3 Marks Question

Question 13 Marks

Consider the following hypothesis test:
$\begin{array}{l} H _0: p \geq 0.75 \\ H _{ a }: p <0.75\end{array}$
A sample of 300 provided a sample proportion of 0.68.
i. Compute the value of the test statistic.
ii. What is the p-value?
iii. At $\alpha=0.05$, what is your conclusion?
iv. What is the rejection rule using critical value? What is your conclusion?

Answer

Given $p _0=0.75, n =300, \bar{p}=0.68$
i. $Z =\frac{\bar{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.68-0.75}{\sqrt{\frac{0.75 \times 0.25}{300}}}$
$=\frac{-0.07 \times 10}{\sqrt{0.25 \times 0.25}}=\frac{-0.07}{0.25}=-2.8$
ii. $\because Z=-2.8<0$
So, p-value of -2.8 = area under the standard normal curve to the left of Z
= 0.0026
$\therefore$ p-value $=0.0026$
iii. Given $\alpha=0.05$
$\because$ p-value $<0.05$
So, reject $H _0$.
iv. Rejection rule using critical value
Reject $H _0$ if $Z \leq-Z_\alpha$
Here, $\alpha=0.05$. So $Z_\alpha= Z _{0.05}=1.645$
$\begin{array}{l}\Rightarrow-Z_\alpha=-1.645 \\ \because-2.8<-1.645\end{array}$
$\Rightarrow Z <-Z_\alpha$
So, reject $H _0$

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Question 23 Marks

For the following data, use the weighted average of price relative method to construct the index number for theyear 2010, taking the year 2005 as the base year.

Commodity	Weight (W)	Price in $2005\left( P _0\right)$	Price in $2007\left( P _1\right)$
E	15	22	30
F	12	15	18
G	8	17	20
H	17	12	15
I	20	25	32

Answer

Commodity	Weight (W)	Price in $2005\left( P _0\right)$	Price in $2010\left(P_1\right)$	$R =\frac{P_1}{P_0} \times 100$	RW
E	15	22	30	136.36	2,045.40
F	12	15	18	120	1440
G	8	17	20	117.64	941.12
H	17	12	15	125	2,125
I	20	25	32	128	2,526

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Question 33 Marks

A traffic engineer records the number of bicycle riders that use a particular cycle track. He records that an average of 3.2 bicycle riders use the cycle track every hour. Given that the number of bicycles that use the cycle track follow a Poisson distribution, what is the probability that
a. 2 or less bicycle riders will use the cycle track within an hour?
b. 3 or more bicycle riders will use the cycle track within an hour?
Also, write the mean expectation and variance for the random variable X.

Answer

Given mean = X = 3.2
Let X be the number of bicycle riders which use the cycle track.
a. Required probability $= P ( X \leq 2)= P ( X =0)+ P ( X =1)+ P ( X =2)$
$\begin{array}{l}=\frac{e^{-3.2}(3.2)^0}{0!}+\frac{e^{-3.2}(3.2)^1}{1!}+\frac{e^{-3.2}(3.2)^2}{2!} \\ =e^{-3.2}[1+3.2+5.12] \\ =0.041 \times 9.32=0.382\end{array}$
b. Required probability $= P ( X \geq 3)=1- P ( X \leq 2)$
= 1 - 0.382 = 0.618
Also, mean expectation and variance of X are $\lambda$ = 3.2

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Question 43 Marks

Find the mean and variance of the number of heads in the two tosses of a coin.

Answer

Let X be a random variable denoting the number of heads in the two tosses of a coin. Therefore, X can take values 0, 1 or 2 such that
P (X = 0) = (Probability of getting no head) = P(TT) = $\frac{1}{4}$
$P ( X =1)=($ Probability of getting one head $)= P ( HT$ or TH $)=\frac{2}{4}=\frac{1}{2}$
and, $P ( X =2)=($ Probability of getting both heads $)= P ( HH )=\frac{1}{4}$
Thus, the probability distribution of X is as follows:

X	0	1	2
P(X)	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

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	$\frac{1}{4}$	0	0
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{1}{4}$	$\frac{1}{2}$	1
		$\Sigma p_i x_i=1$	$\Sigma p_i x_i^2=\frac{3}{2}$

Thus, we have,
$\Sigma p_i x_i=1$ and $\Sigma p_i x_i^2=\frac{3}{2}$
$\therefore \bar{X}=$ Mean $=\Sigma p_i x_i=1$ and $\operatorname{Var}( X )=\Sigma p_i x_i^2-\left(\Sigma p_i x_i\right)^2=\frac{3}{2}-1=\frac{1}{2}$
Hence, Mean $=1$ and Variance $=\frac{1}{2}$

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Question 53 Marks

The marginal revenue function of a commodity is given by $M R=11-3 x+4 x^2$, find the revenue function. Also, find the demand function.

