3 Marks Question

Question 13 Marks

Answer

A group of 5 patients treated with medicine A weigh $10,8,12,6,4 kg$. A second group of 7 patients treated with medicine B weigh $14,12,8,10,6,2,11 kg$. Comment on the rejection of hypothesis with $5 \%$ level of significance.
$\left[\right.$ Given: $\left.t _{(10,0.05)}=1.812\right]$
Consider,
$
\begin{array}{l}
H_0: \mu_1=\mu_2 \text { and } \\
H_1: \mu_1>\mu_2
\end{array}
$
Where $\mu_1$ and $\mu_2$ denotes population means for the given two groups.
for Medicine A
$
\bar{x}=\frac{\sum x}{n}=\frac{40}{5}=8
$

x	10	8	12	6	4	sum x=40
$x-\bar{x}$	2	0	4	-2	-4	0
$(x-\bar{x})^2$	4	0	16	4	16	$\sum(x-\bar{x})^2=40$

For Medicin B
$
\bar{y}=\frac{\sum y}{n}=\frac{63}{7}=9
$

y	14	12	8	10	6	2	11	sum y=63
$y-\bar{y}$	5	3	-1	1	-3	-7	2	0
$(y-\bar{y})^2$	25	9	1	1	9	49	4	$\sum(y-\bar{y})^2=98$

$
\begin{array}{l}
\text { Now, } S^2=\frac{1}{n_1+n_2-2}\left[\sum(x-\bar{x})^2+\sum(y-\bar{y})^2\right] \\
S^2=\frac{1}{5+7-2}[40+98] \\
S^2=\frac{1}{10} \times 138=13.8 \\
S=\sqrt{13.8}=3.71 \\
t=\frac{\bar{x}-\bar{y}}{s \sqrt{\frac{1}{n_1+\frac{1}{n_2}}}} \\
t=\frac{8-9}{3.71 \sqrt{\frac{1}{5}+\frac{1}{7}}} \\
t=\frac{-1}{3.71 \sqrt{\frac{7+5}{35}}} \\
t=\frac{-1}{3.71 \sqrt{\frac{12}{35}}} \\
t=\frac{-1}{3.71 \times 0.58} \\
t=-0.46
\end{array}
$
Given: $t _{(10,0.05)}=1.812$
Since, $t _{\text {cal. }}$, value $< t _{ tab }$ value
Hence null hypothesis $H _0$ may be accepted with $5 \%$ significance.

View full question & answer→

Question 23 Marks

The profit of a paper hag manufacturing company (in lakhs of rupees) during each month of a year are:

Month	Jan	Feb	March	April	May	June	July	August	Sept	Oct	Nov	Dec
Profit	1.2	0.8	1.4	1.6	2	2.4	3.6	4.8	3.4	1.8	0.8	1.2

Plot the given data on a graph sheet. Calculate the four monthly moving averages and plot these on the same
graph sheet.

Answer

Since we are to calculate four monthly moving averages, so the period is even, therefore, we have to calculate centred moving averages.

The dotted curve shows four monthly moving averages.

View full question & answer→

Question 33 Marks

If the sum and the product of the mean and variance of Binomial Distribution are 1.8 and 0.8 respectively, find the
probability distribution and the probability of at least one success.

Answer

According to given, we have
$
\begin{array}{l}
np+npq=1.8 \Rightarrow np(1+q)=\frac{9}{5} ....(i) \\
\text { and } np \cdot npq=0.8 \Rightarrow n^2 p^2 q=\frac{4}{5} ....(ii)
\end{array}
$
Dividing the square of (i) by (ii), we get
$
\begin{array}{l}
\frac{n^2 p^2(1+q)^2}{n^2 p^2 q}=\left(\frac{9}{5}\right)^2 \times \frac{5}{4} \Rightarrow \frac{(1+q)^2}{q}=\frac{81}{20} \\
\Rightarrow 20\left(1+2 q+q^2\right)=81 q \Rightarrow 20 q^2-41 q+20=0 \\
\Rightarrow(5 q-4)(4 q-5)=0 \Rightarrow q=\frac{4}{5}, \frac{5}{4} \text { but } 0 < q < 1 \\
\Rightarrow q=\frac{4}{5} \\
\therefore p=1-q=1-\frac{4}{5}=\frac{1}{5}
\end{array}
$
$
\text { From (i), } n \cdot \frac{1}{5}\left(1+\frac{4}{5}\right)=\frac{9}{5} \Rightarrow n=5
$
Hence, the binomial distribution is $( q + p )^{ n }$ i.e., $\left(\frac{4}{5}+\frac{1}{5}\right)^5$
Probability of atleast one success $=1-P(0)=1-q^5$
$
=1-\left(\frac{4}{5}\right)^2=1-\frac{1024}{3125}=\frac{2101}{3125}
$

