Miscellaneous Problems - Chapter 7 - Probability Distributions - 12th Maths Guide Guide Samacheer Kalvi Solutions

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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:
Let $\mathrm{X}$ be the binomial random variable denoting the number of metal pistons.
Let $p$ be the probability of rejections.
Given that $\mathrm{p}=12 \%==0.12, \mathrm{q}=0.88, \mathrm{n}=10$.
So $\mathrm{X} \sim \mathrm{B}(0.12,10)$. Hence the p.m.f of $\mathrm{X}$ is given by
$
\mathrm{P}(\mathrm{X}=x)={ }^{10} \mathrm{C}_x(0.12)^x(0.88)^{10-x}
$
(a)
$
\begin{aligned}
\mathrm{P}(\mathrm{X} \leq 2) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& ={ }^{10} \mathrm{C}_0(0.12)^0(0.88)^{10}+{ }^{10} \mathrm{C}_1(0.12)^1(0.88)^9+{ }^{10} \mathrm{C}_2(0.12)^2(0.88)^8 \\
& =(0.88)^{10}+10(0.12)(0.88)^9+45(0.12)^2(0.88)^8 \\
& =(0.88)^8+\left[(0.88)^2+(1.2)(0.88)+45(0.12)^2\right] \\
& =(0.35963)[0.7744+1.056+0.648] \\
& =(0.35963)(2.4784) \\
& =0.89131
\end{aligned}
$
Thus out of a batch of 10 pistons, the probability of no more than 2 rejects is $0.89131$
(b)
$
\begin{aligned}
\mathrm{P}(2 \leq \mathrm{X}) & =1-\mathrm{P}(\mathrm{X}<2) \\
& =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \\
& =1-\left[(0.88)^{10}+10(0.12)(0.88)^9\right] \\
& =1-\left[(0.88)^9[0.88+1.2]\right] \\
& =1-\left[(0.88)^9(2.08)\right]
\end{aligned}
$
To find $(0.88)^9$ using calculator.
$
\begin{aligned}
(0.88)^9 & =\left[(0.88)^3\right]^3=[0.681472]^3 \\
& =0.316478 \\
\Rightarrow \quad \mathrm{P}(2 \leq \mathrm{X}) & =1-[(0.316478)(2.08)] \\
& =1-0.65827=0.34173
\end{aligned}
$
Thus out of 10 pistons, the probability that at least 2 will be rejected is $0.34173$

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 $\mathrm{X}$ be the binomial random variable denoting the number of patients.
Let $\mathrm{p}$ be the probability that the patient will recover and $q$ be the probability that patient will die.
According to the problem, $\mathrm{q}=75 \%=0.75$ and $\mathrm{p}=25 \%=0.25$ and $\mathrm{n}=6$
The p.m.f is $\mathrm{P}(\mathrm{X}=x)={ }^6 \mathrm{C}_x(0.25)^x(0.75)^{6-x}$
We want $\mathrm{P}(\mathrm{X}=4)$
$
\begin{aligned}
& ={ }^6 C_4(0.25)^4(0.75)^2 \\
& =15(0.25)^4(0.75)^2 \\
& =0.03295
\end{aligned}
$
Hence the probability that 4 patients will recover out of 6 patients is $0.03295$

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:
Let $\mathrm{X}$ be the poisson random variable. It is given that mean $\lambda=\frac{3}{20} 4 \mathrm{\oplus}, 00+5$ ? $(b)$ Whatpercentofpeopleearnbetween The Poisson probability law, giving $x$ failures per week is given by,
$
\mathrm{P}(\mathrm{X}=x)=\frac{e^{-\lambda} \lambda^x}{x !}=\frac{e^{-0.15}(0.15) x}{x !}, x=0,1,2,3, \ldots .
$
Hence probability that there will not be more than one failure is given by $\mathrm{P}(\mathrm{X} \leq 1)$
$
\begin{aligned}
& =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\
& =\mathrm{e}^{-0.15}[1+0.15] \\
& =\mathrm{e}^{-0.15}(1.15) \\
& =(0.8607)(1.15) \\
& =0.98981
\end{aligned}
$

