3 Marks Question

Question 13 Marks

Find the student's -t for the following variable values in a sample of eight:
-4, -2, -2, 0, 2, 2, 3, 3 taking the mean of the universe to be zero.

Answer

x	$x-\bar{x}$	$(x-\bar{x})^2$
-4	-4.25	18.0625
-2	-2.25	5.0625
-2	-2.25	5.0625
0	-0.25	0.0625
2	1.75	3.0625
2	1.75	3.0625
3	2.75	7.5625
3	2.75	7.5625
$\sum x=2$		$\sum(x-\bar{x})^2=49.5000$

$
\begin{array}{l}
\bar{x}=\text { mean } \\
=\frac{\sum x}{n} \\
=\frac{2}{8} \\
=0.25
\end{array}
$
Now, compute the standard deviation using formula as,
$
\begin{array}{l}
s=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} \\
=\sqrt{\frac{49.5}{7}} \\
=\sqrt{7.071428} \\
=2.659
\end{array}
$
$H _0=$ The mean of universe, $\mu=0$, we get
$t=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}$
$=\frac{0.25-0}{\frac{2.659}{\frac{\sigma}{\sqrt{8}}}}$
$=\frac{0.25}{\frac{2.659}{2.828}}$
$\begin{array}{l}=\frac{0.25}{0.9402} \\ =0.2659\end{array}$

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Question 23 Marks

Construct 5-year Moving averages from the following data of the number of industrial failure in a country during 2003-2018:

Year	No. of Failures	Year	No. of Failure
2003	23	2011	9
2004	26	2012	13
2005	28	2013	11
2006	32	2014	14
2007	20	2015	12
2008	12	2016	9
2009	12	2017	3
2010	10	2018	1

Answer

Year	No. of failures	5-Yearly Moving Totals	5-Yearly Moving Averages
2003	23	-	-
2004	26	-	-
2005	28	129	25.8
2006	32	118	23.6
2007	20	104	20.8
2008	12	86	17.2
2009	12	63	12.6
2010	10	56	11.2
2011	9	55	11.0
2012	13	57	11.4
2013	11	59	11.8
2014	14	59	11.8
2015	12	49	9.8
2016	9	39	7.8
2017	3	-	-
2018	1	-	-

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Question 33 Marks

Let X be a discrete random variable whose probability distribution is defined as follows:
$
P(X=x)=\left\{\begin{array}{cl}
k(x+1) & \text { for } x=1,2,3,4 \\
2 k x & \text { for } x=5,6,7 \\
0 & \text { otherwise }
\end{array}\right.
$
where k is a constant
Find:
i. k
ii. E(X)
iii. Standard deviation of X.

Answer

$
P(X=x)=\left\{\begin{array}{c}
k(x+1) \text { for } x=1,2,3,4 \\
2 k x \text { for } x=5,6,7 \\
0 \text { otherwise }
\end{array}\right.
$
Thus, we have following table:

X	1	2	3	4	5	6	7	otherwise
P(X)	2k	3k	4k	5k	5k	12k	14k	0
XP(X)	2k	6k	12k	20k	20k	72k	98k	0
X²P(X)	2k	12k	36k	80k	80k	432k	686k	0

add $X ^2 P ( X )$ in fourth row and 1st column
i. Since, $\sum P_i=1$
$
\Rightarrow K(2+3+4+5+10+12+14)=1 \Rightarrow k=\frac{1}{50}
$
ii. $\because E(X)=\sum X P(X)$
$
\begin{array}{l}
\therefore E(X)=2 k+6 k+12 k+20 k+50 k+72 k+98 k+0=260 k \\
=260 \times \frac{1}{50}=\frac{26}{5}=5.2\left[\because k=\frac{1}{50}\right] \ldots \text { (i) }
\end{array}
$
iii. We know that,
$\operatorname{Var}( X )=\left[ E \left( X ^2\right)\right]-[ E ( X )]^2=\sum X^2 P(X)-\left[\sum X P(X)\right]^2$
$=[2 k +12 k +36 k +80 k +250 k +432 k +686 k +0]-[5.2]^2 \ldots$ [using Eq. (i)$]$
$=[1498 k ]-27.04=\left[1498 \times \frac{1}{50}\right]-27.04\left[\because k=\frac{1}{50}\right]$
$=29.96-27.4=2.92$
We know that, standard deviation of $X =\sqrt{\operatorname{Var}(X)}=\sqrt{2.92}=1.7088=1.7$ (approx)

