3 Marks Question

Question 13 Marks

Consider the following hypothesis test:
$\mathrm{H}_{0}: \mu=18$
$\mathrm{H}_{\mathrm{a}}: \mu \neq 18$
A sample of 48 provided a sample mean $\bar{x}=17$ and a sample standard deviation $\mathrm{S}=4.5$
i. Compute the value of the test statistic.
ii. Use the t-distribution table to compute a range for the p-value.
iii. At $\alpha=0.05$, what is your conclusion?
iv. What is the rejection rule using the critical value? What is your conclusion?

Answer

Given $\mu_{0}=18, \mathrm{n}=48, \bar{x}=17, \mathrm{~S}=4.5$
$
\text { i. } \begin{aligned}
& \mathrm{t}=\frac{\bar{x}-\mu_{0}}{\frac{S}{\sqrt{n}}}=\frac{17-18}{\frac{4.5}{\sqrt{48}}} \\
&=\frac{-1 \times \sqrt{48}}{4.5}=-1.54 \\
& \therefore \mathrm{t}=-1.54
\end{aligned}
$
and degrees of freedom $=48-1=47$.
ii. $\because t=-1.54<0$
So, p -value of $-1.54=2 \times$ Area under the t -distribution curve to the left of t
$=2 \times$ Area under the t -distribution curve to the right of t
From the $t$-distribution table, we find that $t=1.54$ lies between 1.300 and 1.678 for which area lies between 0.05 and 0.10 , so, p -values lies between $2 \times 0.05$ and $2 \times 0.10$ i.e. between 0.10 and 0.20 .
So, $0.10<$ p-value $<0.20$
iii. $\because$ p-value $>0.05$
So, do not reject $\mathrm{H}_{0}$.

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Question 23 Marks

From the following data calculate the 4-yearly moving averages and determine the trend values.

Years	2012	2013	2014	2015	2016	2017	2018	2019	2020	2021
Value	50	36.5	43	44.5	38.9	38.9	32.6	41.7	41.1	33.8

Answer

Calculation of Trend values by it four yearly Moving Averages:

Year	Value	4-yearly centered Moving Total	4-yearly Moving Average (Trend values)	4-yearly centered Moving Average
2012	50	-
2013	36.5	-
		174	43.5
2014	43	-		42.12
		162.9	40.73
2015	44.5	-		41.03
		165.3	41.33
2016	38.9	-		40.03
		154.9	38.73
2017	38.9	-		38.38
		152.1	38.03
2018	32.6	-		38.31
		154.3	38.58
2019	41.7	-		37.94
		149.2	37.3
2020	41.1	-
2021	33.8	-

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Question 33 Marks

Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation and variance of X .

Answer

Let X be a random variable denoting the number of sixes in throwing a die two times. Then, X can take values $0,1,2$.
Now,
$P(X=0)=P($ six does not appear on any of die $)=\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}$
$\mathrm{P}(\mathrm{X}=1)=\mathrm{P}($ six appears atleast once of the die $)=\left(\frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6}\right)=\frac{10}{36}=\frac{5}{18}$
$P(X=2)=P($ six does appear on both of die $)=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$
Hence, the required probability distribution is,

X	0	1	2
P(X)	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

Computation of mean and variance

$x_i$	$p _{ i }= P \left( X = x _{ i }\right)$	$P _{ i } x _{ i }$	$p_i x_i^2$
0	$\frac{25}{36}$	0	0
1	$\frac{10}{36}$	$\frac{10}{36}$	$\frac{10}{36}$
2	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{4}{36}$
		$\Sigma p_i x_i=\frac{12}{36}$	$\Sigma p_i x_i^2=\frac{14}{36}$

Thus, we have
$\Sigma p_{i} x_{i}=\frac{12}{36}=\frac{1}{3}$ and $\Sigma p_{i} x_{i}^{2}=\frac{7}{18}$
$\therefore \mathrm{E}(\mathrm{X})=\Sigma p_{i} x_{i}=\frac{1}{3}$ and, $\operatorname{Var}(\mathrm{X})=\Sigma p_{i} x_{i}^{2}-\left(\Sigma p_{i} x_{i}\right)^{2}=\frac{7}{18}-\frac{1}{9}=\frac{5}{18}$
Hence, $\mathrm{E}(\mathrm{X})=\frac{1}{3}$ and $\operatorname{Var}(\mathrm{X})=\frac{5}{18}$

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Question 43 Marks

From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the items in the sample are drawn one by one without replacement, find:
i. The probability distribution of X
ii. Mean of X
iii. Variance of X

