3 Marks Question

Question 13 Marks

A random sample of 17 values from a normal population has a mean of 105 cm and the sum of the squares of deviations from this mean is $1225 cm^2$. Is the assumption of a mean of 110 cm for the normal population reasonable? Test under 5% and 1% levels of significance. Also, obtain the 95% and 99% confidence limits. (Given $t _{16}(0.05)=2.12$ and $\left.t _{16}(0.01)=2.921\right)$

Answer

We have,
$\mu=$ Population mean $=110, \bar{X}=$ Sample mean $=105$
$\begin{array}{l} n =\text { Sample size }=17 \text { and, } \sum_{i=1}^{17}\left(x_i-\bar{X}\right)^2=1225 \\ \therefore s^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{X}\right)^2 \\ \Rightarrow s^2=\frac{1225}{17}=72.0588 \Rightarrow s=\sqrt{72.0588}=8.4887\end{array}$
We define, Null Hypothesis $H _0$ : There is no significant difference between the sample mean and population means i.e. assumption that mean of the population is 110 cm is valid.
Alternate hypothesis $H _1$ :Assumption that mean of the population is 110 cm is not valid. Let t be the test statistic given by
$\begin{array}{l} t =\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n-1}}} \Rightarrow t =\frac{105-110}{8.4887} \times \sqrt{17-1}=\frac{-5 \times 4}{8.4887}=-2.3561 \\ \Rightarrow| t |=2.3561\end{array}$
The sample statistic follows Student's t -distribution with v = (17 - 1) = 16 degrees of freedom.
We shall now compare this calculated value with the tabulated value of t for 16 degrees of freedom at 5% and 1% levels of significance.
At 5% level of significance: It is given that $t _{16}(0.05)=2.12$
We find that Calculated $| t |=2.3561>2.12= t _{16}(0.05)$
i.e. Calculated $| t |>$ Tabulated $t _{16}(0.05)$
So, we reject the null hypothesis at 5% level of significance. Hence, the assumption that the population has a mean of 110 cm is not correct.
The confidence limits at 5% level of significance are
$\begin{array}{l}\bar{X}-\frac{s}{\sqrt{n-1}} t _{16}(0.05) \text { and } \bar{X}+\frac{s}{\sqrt{n-1}} t _{16}(0.05) \\ \text { or } 105-\frac{8.4887}{4} \times 2.12 \text { and } 105+\frac{8.4887}{4} \times 2.12\end{array}$
or, 105 - 4.499 = 100.501 and 105 + 4.499 = 109.499
The confidence interval is [100.501,109.499]
At 1% level of significance: It is given that $t _{16}(0.01)=2.921$
Clearly, calculated |t| < tabulated $t _{16}(0.01)$
So, we accept the null hypothesis at 1% level of significance. Hence, the assumption that the mean of the population is 110 cm is valid.
The confidence limits at 1% level of significance are
$\begin{array}{l}\bar{X}-\frac{s}{\sqrt{n-1}} t _{16}(0.01) \text { and } \bar{X}+\frac{s}{\sqrt{n-1}} t _{16}(0.01) \\ \text { or, } 105-\frac{8.4887}{4} \times 2.921 \text { and } 105+\frac{8.4887}{4} \times 2.921\end{array}$
or, 105 - 6.199 = 98.801 and 105 + 6.199 = 111.199
The confidence interval at 1% level of significance or at 99% confidence level is [98.801, 111.199]

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Question 23 Marks

Given below are the consumer price index numbers (CPI) of the industrial workers.

Year	2014	2015	2016	2017	2018	2019	2020
Index number	145	140	150	190	200	220	230

Find the best fitted trend line by the method of least squares and tabulate the trend values.