Question 4.
Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
(a) Find the probability that none passes in a given minute.
(b) What is the expected number passing in two minutes?
Solution:
Let $\mathrm{X}$ be the Poisson random variable. It is given that mean $\lambda=300 /$ hour $=300 / 60$ minutes $=5$ per minute.
(a) The Poisson law giving $x$ vehicles passing on a road in one minute is given by
$
\mathrm{P}(\mathrm{X}=x)=\frac{e^{-\lambda} \lambda^x}{x !}=\frac{e^{-5}(5)^x}{x !} x=0,1,2,3, \ldots .
$
Now the probability that no vehicles pass in a given minute is given by,
$
\mathrm{P}(\mathrm{X}=0)=\frac{e^{-5} 5^0}{0 !}=e^{-5}=0.0067379
$
(b) $\mathrm{E}(\mathrm{X})=\lambda$ (i.e) No of vehicles passing per minute, since $\lambda=5$, the expected number passing in two minutes is 10 .

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 $\mathrm{X}$ be the normal random variable denoting the scores of the students. Given that mean $\mu=500$ and s.d $\sigma=100$. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100 , we obtain percentages.

When $\mathrm{X}=585, \mathrm{Z}=\frac{585-500}{100} 65,0085 ?(c)$ Whatpercento fpeople $*$ earnmorethan
The proportion of students who scored below 585 is given by $P$ [area to the left of $Z=0.85$ ] (i.e.) $P(Z<0.85)=0.5+P(0 $=0.8023$
$=80.23 \%$
Raghul scored better than $80.23 \%$ of the students who took the test and he will be admitted to this 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?
(a) less than $19.5$ hours?
(b) between 20 and 22 hours?
Solution:
Let $\mathrm{X}$ be the normal random variable denoting the time taken to assemble a car.
Given that mean $\mu=20$ and s.d $\sigma=2$
(a) Probability that car is assembled in less than $19.5$ hours
$
\begin{aligned}
& \mathrm{P}(\mathrm{X}<19.5)=\mathrm{P}\left(\mathrm{Z}<\frac{19.5-20}{2}\right) 5,000 \text { and } \\
& \mathrm{P}(\mathrm{Z}<-0.25)=\mathrm{P}(\mathrm{Z}>0.25) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<0.25) \\
& =0.5-0.0987 \\
& =0.4013
\end{aligned}
$

(b) The probability that a car can be assembled between 20 and 22 hours is given by

$
\begin{aligned}
& \mathrm{P}(0<\mathrm{Z}<1)=0.3413 \\
&
\end{aligned}
$
Question 7.
The annual salaries of employees in a large company are approximately normally distributed with a mean of 20,000 .
(a) What per cent of people earn less than 45,000 and 70,000 ?
Solution:
Let $\mathrm{X}$ be the normal variable denoting the annual salaries of employees.
Given the mean $\mu=50,000$ and s.d $\sigma=20,000$
(a) Probability of people ehming less than $\$ 40,000$ is given by $P(X<40,000)$
$=\mathrm{P}(\mathrm{X}<40,000)$
$=\mathrm{P}\left(\mathrm{Z}<\frac{40000-50000}{20,000}\right)$
$=\mathrm{P}(\mathrm{Z}<-0.5)$
$=\mathrm{P}(0.5<\mathrm{Z})$ (By symmetry)
$=0.5-\mathrm{P}(0<\mathrm{Z}<0.5)$
$=0.5-0.1915$
$=0.3085$
Hence people who earn less than $\$ 40,000$ is $0.3085 \times 100=30.85 \%$

(b) Probability of people earning between 65,000 is $\mathrm{P}(45000<\mathrm{X}<65000)$
$
\begin{aligned}
& =\mathrm{P}\left(\frac{45000-50000}{20000}<\mathrm{Z}<\frac{65000-50000}{20000}\right) \\
& =\mathrm{P}(-0.25<\mathrm{Z}<0.75) \\
& =\mathrm{P}(-0.25<\mathrm{Z}<0)+\mathrm{P}(0<\mathrm{Z}<0.75) \\
& =\mathrm{P}(0<\mathrm{Z}<0.25)+\mathrm{P}(0<\mathrm{Z}<0.75) \\
& =0.0987+0.2734 \\
& =0.3721
\end{aligned}
$
Hence percent of people who earn between 65,000 is $0.3721 \times 100=37.21 \%$