Answer

It is clear that $X$ can assume values $0,1,2,3$ such that,
$\mathrm{P}(\mathrm{X}=0)=\frac{{ }^{7} C_{4}}{{ }^{10} C_{4}}=\frac{1}{6}, \mathrm{P}(\mathrm{X}=1)=\frac{{ }^{3} C_{1} \times{ }^{7} C_{3}}{{ }^{10} C_{4}}=\frac{1}{2}$
$\mathrm{P}(\mathrm{X}=2)=\frac{{ }^{3} C_{2} \times{ }^{7} C_{2}}{{ }^{10} C_{4}}=\frac{3}{10}$, and $\mathrm{P}(\mathrm{X}=3)=\frac{{ }^{3} C_{3} \times{ }^{7} C_{1}}{{ }^{10} C_{4}}=\frac{1}{30}$
Therefore, the probability distribution of X is as follows:

X	0	1	2	3
P(X)	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Computation of mean and variance :

$x _{ i }$	$P \left( X = x _{ i }\right)= p _{ i }$	$p_i x_i$	$p_i x_i^2$
0	$\frac{1}{6}$	0	0
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{3}{10}$	$\frac{3}{5}$	$\frac{6}{5}$
3	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{3}{10}$
		$\Sigma p_i x_i=\frac{12}{10}$	$\Sigma p_i x_i^2=2$

Thus, we have $\Sigma p_{i} x_{i}=\frac{12}{10}=\frac{6}{5}$ and $\Sigma p_{i} x_{i}^{2}=2$
$\therefore \bar{X}=$ Mean $=\Sigma p_{i} x_{i}=\frac{6}{5}$
and, $\operatorname{Var}(\mathrm{X})=\Sigma p_{i} x_{i}^{2}-\left(\Sigma p_{i} x_{i}\right)^{2}=2-\left(\frac{6}{5}\right)^{2}=2-\frac{36}{25}=\frac{14}{25}$
Hence, Mean $=\frac{6}{5}$ and Variance $=\frac{14}{25}$

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Question 53 Marks

The marginal cost function of a product is given by MC $=\frac{x}{\sqrt{x^{2}+400}}$. Find the total cost and the average cost if the fixed cost is ₹ 1000 .

Answer

Let $\mathrm{C}(\mathrm{x})$ be the total cost of x units of the product and MC be the marginal cost, then
$\mathrm{MC}=\frac{x}{\sqrt{x^{2}+400}}$ (given)
As MC $=\frac{d}{d x}(\mathrm{C}(\mathrm{x}))$, so $\frac{d}{d x}(\mathrm{C}(\mathrm{x}))=\frac{x}{\sqrt{x^{2}+400}}$
$\therefore \mathrm{C}(\mathrm{x})=\int \frac{x}{\sqrt{x^{2}+400}} d x$ (put $\sqrt{x^{2}+400}=\mathrm{t}$ i.e. $\mathrm{x}^{2}+400=\mathrm{t}^{2} \Rightarrow 2 \mathrm{x} \mathrm{dx}=2 \mathrm{tdt}$ i.e. $\mathrm{x} \mathrm{dx}=\mathrm{t} \mathrm{dt}$ )
$=\int \frac{t d t}{t}=\int 1 \mathrm{dt}=\mathrm{t}+\mathrm{k}, \mathrm{k}$ is constant of integration
$\Rightarrow \mathrm{C}(\mathrm{x})=\sqrt{x^{2}+400}+\mathrm{k}$.
Given fixed cost (in ₹) $=1000$ i.e. when $x=0, C(x)=1000$
$\Rightarrow 1000=\sqrt{0^{2}+400}+\mathrm{k} \Rightarrow 1000=20+\mathrm{k} \Rightarrow \mathrm{k}=980$
$\therefore \mathrm{C}(\mathrm{x})=\sqrt{x^{2}+400}+980$
Average cost $=\frac{C(x)}{x}=\frac{\sqrt{x^{2}+400}}{x}+\frac{980}{x}$.

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Question 63 Marks

Consider a bond with a coupon rate of $10 \%$ charged annually. The par value is ₹ 2,000 and the bond has 5 years of maturity. The yield to maturity is $11 \%$. What is the value of the bond.

Answer

Face value $\mathrm{C}=$₹ $2,000$
Coupon rate $i_{d}=10 \%$ annually or 0.1
Therefore $\mathrm{R}=\mathrm{C} \times \mathrm{i}_{\mathrm{d}}=2,000 \times 0.1=$₹ $200$
No. of periods before redemption ( n ) $=5$
Yield rate $\mathrm{i}=11 \%$ or 0.11
Therefore,
$\mathrm{V}=R\left|\frac{1-(1+i)^{-n}}{i}\right|+\mathrm{C}(1+i)^{-n}$
$=200\left[\frac{1-(1+0.11)^{-5}}{0.11}\right]+2000(1+0.11)^{-5}$
$=200\left[\frac{1-(1.11)^{-5}}{0.11}\right]+2000(1.11)^{-5}$
$=200\left|\frac{1-0.593451}{0.11}\right|+2000(0.593451)$
$=200(3.6959)+1186.902-739.18+1186.902$
$=1926.08$
Therefore, the value of the bond is ₹ 1,927 .

