2 Marks Questions

Question 12 Marks

For some symmetrical distribution, $Q_1=36$ and $Q_3$ $=63$. Using Bowely's measures of skewness, find the median.

Answer

For symmetrical distribution, skewness $=0$
Given, $Q_1=36$ and $Q_3=63$
$\begin{array}{ll}\text { Since, } & S_k=\frac{Q_3+Q_1-2 Q_2}{Q_3-Q_1} \\ \Rightarrow & 0=\frac{63+36-2 Q_2}{63-36}\end{array}$
$\begin{array}{ll}\Rightarrow & 2 Q_2=99 \\ \Rightarrow & Q_2=\frac{99}{2}=49.5 \\ & Q_2=\text { median }=49.5 .\end{array}$

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Question 22 Marks

Compute $S_k$ if
(i) Mean $=108$, Mode $=99, \sigma=5$
(ii) $\sigma=4$, Mean $=20.5$, Mode $=22$
Also, mention type of skewness.

Answer

$S_k=\frac{\text { Mean }- \text { Mode }}{\sigma}$
(i) $S_k=\frac{108-99}{5}=\frac{9}{5}=1.8>0$
i.e., $S_k>0$, therefore, it is positively skewed.
(ii) $S_k=\frac{20.5-22}{4}=\frac{-1.5}{4}=-0.375<0$
Here, $S_k<0$, therefore, it is negatively skewed.

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Question 32 Marks

If the central moments $\mu_1, \mu_2, \mu_3$ and $\mu_4$ are $0,6,31$ and 108, respectively. Find coefficient of kurtosis.

Answer

Since, coefficient of kurtosis, $\beta_2=\frac{\mu_4}{\mu_2^2}=\frac{108}{(6)^2}$
$
=\frac{108}{36}=3
$
Also, here $\beta_2=3$, then curve is mesokurtic.

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Question 42 Marks

Karl Pearson's coefficient of skewness is 1.28 , its mean is 164 and mode 100, find standard deviation.

Answer

Using the formula, we have
$
\begin{aligned}
S_k & =\frac{\text { Mean }- \text { Mode }}{\sigma} \\
\text { i.e., } \quad 1.28 & =\frac{164-100}{\sigma} \\
\text { or, } \quad \sigma & =\frac{64}{1.28}=50 .
\end{aligned}
$

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Question 52 Marks

If $Q_1=26, Q_2=46$ and $Q_3=76$, find coefficient of skewness.

Answer

Here, we use method of Bowley's coefficient of skewness
$
ie., S_k=\frac{\left(Q_3-Q_2\right)-\left(Q_2-Q_1\right)}{\left(Q_3-Q_1\right)}
$
$S_k=\frac{(76-46)-(46-26)}{76-26}$
or $\qquad S_k=\frac{30-20}{50}=\frac{10}{50}=0.20$.

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Question 62 Marks

If the first moment of a distribution is 2 about the value 2 find the mean.

Answer

Given, $\mu_1^{\prime}=2$ and $A=2$.
$
\begin{array}{ll}
\because & \bar{x}=\mu_1^{\prime}+A \\
\therefore & \bar{x}=2+2=4 .
\end{array}
$

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Question 72 Marks

For a group of 60 boys students, the mean and S.D. of statistics marks are 45 and 2 respectively. The same figures for a group of 40 girls students are 55 and 3 respectively. What is the mean and S.D. of marks if the two groups are pooled together?

Answer

As given $n_1=60, \bar{x}_1=45, \sigma_1=2, n_2=40, \bar{x}_2=53$, $\sigma_2=3$
Thus, combined mean is given by
$
\begin{aligned}
\bar{x} & =\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2} \\
& =\frac{60 \times 45+40 \times 55}{60+40} \\
& =49
\end{aligned}
$
Thus,
$
\begin{array}{l}
d_1=\bar{x}_1-\bar{x}=45-49=-4 \\
d_2=\bar{x}_2-\bar{x}=55-49=6
\end{array}
$
Combined S.D. $=\sqrt{\frac{n_1 \sigma_1^2+n_2 \sigma_2^2+n_1 d_1^2+n_2 d_2^2}{n_1+n_2}}$
$\begin{array}{l}=\sqrt{\frac{60 \times(2)^2+40 \times(3)^2+60 \times(-4)^2+40 \times(6)^2}{60+40}} \\ =\sqrt{30}=5.48\end{array}$

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Question 82 Marks

The sum of the squares of deviations for 10 observations taken from their mean 50 is 250 . Find the co-efficient of variation.