(c) Probability of people earning more than $\$ 70,000$ is $\mathrm{P}(\mathrm{X}>70,000)$
$
\begin{aligned}
& =\mathrm{P}\left(\mathrm{Z}>\frac{70000-50000}{20000}\right) \\
& =\mathrm{P}(\mathrm{Z}>1) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<1) \\
& =0.5-0.3413
\end{aligned}
$
$
=0.1587
$
Hence percent of people who earn more than $\$ 70,000$ is $0.1587 \times 100=15.87 \%$

Question 8.
$\mathrm{X}$ is a normally distributed variable with mean $\mu=30$ and standard deviation $\sigma=4$. Find
(a) $\mathrm{P}(\mathrm{X}<40$ ) (b) $\mathrm{P}(\mathrm{X}>21)$
(c) $\mathrm{P}(30<\mathrm{X}<35)$.
Solution:
Given $X \sim N\left(\mu, \sigma^2\right)$
$\mu=30, \sigma=4$
(a) $\mathrm{P}(\mathrm{X}<40)=\mathrm{P}\left(\mathrm{Z}<\frac{40-30}{4}\right)$
$=\mathrm{P}(\mathrm{Z}<2.5)$
$=0.5+\mathrm{P}(0<\mathrm{Z}<2.5)$
$=0.5+0.4938$
$=0.9938$

(b) $\mathrm{P}(\mathrm{X}>21)=\mathrm{P}\left(\mathrm{Z}>\frac{21-30}{4}\right)$
$=\mathrm{P}(\mathrm{Z}>-2.25)$
$=0.5+\mathrm{P}(-2.25<\mathrm{Z}<0)$
$=05+\mathrm{P}(0<\mathrm{Z}<2.25)$
$=0.5+0.4878$
$=0.9878$

(c) $\mathrm{P}(30<\mathrm{X}<35)$
$=\mathrm{P}\left(\frac{30-30}{4}<\mathrm{Z}<\frac{35-30}{4}\right)$

$
\begin{aligned}
& =\mathrm{P}(0<\mathrm{Z}<1.25) \\
& =0.3944
\end{aligned}
$
Question 9.
The birth weight of babies is normally distributed with mean 3,500g and standard deviation $500 \mathrm{~g}$, What is the probability that a baby is born that weight less than $3,100 \mathrm{~g}$ ?
Solution:
Let $\mathrm{X}$ be the normal variable denoting the birth weights of babies. Given that the mean $\mu=3500$ and s.d $\sigma=500$

Probability that a baby is bom with weight less than $3100 \mathrm{~g}=\mathrm{P}(\mathrm{X}<3100)$
$
\begin{aligned}
& =\mathrm{P}\left(\mathrm{Z}<\frac{3100-3500}{500}\right) \\
& =\mathrm{P}(\mathrm{Z}<-0.8)=\mathrm{P}(\mathrm{Z}>0.8) \text { (by symmetry) } \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<0.8) \\
& =0.5-0.2881 \\
& =0.2119
\end{aligned}
$
Question 10.
People's monthly electric bills in Chennai are normally distributed with a mean of ₹ 225 and a standard deviation of ₹ 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 ₹ 100 or less?
Solution:
Let $\mathrm{X}$ be the normal variable denoting the monthly bills in rupees.

Given mean $\mu=225$ and s.d $\sigma=55$
Now the probability that the bill will be ₹100 or less is $\mathrm{P}(\mathrm{X} \leq 100)$
$
\begin{aligned}
& =\mathrm{P}\left(\mathrm{Z} \leq \frac{100-225}{55}\right) \\
& =\mathrm{P}(\mathrm{Z} \leq-2.27) \\
& =0.5-\mathrm{P}(-2.27<\mathrm{Z}<0) \\
& =0.5-\mathrm{P}(0<\mathrm{Z}<2.27) \\
& =0.5-0.4884
\end{aligned}
$

$
=0.0116
$
Thus, in a group of 500 customers, we expect to have $500 \times 0.0116=5.8 \sim 6$ customers whose electric bills will be ₹ 100 or less.