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Question 73 Marks

Solve the differential equation: $\mathrm{x} \log \mathrm{x} \frac{d y}{d x}+\mathrm{y}=\frac{2}{x} \log \mathrm{x}$

Answer

The given differential equation is
$\mathrm{x} \log \mathrm{x} \frac{d y}{d x}+\mathrm{y}=\frac{2}{x} \log \mathrm{x}$
$\Rightarrow \frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2}{x^{2}}$
This is a linear differential equation of the form
$\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$, where $\mathrm{P}=\frac{1}{x \log x}$ and $\mathrm{Q}=\frac{2}{x^{2}}$
$\therefore$ I.F. $=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x}=e^{\int \frac{1}{t} d t}$, where $\mathrm{t}=\log \mathrm{x}$
$\Rightarrow$ I.F. $=\mathrm{e}^{\log \mathrm{t}}=\mathrm{t}=\log \mathrm{x}$
Multiplying both sides of (i) by I.F. $=\log x$, we get
$\log \mathrm{x} \frac{d y}{d x}+\frac{1}{x} \mathrm{y}=\frac{2}{x^{2}} \log \mathrm{x}$
Integrating both sides with respect to $x$, we get
$y \log x=\int \frac{2}{x^{2}} \log \mathrm{xdx}+\mathrm{C}$ [Using: $\mathrm{y}($ I.F. $)=\int \mathrm{Q}$ (I.F.) $\mathrm{dx}+\mathrm{c}$ ]
$\Rightarrow \mathrm{y} \log \mathrm{x}=2 \iint_{I}^{\log } x x_{I I}^{-2} d x+C$
$\Rightarrow \mathrm{y} \log \mathrm{x}=2\left\{\log x\left(\frac{x^{-1}}{-1}\right)-\int \frac{1}{x}\left(\frac{x^{-1}}{-1}\right) d x\right\}+\mathrm{C}$
$\Rightarrow \mathrm{y} \log \mathrm{x}=2\left\{-\frac{\log x}{x}+\int x^{-2} d x\right\}+\mathrm{C}$
$\Rightarrow \mathrm{y} \log \mathrm{x}=2\left\{-\frac{\log x}{x}-\frac{1}{x}\right\}+\mathrm{C}$
$\Rightarrow \mathrm{y} \log \mathrm{x}=-\frac{2}{x}(1+\log \mathrm{x})+\mathrm{C}$, which gives the required solution.

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Question 83 Marks

It is given that radium decomposes at a rate proportional to the amount present. If $p \%$ of the original amount of radium disappears in 1 year. What percentage of it will remain after 21 years?

Answer

Let $A_{0}$ be the original amount of radium and $A$ be the amount of radium at any time $t$. Then, the rate of decomposing of radium is $\frac{d A}{d t}$. It is given that
$\frac{d A}{d t} \propto \mathrm{~A}$
$\Rightarrow \frac{d A}{d t}=-\lambda A$, where $\lambda$ is a positive constant
$\Rightarrow \frac{d A}{A}=-\lambda d t$
$\Rightarrow \log \mathrm{A}=-\lambda \mathrm{t}+\mathrm{C} . . .(\mathrm{i})$
At $t=0$, we have $A=A_{0}$. Putting $t=0$ and $A=A_{0}$ in (i), we get
$\log A_{0}=0+C \Rightarrow C=\log A_{0}$
Putting $\mathrm{C}=\log \mathrm{A}_{0}$ in (i), we get
$\log \mathrm{A}=-\lambda \mathrm{t}+\log \mathrm{A}_{0}$
$\Rightarrow \log \left(\frac{A}{A_{0}}\right)=-\lambda t \quad \ldots (ii) $
It is given that $\mathrm{p} \%$ of the original amount of radium disintegrates in l years. This means that the amount of radium present att $=\mathrm{l}$ is $\left(A_{0}-\frac{p}{100} \times A_{0}\right)=\left(1-\frac{p}{100}\right) A_{0}$. Putting A $=\mathrm{A}_{0}\left(1-\frac{p}{100}\right)$ and $\mathrm{t}=\mathrm{l}$ in (ii), we get
$\log \left(1-\frac{p}{100}\right)=-\lambda l \Rightarrow \lambda=-\frac{1}{l} \log \left(1-\frac{p}{100}\right) $
Substituting the value of $\lambda$ in (ii), we get
$\log \left(\frac{A}{A_{0}}\right)=\frac{t}{l} \log \left(1-\frac{p}{100}\right)\quad \ldots (iii)$
Let A be the amount of radium available after 21 years.
Putting $\mathrm{t}=21$ in (iii), we get
$\log \left(\frac{A}{A_{0}}\right)=2 \log \left(1-\frac{p}{100}\right)$
$\Rightarrow \frac{A}{A_{0}}=\left(1-\frac{p}{100}\right)^{2}$
$\Rightarrow \frac{A}{A_{0}} \times 100=\left(1-\frac{p}{100}\right)^{2} \times 100$ [Multiplying both sides by 100]
$\Rightarrow \frac{A}{A_{0}} \times 100=\left(10-\frac{p}{10}\right)^{2}$

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Applied Maths 3 Marks Question

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