Answer

Here, $N=10, \bar{x}=50$
and, $\quad \sum_{i=1}^{10}\left(x_i-50\right)^2=250$
$\begin{array}{rlrl}\Rightarrow & & \sum_{i=1}^{10}\left(x_i-50\right)^2 \\ \Rightarrow & & =25 \\ \Rightarrow & & \sigma^2 & =25 \\ \Rightarrow & & \sigma & =5\end{array}$
$\therefore$ Co-efficeint of variation,
$
\begin{aligned}
\text { C.V. } & =\frac{\sigma}{\bar{x}} \times 100 \\
& =\frac{5}{50} \times 100 \\
& =\frac{500}{50} \\
& =10
\end{aligned}
$
$\therefore 10$ is the co-efficient of variation

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Question 92 Marks

The mean of 100 observations is 50 and their standard deviations is 5 . Find the sum of all squares of the observations.

Answer

Given,
$
\begin{aligned}
\bar{x} & =50, \\
n & =100 \text { and } \sigma=5 \\
\sigma^2 & =\frac{\sum x_i^2}{n}-(\bar{x})^2
\end{aligned}
$
$\begin{aligned} \frac{\sum x_i^2}{n} & =\sigma^2+(\bar{x})^2 \\ \Sigma x_i^2 & =n\left[\sigma^2+(\bar{x})^2\right] \\ & =100\left[5^2+(50)^2\right] \\ & =100(25+2500) \\ & =100 \times 2525 \\ & =252500\end{aligned}$
Hence, the sum of all squares of all the observations is 252500 .

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Question 102 Marks

Show that for any numbers $a$ and $b$ (S.D.) is given by $\frac{|a-b|}{2}$.

Answer

For two numbers $a$ and $b$, given by $\bar{x}=\frac{(a+b)}{2}$
The variance is $\sigma^2=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{2}$
$
\begin{array}{l}
=\frac{\left(a-\frac{a+b}{2}\right)^2+\left(b-\frac{a+b}{2}\right)^2}{2} \\
=\frac{\frac{(a-b)^2}{4}+\frac{(a-b)^2}{4}}{2} \\
=\frac{(a-b)^2}{4}
\end{array}
$
$\therefore$ Standard Deviation $=\frac{|a-b|}{2}$
(The absolute sign is taken, as SD cannot be negative)

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Question 112 Marks

State two merits of quartile deviation.

Answer

(i) Quartile deviation is the best measure of dispersion for open-end classification as quartile deviation does not taken into account the first twenty five percent and the last twenty five percent of the observations.
(ii) It is independent of change of data.

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Question 122 Marks

Find the unknown value in the following table :

Class Interval	Frequency	Cumulative frequency
0-10	5	5
10-20	7	$x_1$
20-30	$x_2$	18
30-40	5	$x_3$
40-50	$x_4$	30

Answer

$\begin{aligned} x_1 & =5+7=12 \\ x_2=18-x_1 & =18-12=6 \\ x_3 & =18+5=23\end{aligned}$
and $\quad x_4=30-x_3=30-23=7$.

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Question 132 Marks

The mean and median of 100 observations are 50 and 52 respectively. The value of the larges observation is $1 0 0$. It was later found that it is $1 1 0$ not 100 . Find the true mean and median.

Answer

Mean $=\frac{\Sigma f x}{\Sigma f}$
$\Rightarrow \quad 50=\frac{\Sigma f x}{100}$
$
\begin{aligned}
\Rightarrow \quad \Sigma f x & =5000 \\
\text { Correct } \Sigma f x & =5000-100+110=5010 \\
\therefore \quad \text { Correct mean } & =\frac{5010}{100} \\
& =50.1
\end{aligned}
$
Median will remain same i.e., median $=52$.

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Question 142 Marks

The monthly income ₹ 1000 of 8 persons working in a factory. Find $P_{30}$ income value 17, 21, 14, 36, 10, 25,15,29.

Answer

Arrange the data in the increasing order :
$
10,14,15,17,21,25,29,36
$
Here,
$\begin{aligned}
n & =8 \\
P_{30} & =\left[\frac{30(n+1)}{100}\right] \text { th item } \\
& =\frac{30(8+1)}{100} \text { th item } \\
& =\frac{30 \times 9}{100} \text { th item }=2.7 \text { th item } \\
& =2^{\text {nd }} \text { item }+0.7\left(3^{\text {rd }} \text { item }-2^{\text {nd }} \text { item }\right) \\
& =14+0.7(15-14) \\
& =14+0.7 \\
& =14.7 .
\end{aligned}
$

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