View full question & answer→

Question 43 Marks

The random variable $X$ can take only the values $0,1,2,3$. Given that $P(X=0)=P(X=1)=p$ and $P(X=2)=$ $P ( X =3)$ such that $\Sigma p_i x_i^2=2 \Sigma p_i x_i$, find the value of p .

Answer

Given $P(X=0)=P(X=1)=p$ and $P(X=2)=P(X=3)=k($ say $)$
The probability distribution of the random variable X is

X	0	1	2	3
P(X)	p	p	k	k

We know that $\sum p _{ i }=1$
$
\begin{array}{l}
\Rightarrow p+p+k+k=1 \Rightarrow 2 p+2 k=1 \\
\Rightarrow p+k=\frac{1}{2} \Rightarrow k=\frac{1}{2}-p
\end{array}
$
We construct the following table:

x_i	P_i	P_ix_i	P_ix²_i
0	p	0	0
1	p	p	p
2	$\frac{1}{2}=p$	1-2p	2-4p
3	$\frac{1}{2}=p$	$\frac{3}{2}-3 p$	$\frac{9}{2}-9 p$
Total		$\frac{5}{2}=4 p$	$\frac{13}{2}-12 p$

$
\begin{array}{l}
\text { Given } \Sigma p_i x_i^2=2 \Sigma p_i x_i \\
\Rightarrow \frac{13}{2}-12 p=2\left(\frac{5}{2}-4 p\right) \Rightarrow \frac{13}{2}-12 p=5-8 p \Rightarrow-4 p=5-\frac{13}{2} \\
\Rightarrow-4 p=-\frac{3}{2} \Rightarrow p=\frac{3}{8}
\end{array}
$
Hence, the value of $p$ is $\frac{3}{8}$.

View full question & answer→

Question 53 Marks

Suppose when $x$ units of a commodity are produced, the demand is $p=45-x^2$ rupees per unit, and the marginal cost is MC $=6+\frac{1}{4} x^2$, Assume there is no overhead i.e. $C (0)=0$. Find:
i. the total revenue and the marginal revenue.
ii. the value of $x$ (to the nearest unit) that maximizes profit.
iii. the consumer's surplus at the value of $x$ where profit is maximized (use the exact value of $x$ ).

Answer

i. Let R be the total revenue. Then,
$
\begin{array}{l}
R=px \\
\Rightarrow R=\left(45-x^2\right) x \\
\Rightarrow R=45 x-x^3 \text { and } \frac{d R}{d x}=45-3 x^2 \\
\Rightarrow R=45 x-x^2, \text { and } MR=45-3 x^2
\end{array}
$
ii. Let P be the profit function. Then,
$
\begin{array}{l}
\frac{d P}{d x}=MR-MC \\
\left.\Rightarrow \frac{d P}{d x}=45-3 x^2\right)-\left(6+\frac{1}{4} x^2\right) \\
\Rightarrow \frac{d P}{d x}=39-\frac{13}{4} x^2 \text { and } \frac{d^2 P}{d x^2}=-\frac{13}{2} x
\end{array}
$
For maximum profit, we must have
$
\frac{d P}{d x}=0 \Rightarrow 39-\frac{13}{4} x^2=0 \Rightarrow x^2=12 \Rightarrow x=2 \sqrt{3}=3.46 \simeq 3.5
$
Clearly, $\left(\frac{d^2 P}{d x^2}\right)_{x=35}=-\frac{13}{2} \times 3.5<0$
Hence, P is maximum when $x =3.5$. Putting $x =2 \sqrt{3}$ in $p =45- x ^2$, we obtain $p =45-12=33$. Thus, we obtain $p _0=33$ and $x _0=2 \sqrt{3}$.
iii. The consumer's surplus at $x_0=2 \sqrt{3}$ is given by
$
CS=\int_0^{x_0} p d x-p_0 x_0
$
$\begin{array}{l}\Rightarrow CS =\int_0^{2 \sqrt{3}}\left(45-x^2\right) d x-33 \times 2 \sqrt{3} \\ \Rightarrow CS =\left[45 x-\frac{x^3}{3}\right]_0^{2 \sqrt{3}}-66 \sqrt{3}=(90 \sqrt{3}-8 \sqrt{3}-66 \sqrt{3})=16 \sqrt{3}\end{array}$
Hence, the consumer's surplus at$x _0=2 \sqrt{3}$ is ₹$16 \sqrt{3} \approx$₹ 28

