =\mathrm{P}\left(\mathrm{Z}<\frac{2-3.2}{1.1}\right) \\
& =\mathrm{P}(\mathrm{Z}<-1.09)=\mathrm{P}(\mathrm{Z}>1.09) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<1.09) \\
& =0.5-0.3621 \\
& =0.138
\end{aligned}
$
Question 23.
Monthly expenditure on their credit cards, by credit card holders from a certain bank, follows a normal distribution with a mean of ₹ $1,295.00$ and a standard deviation of ₹ 750.00. What proportion of credit card holders spend more than ₹ $1,500.00$ on their credit cards per month?
(a) $0.487$

(b) $0.394$
(c) $0.500$
(d) $0.791$
Answer:
(b) $0.394$
Hint:
$
\begin{aligned}
& \mu=1295, \sigma=750 \\
& \mathrm{P}(\mathrm{X}>1500) \\
& =\mathrm{P}\left(\mathrm{Z}>\frac{1500-1295}{750}\right) \\
& =\mathrm{P}(\mathrm{Z}>0.27) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<0.27) \\
& =0.5-0.1064 \\
& =0.3936 \sim 0.394
\end{aligned}
$
Question 24.
Let $\mathrm{z}$ be a standard normal variable. If the area to the right of $\mathrm{z}$ is $0.8413$, then the value of $\mathrm{z}$ must be:
(a) $1.00$
(b) $-1.00$
(c) $0.00$
(d) $-0.41$
Answer:
(b) $-1.00$
Hint:

$
\begin{aligned}
& \mathrm{P}(\mathrm{Z}>-\mathrm{Z})=0.8413 \\
& \Rightarrow \mathrm{P}(-\mathrm{Z}<\mathrm{Z}<0)+0.5=0.8413 \\
& \mathrm{P}(0<\mathrm{Z}<\mathrm{z})=0.8413-0.5=0.3413
\end{aligned}
$
from normal tables, $\mathrm{z}=1$.
The required value is $-\mathrm{z}=-1$
Question 25.
If the area to the left of a value of $z$ ( $z$ has a standard normal distribution) is $0.0793$, what is the value of $z$ ?
(a) $-1.41$
(b) $1.41$.
(c) $-2.25$
(d) $2.25$
Answer:
(a) $-1.41$
Hint:

$\mathrm{P}(\mathrm{Z}<-\mathrm{z})=0.0793$ By symmetry, $\mathrm{P}(\mathrm{Z}>\mathrm{z})=0.0793$
(i.e) $0.5-\mathrm{P}(0<\mathrm{Z}<\mathrm{z})=0.0793$
$\mathrm{P}(0<\mathrm{Z}<\mathrm{z})=0.4207$ from normal tables, $\mathrm{z}=1.41$
Thus the required value is $-1.41$ (since it is left of $Z=0$ )
Question 26.
If $\mathrm{P}(\mathrm{Z}>\mathrm{z})=0.8508$ what is the value of $\mathrm{z}$ ( $\mathrm{z}$ has a standard normal distribution)?
(a) $-0.48$
(b) $0.48$
(c) $-1.04$
(d) $-0.21$
Answer:
(c) $-1.04$
Hint:

$
\mathrm{P}(\mathrm{Z}>\mathrm{z})=0.8508
$
Since the given value is more than we take $z$ to the left of $Z=0$ axis.
$
\begin{aligned}
& \mathrm{P}(\mathrm{Z}>-\mathrm{Z})=0.8508 \\
& \mathrm{P}(-\mathrm{Z}<\mathrm{Z}<0)+0.5=0.8508 \\
& \mathrm{P}(0<\mathrm{Z}<\mathrm{z})=0.3508
\end{aligned}
$
From normal tables, $\mathrm{z}=1.04$
So required value is $-1.04$
Question 27.
If $\mathrm{P}(\mathrm{Z}>\mathrm{z})=0.5832$ what is the value of $\mathrm{z}$ ( $\mathrm{z}$ has a standard normal distribution)?
(a) $-0.48$
(b) $0.48$
(c) $1.04$
(d) $-0.21$
Answer:
(d) $-0.21$
Hint:

$
\mathrm{P}(0<\mathrm{Z}<\mathrm{z})=0.5832-0.5=0.0832
$

From tables, $\mathrm{z}=0.21$
Since $\mathrm{z}$ is to the left of $Z=0$, the required value is $-0.21$
Question 28.
In a binomial distribution, the probability of success is twice as that of failure. Then out of 4 trials, the probability of no success is
(a) $\frac{16}{81}$
(b) $\frac{1}{16}$
(c) $\frac{2}{27}$
(d) $\frac{1}{81}$
Answer:
(d) $\frac{1}{81}$
Hint:
$
\begin{aligned}
& p=2 q \Rightarrow p+q=1 \Rightarrow q=\frac{1}{3}, p=\frac{2}{3} \\
& \mathrm{P}(\mathrm{X}=0)={ }^4 \mathrm{C}_0 p^0 q^4=\left(\frac{1}{3}\right)^4=\frac{1}{81}
\end{aligned}
$