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Question 43 Marks

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is What is the probability that he will win a prize $\frac{1}{100}$.
i. at least once
ii. exactly once
iii. at least twice?

Answer

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.
Clearly, X has a binomial distribution with $n =50$ and $p =\frac{1}{100}$
$
\begin{array}{l}
\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100} \\
\therefore P(X=x)={ }^n C_{x} q^{n-x P^{x}={ }^{50} C_{x}\left(\frac{99}{100}\right)^{50-x} \cdot\left(\frac{1}{100}\right)^x}
\end{array}
$
$
\text { i. } P(\text {winning at least once})=P(X \geq 1)
$
$
\begin{array}{l}
=1-P(X<1) \\
=1-P(X=0) \\
=1-{ }^{50} C_0\left(\frac{99}{100}\right)^{50} \\
=1-1 \cdot\left(\frac{99}{100}\right)^{50} \\
=1-\left(\frac{99}{100}\right)^{50}
\end{array}
$
ii. $P ($winning exactly once$)= P ( X =1)$
$
\begin{array}{l}
={ }^{50} C_1\left(\frac{99}{100}\right)^{49} \cdot\left(\frac{1}{100}\right)^1 \\
=50\left(\frac{1}{100}\right)\left(\frac{99}{100}\right)^{49} \\
=\frac{1}{2}\left(\frac{99}{100}\right)^{49}
\end{array}
$
iii. $P$ (at least twice$)=P(x \geq 2)$
$\begin{array}{l}=1- P ( X <2) \\ =1- P ( X \leq 1) \\ =1-[ P ( X =0)+ P ( X =1)] \\ =[1- P ( X =0)- P ( X =1)] \\ =1-\left(\frac{99}{100}\right)^{50}-\frac{1}{2} \cdot\left(\frac{99}{100}\right)^{49} \\ =1-\left(\frac{99}{100}\right)^{49}\left(\frac{99}{100}+\frac{1}{2}\right] \\ =1-\left(\frac{99}{100}\right)^{49} \cdot\left(\frac{149}{100}\right) \\ =1-\left(\frac{149}{100}\right)\left(\frac{99}{100}\right)^{49}\end{array}$

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Question 53 Marks

The demand function for a commodity is p = 20 e^-x/10. Find the consumer's surplus at equilibrium price p = 2. (Given log₁₀ e = 0.4343)

Answer

Given, the demand function is
$
p=20 e^{-x / 10} ...(i)
$
and the equilibrium price $p _0=2$.
Substituting this value of $p _0=2$ in (i), we get
$\begin{array}{l}
2=20 e^{-x_0 / 10} \Rightarrow e^{-x_0 / 10}=\frac{1}{10} \quad \ldots ...(ii) \\
\Rightarrow e^{x_0 / 10}=10 \Rightarrow \log _e 10=\frac{x_0}{10} \\
\Rightarrow x_0=10 \log _{e} 10=\frac{10}{\log _{10} e}=\frac{10}{0.4343}=\frac{100000}{4343} \\
\Rightarrow x_0=23.03 \ldots(iii) \\
\therefore CS=\int_0^{x_0} 20 e^{-x / 10} d x-x_0 \times p_0 \\
=20\left[\frac{e^{-x / 10}}{-\frac{1}{10}}\right]_0^{x_0}-23.03 \times 2 \text { (using (iii)) } \\
=-200\left[\left[e^{-x_0 / 10}-e^0\right]-46.06\right. \\
=-200\left[\frac{1}{10}-1\right]-46.06 \text { (using (ii)) } \\
=180-46.06=133.94
\end{array}$
Hence, consumer's surplus is 133.94