Answer

Note that the number of years is Odd
$\Rightarrow n = odd$
Procedure:
i. Take middle year values as As i.e. A = 2017
ii. Find $X=x_i-A$
iii. Find $X^2$ and $X Y$

Year	Index number (Y)	X=x_i-A=x_i-2017	X²	XY	Trend value Y_t=a+bX
2014	145	-3	9	-435	182.1+(-3)xx16.6=132.3
2015	140	-2	4	-280	182.1+(-2)xx16.6=148.9
2016	150	-1	1	-150	182.1+(-1)xx16.6=165.5
2017	190	0	0	0	182.1+(0)xx16.6=182.1
2018	200	1	1	200	182.1+(1)xx16.6=198.7
2019	220	2	4	440	182.1+(2)xx16.6=215.3
2020	230	3	9	690	182.1+(3)xx16.6=231.9
n=7	$\sum Y$ =1275	$\sum X$=0	$\sum X^2$=28	$\sum X Y$=465	$\sum Y_t$=1274.7

$\begin{array}{l}a=\frac{\sum Y}{n}=\frac{1275}{7}=182.14 \\ \text { and } b=\frac{\sum X Y}{\sum X^2}=\frac{465}{28}=16.6\end{array}$
Therefore, the required equation of the straight-line trend is given by
y = a + bX
$\Rightarrow y =182.1+16.6 X$

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Question 33 Marks

In a binomial distribution the sum and product of the mean and the variance are $\frac{25}{3}$ and $\frac{50}{3}$ respectively. Find the distribution.

Answer

We have,
Sum of the mean and variance = $\frac{25}{3}$
$\begin{array}{l}\Rightarrow np + npq =\frac{25}{3} \\ \Rightarrow np (1+ q )=\frac{25}{3} \ldots( i )\end{array}$
Product of the mean and variance $=\frac{50}{3}$
$\Rightarrow np ( npq )=\frac{50}{3} \ldots$ (ii)
Dividing eq. (ii) by eq. (i), we have,
$\frac{ np ( npq )}{ np (1+ q )}=\frac{50}{3} \times \frac{3}{25}$
$\begin{array}{l}\Rightarrow \frac{n p q}{1+q}=2 \\ \Rightarrow npq =2(1+ q ) \\ \Rightarrow np (1- p )=2(2- p ) \\ \Rightarrow n p=\frac{2(2-p)}{(1-p)}\end{array}$
Substituting this value in np + npq = $\frac{25}{3}$,we have,
$\begin{array}{l}\frac{2(2-p)}{(1-p)}(2-p)=\frac{25}{3} \\ \Rightarrow 6\left(4-4 p + p ^2\right)=25-25 p \\ \Rightarrow 6 p ^2+ p -1=0 \\ \Rightarrow(3 p -1)(2 p +1)=0 \\ \Rightarrow p=\frac{1}{3} \text { or } \frac{-1}{2}\end{array}$
As p cannot be negative, therefore possible value of p is $\frac{1}{3}$
$\begin{array}{l} q =1- p =\frac{2}{3} \\ \Rightarrow np + npq =\frac{25}{3} \\ \Rightarrow n\left(\frac{1}{3}\right)\left(1+\frac{2}{3}\right)=\frac{25}{3} \\ \Rightarrow n =15 \\ \therefore P ( X = r )={ }^{15} C_r\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{15-r}, r =0,1,2 \ldots .15\end{array}$

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Question 43 Marks

The income of a group of 10,000 persons was found to be normally distributed with mean ₹ 750 p.m. and standard deviation ₹ 50. Show that of this group about 95% had income exceeding ₹ 668 and only 5% had income exceeding ₹ 832. What was the lowest income among the richest 100?