View full question & answer→

Question 63 Marks

Riya invested ₹ 20,000 in a mutual fund in year 2016. The value of mutual fund increased to ₹ 32,000 in year 2021. Calculate the compound annual growth rate of her investment. [Given, $\log (1.6)=0.2041$, antilog $(0.04082)=1.098]$

Answer

Given beginning value of investment $=$₹ 20,000
Final value of the investment = ₹ 32,000 No. of years = 5
$
\begin{array}{l}
\text { So, CAGR }=\left(\frac{\text { End Value }}{\text { Beginning Value }}\right)^{\frac{1}{n}}-1 \\
=\left(\frac{32000}{20000}\right)^{\frac{1}{5}}-1 \\
=(1.6)^{\frac{1}{5}}-1 \\
x=(1.6)^{\frac{1}{5}}
\end{array}
$
Let,
Taking log both sides, we get
$
\begin{array}{l}
\log x=\frac{1}{5} \log (1.6) \\
\Rightarrow \log x=\frac{1}{5} \times 0.2041 \\
\Rightarrow \log x=0.04082 \\
\Rightarrow x=\text { antilog }(0.04802) \\
=1.098 \\
\text { CAGR }=1.098-1=0.098 \\
=9.8 \%
\end{array}
$

View full question & answer→

Question 73 Marks

Determine the order and the degree (when defined) differential equations:
$
5 \frac{d^2 y}{d x^2}=\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{1}{4}}
$

Answer

The given differential equation can be written as
$
625\left(\frac{d^2 y}{d x^2}\right)^4=1+\left(\frac{d y}{d x}\right)^2
$
It is of order 2 and degree 4 .

View full question & answer→

Question 83 Marks

Solve the initial value problem: $e^{\frac{d y}{d z}}= x +1 ; y (0)=3$

Answer

The given differential equation is,
$
e^{\frac{d_y}{d x}}=x+1
$
Taking log on both sides, we get,
$
\begin{array}{l}
\frac{d y}{d x} \log e=\log (x+1) \\
\Rightarrow \frac{d y}{d x}=\log (x+1) \\
\Rightarrow dy=\{\log (x+1)\} dx
\end{array}
$
Integrating both sides, we get
$
\begin{aligned}
\int dy & =\int\{\log (x+1) dx \\
\Rightarrow y & =\int{\underset{I I}{ } \times \log (x+1) d x}_I^1
\end{aligned}
$
$
\begin{array}{l}
\Rightarrow y=\log (x+1) \int 1 dx-\int\left[\frac{d}{d x}(\log x+1) \int 1 dx\right] dx \\
\Rightarrow y=x \log (x+1)-\int \frac{x}{x+1} d x \\
\Rightarrow y=x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) d x \\
\Rightarrow y=x \log (x+1)-x+\log (x+1)+C \ldots \text { (i) }
\end{array}
$
It is given that $y(0)=3$
$
\begin{array}{l}
\therefore 3=0 \times \log (0+1)-0+\log (0+1)+C \\
\Rightarrow C=3
\end{array}
$
Substituting the value of $C$ in (i), we get
$
\begin{array}{l}
y=x \log (x+1)+\log (x+1)-x+3 \\
\Rightarrow y=(x+1) \log (x+1)-x+3
\end{array}
$
Hence, $y=(x+1) \log (x+1)-x+3$ is the solution to the given differential equation.

View full question & answer→

Applied Maths 3 Marks Question

Test yourself on this topic