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Question 63 Marks

A firm anticipates an expenditure of ₹ 50,0000 for plant modernization at end of 10 years from now. How much should the company deposit at the end of year into a sinking fund earning interest 5% per annum. [Given log 1.05 = 0.0212, antilog (0.2120) = 1.629]

Answer

Given, A = ₹ 5,00,000, r -5% and $n =10$
Using formula, $A =p\left[\frac{(1+i)^n-1}{i}\right]$
where $i =\frac{r}{100}$
$
\begin{array}{l}
\Rightarrow 500000=p\left[\frac{(1+0.05)^{10}-1}{0.05}\right] \\
p=\frac{500000 \times 0.05}{(1.05)^{10}-1}
\end{array}
$
Now, let $x=(1.05)^{10}$
Taking log both sides, we get
$
\begin{array}{l}
\log x=10 \log (1.05) \\
=10 \times 0.0212 \\
=0.2120 \\
\Rightarrow x=\text { antilog }(0.2120) \\
=1.629
\end{array}
$
Thus, $(1.05)^{10}=1.629$
Now, $p=\frac{500000 \times 0.05}{1.629-1}$
$
\begin{array}{l}
=\frac{25000}{0.629} \\
=39745.63
\end{array}
$
Hence, the company should deposit X 39745.63 every year into the sinking fund.

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Question 73 Marks

Show that the differential equation representing one parameter family of curves $\left(x^2-y^2\right)=c\left(x^2+y^2\right)^2$ is $\left(x^2-3 x y^2\right)$ $ d x=\left(y^2-3 x^2 y\right) d y $

Answer

The given equation of one parameter family of curves is
$
x^2-y^2 c\left(x^2+y^2\right)^2
$...(i)
Differentiating (i) with respect to $x$, we get
$
\begin{array}{l}
2 x-2 y \frac{d y}{d x}=2 c\left(x^2+y^2\right)\left(2 x+2 y \frac{d y}{d x}\right) \\
\Rightarrow\left(x-y \frac{d y}{d x}\right)=2 c\left(x^2+y^2\right)\left(x+y \frac{d y}{d x}\right)...(ii)
\end{array}
$
On substituting the value of c obtained from (i) in (ii), we get,
$
\begin{array}{l}
\left(x-y \frac{d y}{d x}\right)=\frac{2\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)^2}\left(x+y \frac{d y}{d x}\right) \\
\Rightarrow\left(x^2+y^2\right)\left(x-y \frac{d y}{d x}\right)=2\left(x^2-y^2\right)\left(x+y \frac{d y}{d x}\right) \\
\Rightarrow\left\{x\left(x^2+y^2\right)-2 x\left(x^2-y^2\right)\right\}=\frac{d y}{d x}\left\{2 y\left(x^2-y^2\right)+y\left(x^2+y^2\right)\right\} \\
\Rightarrow\left(3 xy^2-x^3\right)=\frac{d y}{d x}\left(3 x^2 y-y^3\right)
\end{array}
$
$\Rightarrow\left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y$, which is the given differential equation.

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Question 83 Marks

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

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
$\int d y=\int\{\log (x+1) d x$
$\Rightarrow$ y = $\int \frac{1}{I I}$ $\times$ log$(\underset{I}{x}+1)$dx
$\begin{array}{l}\Rightarrow y =\log ( x +1) \int 1 dx -\int\left[\frac{d}{d x}(\log x +1) \int 1 dx \right] dx \\ \Rightarrow y = x \log ( x +1)-\int \frac{x}{x+1} d x \\ \Rightarrow y = x \log ( x +1)-\int\left(1-\frac{1}{x+1}\right) d x \\ \Rightarrow y = x \log ( x +1)- x +\log ( x +1)+ C \ldots( i )\end{array}$
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.

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Applied Maths 3 Marks Question

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