Answer

Let X denote the income. Then X is normally distributed with mean $\mu$ = ₹ 750 and standard deviation $\sigma$ = ₹ 50. Let Z be the
standard normal variate. Then,
$Z =\frac{X-\mu}{\sigma}$ or, $Z =\frac{X-750}{50}$
When X = 668, we obtain: $Z =\frac{668-750}{50}=-\frac{82}{50}=-1.64$
Now, P(X > 668)
= P(Z > -1.64)
$= P (-1.64< Z \leq 0)+ P ( Z \geq 0)= P (0 \leq Z <1.64)+0.5=0.4495+0.5=0.9495$
Thus, 94.95% persons had income exceeding ₹ 668
When X = 832, we obtain: $Z =\frac{832-750}{50}=1.64$
$\begin{array}{l}\therefore P ( X >832) \\ = P ( Z >1.64) \\ = P ( Z \geq 0)- P (0 \leq Z <1.64)=0.5-0.4495=0.0505\end{array}$
Thus, 5.05 % persons had income exceeding ₹ 832
Now, probability of selecting a person out of richest 100 persons $=\frac{100}{10000}=0.01$
In order to find the lowest income among the richest 100, we have to find the value k of X such that $P(X \geq k)=0.01$
When X = k, we obtain $Z =\frac{k-750}{50}= Z _1$ (Say)
Now, P(X > k) = 0.01
$\begin{array}{l}= P \left( Z \geq Z _1\right)=0.01 \\ =0.5- P \left(0 \leq Z \leq Z _1\right)=0.01 \\ = P \left(0 \leq Z \leq Z _1\right)=0.49 \\ = Z _1=2.33\end{array}$
$=\frac{k-750}{50}=2.33 \Rightarrow k =750+50 \times 2.33 \Rightarrow k =866.5$
Hence, the lowest income among the richest 100 was ₹ 866.50

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Question 53 Marks

A company suffers a loss of ₹1,000 if its product does not sell at all. Marginal revenue and Marginal cost functions for the product are given by MR = 50 - 4x and MC = -10 + x respectively. Determine the total profit function, break-even points and the profit maximization level of output

Answer

Let P denote the profit function. Then,
$\begin{array}{l}\frac{d P}{d x}=\text { MR }- \text { MC } \\ \Rightarrow \frac{d P}{d x}=(50-4 x )-(-10+ x ) \\ \Rightarrow \frac{d P}{d x}=60-5 x \text { and } \frac{d^2 P}{d x^2}=-5\end{array}$
For maximum value of P, we must have
$\frac{d P}{d x}=0 \Rightarrow 60-5 x =0 \Rightarrow x =12$
Clearly, $\frac{d^2 P}{d x^2}=-5<0$ for all x.
So, profit P is maximum when 12 units are produced. Thus, the profit maximization level of output is 12 units.
Now, $\frac{d P}{d x}=60-5 x$
$\begin{array}{l}\Rightarrow P =\int(60-5 x) d x+k \ldots \text { [On intergrating] } \\ \Rightarrow P =60 x -\frac{5}{2} x^2+ k \ldots \text { (i) }\end{array}$
where k is the constant of integration
It is given that the company suffers a loss of ₹ 1000, if its product does not sell at all i.e. P = -1000 at x = 0. Substituting these values in (i), we obtain k = -1000.
Putting k = -1000 in (i), we obtain:
$P =60 x -\frac{5}{2} x^2+1000$
This is the total profit function. For break-even points
$P =0 \Rightarrow 60 x -\frac{5}{2} x^2+1000=0 \Rightarrow 5 x ^2-120 x +2000=0$
$\Rightarrow x^2-24 x+400=0$
This equation does not give real values of x. So, there is no break-even point.

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Question 63 Marks

Find the purchase price of a ₹600, 8% bond, dividends payable semi-annually redeemable at par in 5 years, if the yield rate is to be 8% compounded semi-annually.

Answer

Face value of the bond C = ₹600
Nominal rate of interest i = 8% or 0.08
As dividends are paid semi-annually
Therefore, Rate of interest per period $i _{ d }=\frac{0.08}{2}=0.04$
Therefore, periodic dividend payment $R = C \times i _{ d }=600 \times 0.04=24$
So, semi-annual dividend R is ₹24
Yield rate is 8% = 0.08, compounded semi annually
Therefore $i =\frac{0.08}{2}=0.04$
No. of years n = 5
Therefore, no. of dividend periods (n) = 5 * 2 = 10
Purchase price (V) of the bond is given by
$\begin{array}{l} V =R\left|\frac{1-(1+i)^{-n}}{i}\right|+C(1+i)^{-n} \\ =24\left|\frac{1-(1+0.04)^{-10}}{0.04}\right|+600(1+0.04)^{-10} \\ =24\left|\frac{1-(1.04)^{-10}}{0.04}\right|+600(1.04)^{-10}\end{array}$
$\begin{array}{l}=24\left[\left.\frac{1-0.6755}{0.04} \right\rvert\,+600(0.6755)\right. \\ =194.7+405.3=600\end{array}$
Therefore, purchase price of bond is ₹600.

