Fill in the blanks.

Question 11 Mark

In symmetric distribution, the third quartile and first quartile of data in distribution must be at ____________ distance.

Answer

equal, because
For, symmetric distribution, $Q_1$ and $Q_2$ are at equal distance.

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Question 21 Mark

When the frequency curve is more peaked than normal curve it is called __________

Answer

Leptokurtic, because
The curves with greater peakness than the normal curve are called leptokurtic

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Question 31 Mark

In case of negatively skewed distribution, the extreme value lies in ___________

Answer

Left tail, because
When distribution is negatively skewed, values lie on left side (or left tail).

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Question 41 Mark

For a positively skewed distribution mean is always ___________

Answer

greater than mode, because
For positively skewed distribution mean $>$ mode, for negatively skewed distribution mean $<$ mode

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Question 51 Mark

The sum of square of the deviations of values of variables is ___________ , when taken about their arithmetic mean.

Answer

minimum, because
The sum of the squares of the deviations of the value of variable is minimum when taken about their arithmetic mean.

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Question 61 Mark

The standard deviations is ____________ to mean deviations taken from the arithmetic mean.

Answer

Greater then or equal, because
The standard deviation is greater than or equal to the mean deviations taken from the arithmetic mean.

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Question 71 Mark

The standard deviation of data is _____________ of any change in origin but is _____________ in change of scale.

Answer

independent; dependent, because
Since, the standard deviation of any data is independent of any change in origin but is dependent of any change of scale.

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Question 81 Mark

If the variance of data is 121 , then the standard deviation of data is ___________.

Answer

11, because
We know that,
$
\begin{aligned}
\text { S.D. } & =\sqrt{\text { variance }} \\
& =\sqrt{121} \\
& =11 .
\end{aligned}
$

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Question 91 Mark

Coefficient of variation $=\frac{\cdots \cdots \cdots \cdots}{\text { Mean }} \times 100$

Answer

S.D. (Standard Deviation), because
$
\text { C.V. }=\frac{\text { S.D. }}{\text { Mean }} \times 100
$

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Question 101 Mark

When tested the lines (in hours) of 5 bulbs were noted as follows : 1357, 1090, 1666, 1494, 1623; then mean deviation (in hours) from their mean is _________

Answer

178 , because
The lines of 5 bulbs are give as :
1357,1090,1666,1494,1623
$\therefore$ Mean, $\bar{x}=\frac{1357+1090+1666+1494+1623}{5}$
$=\frac{7230}{5}=1446$

$x_i$	$d_i=\left\|x_i-\bar{x}\right\|$
1357 1090 1666 1494 1623	89 356 220 48 177
Total	$\Sigma d_i=890$

Mean deviation, $MD =\frac{\Sigma d_i}{n}=\frac{890}{5}=178$.

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Question 111 Mark

Following are the wages of 8 workers expressed i
₹: 82,96,52,75,65,70,50,70 .
The range is _________ and coefficient of range is ________

Answer

₹ 46 and 315, because
The largest and smallest wages are L= ₹ 96 and S= ₹ 50
Thus, range = ₹ 96- ₹ 50= ₹ 46
and coefficient of range
$\begin{array}{l}=\frac{L-S}{L+S} \times 100 \\ =\frac{96-50}{96+50} \times 100 \\ =\frac{46}{146} \times 100=0.315 \times 100 \\ =31.5\end{array}$

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Question 121 Mark

$\left[\frac{\left(Q_3-Q_1\right)}{\left(Q_3+Q_1\right)}\right] \times 100$ is _________

Answer

Coefficient of quartile deviation, because. A relative measure of dispersion using quartiles is given by coefficient of quartile deviation which is Coefficient of quartile deviation $=\frac{Q_3-Q_1}{Q_3+Q_1} \times 100$

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Question 131 Mark

For the data : $5,24,36,12,20$, and 8 , value of $D_5$ is $\ldots$

Answer

16, because
Arranging the given data in ascending order, we get
$
\begin{aligned}
& 5,8,12,20,24,36 \\
D_5 & =\frac{5(6+1)}{10} \text { th value } \\
& =3.5^{th} \text { value }
\end{aligned}
$
$\begin{array}{l}=3^{\text {rd }} \text { value }+\frac{1}{2}\left(4^{\text {th }} \text { value }-3^{\text {rd }} \text { value }\right) \\ =12+\frac{1}{2}(20-12) \\ =12+4=16 .\end{array}$

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Question 141 Mark

The third quartile of the data set $33,25,42,25,31$, $37,46,29,38$, is .....

Answer

40 , because
On arranging the given data in ascending order, we get
$
\begin{array}{l}
25,25,29,31,33,37,38,42,46 \\
\qquad \begin{aligned}
Q_3 & =\frac{3(n+1)}{4} \text { th value } \\
& =\frac{3(9+1)}{4} \text { the value } \\
& {[\because n=9] }
\end{aligned}
\end{array}
$
$=\frac{30}{4}$ th value i.e., 7.5 value
$\begin{aligned} \therefore \quad Q_3 & =\frac{7^{\text {th }} \text { value }+8^{\text {th }} \text { value }}{2} \\ & =\frac{38+42}{2}=\frac{80}{2}=40 .\end{aligned}$

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