Answer

Let R(x) be the revenue function of x units of the product and MR be the marginal revenue function, then
$MR =11-3 x +4 x ^2$
As $MR =\frac{d}{d x}( R ( x ))$, so $\frac{d}{d x}( R ( x ))=11-3 x +4 x ^2$
$\therefore R ( x )=\int\left(11-3 x +4 x ^2\right) dx$
$=11 x -3 \cdot \frac{x^2}{2}+4 \cdot \frac{x^3}{3}+ k$, where k is constant of integration.
When x = 0, R(x) = 0
$\begin{array}{l}\Rightarrow 0=11 \times 0-\frac{3}{2} \times 0+\frac{4}{3} \times 0+k \Rightarrow k=0 . \\ \therefore R(x)=11 x-\frac{3}{2} x^2+\frac{4}{3} x^3 .\end{array}$
If p is the price per unit when x units of the product are sold, then
R(x) = $p \cdot x$
$\begin{array}{l}\Rightarrow p x=11 x-\frac{3}{2} x^2+\frac{4}{3} x^3 \\ \Rightarrow p=11-\frac{3}{2} x+\frac{4}{3} x^2 \text {, which is the corresponding demand function. }\end{array}$

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Question 63 Marks

A loan of ₹ 250000 at the interest rate of 6 % p.a. compounded monthly is to be amortized by equal payments at the end of each month for 5 years, find
i. the size of each monthly payment.
ii. the principal outstanding at beginning of 40th month.
iii. interest paid in 40th payment.
iv. principal contained in 40th payment and
v. total interest paid. $\left(\right.$ Given $\left.(1.005)^{60}=1.3489,(1.005)^{21}=1.1104\right)$

Answer

Given, $P =₹ 250000, i =\frac{6}{12 \times 100}=0.005$ and $n =5 \times 12=60$.
$\begin{array}{l}\text { i. } EMI =\frac{250000 \times 0.005 \times(1.005)^{60}}{(1.005)^{60}-1} \\ \quad=\frac{250000 \times 0.005 \times 1.3489}{0.3489}=₹ 4832.69\end{array}$
ii. Principal outstanding at beginning of 40 th month
$\begin{array}{l}=\frac{\operatorname{EMI}\left[(1+i)^{60-40+1}-1\right]}{i(1+i)^{60-40+1}}=\frac{4832.69 \times\left[(1.005)^{21}-1\right]}{0.005 \times(1.005)^{21}} \\ =\frac{4832.69 \times[1.1104-1]}{0.005 \times 1.1104}=\frac{4832.69 \times 0.1104}{0.005 \times 1.1104}=₹ 96096.72\end{array}$
iii. Interest paid in 40th payment = $\frac{ EMI \left[(1+i)^{60-40+1}-1\right]}{(1+i)^{60-40+1}}$
$=\frac{4832.69 \times\left[(1.005)^{21}-1\right]}{(1.005)^{21}}=\frac{4832.69 \times 0.1104}{1.1104}=₹ 480.48$
iv. Principal paid in 40 th payment = EMI - Interest paid in 40 th payment
= 4832.69 - 480.48 = ₹ 4352.21
v. Total interest paid = n * EMI - P = 60 * 4832.69 - 250000
= 289961.40 - 250000 = ₹39961.40

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Question 73 Marks

A radioactive substance disintegrates at a rate proportional to the amount of substance present. If 50% of the given amount disintegrates in 1600 years. What percentage of the substance disintegrates in 10 years?(Take $e^{\frac{-\log 2}{160}}=$0.9957)

Answer

Let A denote the amount of the radioactive substance present at any instant t and let $A_0$ be the initial amount of the substance.
It is given that
$\frac{d A}{d t} \alpha A \Rightarrow \frac{d A}{d t}=-\lambda A \ldots$
where $\lambda$ is the constant of proportionality such that $\lambda>0$. Here, the negative sign indicates that A decreases with the increase in t .
Now,
$\frac{d A}{d t}=-\lambda A$
$\begin{array}{l}\Rightarrow \frac{1}{A} d A=-\lambda d t \\ \Rightarrow \int \frac{1}{A} d A=-\lambda \int 1 . d t \\ \Rightarrow \log A=-\lambda t + C \ldots \text { (ii) }\end{array}$
Initially i.e. at $t=0$, we have $A=A_0$. Putting $t=0$ and $A=A_0$ in (ii), we get
$\log A_0=0+C \Rightarrow C=\log A_0$
Putting $C=\log A_0$ in (ii), we get
$\log A=-\lambda t+\log A_0$
$\Rightarrow \log \left(\frac{A}{A_0}\right)=-\lambda t \ldots( iii )$
It is given that $A =\frac{A_0}{2}$ at $t =1600$ years. Putting $A =\frac{A_0}{2}$ and $t =1600$ in (iii), we get
$\log \left(\frac{1}{2}\right)=-1600 \lambda \Rightarrow \lambda=\frac{1}{1600} \log 2$
Substituting the value of $\lambda$ in (iii), we get
$\log \left(\frac{A}{A_0}\right)=-\left(\frac{1}{1600} \log 2\right) t$
$\begin{array}{l}\Rightarrow \frac{A}{A_0}=e^{-\frac{\log 2}{1600} t} \\ \Rightarrow A=A_0 e^{-\frac{\log 2}{1600} t}\end{array}$
Putting t = 10, we obtain the amount of the radioactive substance present after 10 years and is given by
$A = A _0(0.9957)\left[\because e^{-\frac{\log 2}{160}}=0.9957\right]$
$\therefore$ Amount that disintegrates in 10 years $= A _0- A = A _0-0.9957 A_0-0.0043$
percentage of the amount disintegrated in 10 years $=\frac{0.0043 A_0}{A_0} \times 100=0.43$
Hence, 0.43% of the original amount disintegrates in 10 years.

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Question 83 Marks

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

Answer

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

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Applied Maths 3 Marks Question

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