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Question 73 Marks

Solve: $\left( x ^2+1\right) \frac{d y}{d x}+2 xy -4 x ^2=0$ subject to the initial condition $y (0)=0$.

Answer

The given differential equation can be written as
$\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{4 x^2}{1+x^2} \ldots$ (i)
This is a linear differential equation of the form $\frac{d y}{d x}+ Py = Q$, where
$P =\frac{2 x}{1+x^2}$ and $Q =\frac{4 x^2}{1+x^2}$
$\therefore$ I.F. $=e^{\int P d x}=e^{\int \frac{2 x}{\left(1+x^2\right) d x}}=e^{\log \left(1+x^2\right)}=1+ x ^2$
Multiplying both sides of (i) by I.F. $=\left(1+x^2\right)$, we get $\left(1+ x ^2\right) \frac{d y}{d x}+2 xy =4 x ^2$
Integrating both sides with respect to x, we get
$\begin{array}{l}\left.y\left(1+x^2\right)=\int 4 x^2 d x+C \text { [Using: } y(\text { I.F. })=\int Q \text { (I.F.) } d x+C\right] \\ \Rightarrow y\left(1+x^2\right)=\frac{4 x^3}{3}+C \ldots \text { (ii) }\end{array}$
It is given that y = 0, when x = 0. Putting x = 0 and y = 0 in (i), we get $0=0+C \Rightarrow C=0$
Substituting C = 0 in (ii), we get y = $\frac{4 x^3}{3\left(1+x^2\right)}$, which is the required solution.

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Question 83 Marks

Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of radium to decompose? [Given $\log _e 0.989=0.01106$ and $\log _e 2=0.6931$]

Answer

Let A be the quantity of radium present at time t and $A_0$ be the initial quantity of radium.
Therefore, we have,
$\begin{array}{l}\frac{d A}{d t} \propto A \\ \frac{d A}{d t}=-2 A \\ \frac{d A}{A}=-2 dt \\ \int \frac{d A}{A}=-\lambda t d t\end{array}$
$\begin{array}{l}\log A =-\lambda t+ c \ldots( i ) \\ \text { Now, } A = A _0 \text { when } t =0 \\ \log A_0=0+ c \\ c =\log A _0\end{array}$
Put value of c in equation
$\begin{array}{l}\log A =-\lambda t +\log A _0 \\ \log \left(\frac{A}{A_0}\right)=-\lambda t \ldots( ii )\end{array}$
Given that, In 25 years, bacteria decomposes 1.1 %, so
$A=(100-1.1) \%=98.996 \%=0.989 A_0, t=25$
Therefore, (ii) gives,
$\begin{array}{l}\log \left(\frac{0.989 A_0}{A_0}\right)=-25 \lambda \\ \log (0.989)=-25 \lambda \\ \lambda=-\frac{1}{25} \log (0.989)\end{array}$
Now, equation (ii) becomes,
$\log \left(\frac{A}{A_0}\right)=\left\{\frac{1}{25} \log (0.989)\right\} t$
Now $A =\frac{1}{2} A_0$
$\log \left(\frac{A}{2 A}\right)=\frac{1}{25} \log (0.989) t$
$\frac{-\log 2 \times 25}{\log (0.989)}= t$
$-\frac{0.6931 \times 25}{0.01106}=t$
t = 1567 years
Required time = 1567 years

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Applied Maths 3 Marks Question

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