Measures Of Dispersion(11th Grade > Statistics ) Questions and answers for exam Preparation

11th Grade > Statistics

MEASURES OF DISPERSION MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11.

The square root of ___ is called standard deviation

Discuss Question

Answer: Option A. ->
:
Standard deviation is the square root of variance.

Question 12.

The distribution of weights of 50 students in a class is as follows.
$\begin{matrix} W e i g h t (k g) & 40 - 45 & 45 - 50 & 50 - 55 & 55 - 60 & 60 - 65 & 65 - 70 N o . o f s t u d e n t s & 3 & 7 & 11 & 12 & 10 & 7 \end{matrix}$
Calculate the standard deviation using the step deviation method.

5.86
6.34
6.92
7.14

Discuss Question

Answer: Option D. -> 7.14
:
D
Let the assumed mean be 52.5 i.e. A = 52.5 and the dividing factor (c) be 5.

\begin{matrix} X & F & D & D^{'} = \frac{D}{c} & F D^{'} & F D^{' 2} 42.5 & 3 & - 10 & - 2 & - 6 & 12 47.5 & 7 & - 5 & - 1 & - 7 & 7 52.5 & 11 & 0 & 0 & 0 & 0 57.5 & 12 & 5 & 1 & 12 & 12 62.5 & 10 & 10 & 2 & 20 & 40 67.5 & 7 & 15 & 3 & 21 & 63 \sum F = 50 & \sum F D^{'} = 40 & \sum F D^{' 2} = 134 \end{matrix}

σ = \sqrt{\frac{\sum F D^{' 2}}{\sum F} - {(\frac{\sum F D^{'}}{\sum F})}^{2}} \times c = \sqrt{\frac{134}{50} - {(\frac{40}{50})}^{2}} \times 5 = \sqrt{2.68 - 0.64} \times 5 = \sqrt{2.04} \times 5 = 1.428 \times 5 = 7.14

Question 13.

Relative measures of dispersion have the same units as that of the observations in the data. State true or false.

True
False
6.92
7.14

Discuss Question

Answer: Option B. -> False
:
B
Relative measures are ratios and have no units.

Question 14.

Find the coefficient of mean deviation about median for the following cumulative frequency distribution:
$\begin{matrix} x & frequency & Cumulative frequency 22 & 1 & 1 34 & 2 & 3 39 & 2 & 5 45 & 8 & 13 54 & 8 & 21 56 & 9 & 30 68 & 4 & 34 78 & 4 & 38 84 & 2 & 40 \end{matrix}$

0.187
0.203
0.215
0.224

Discuss Question

Answer: Option A. -> 0.187
:
A

Here, the sum of frequencies, n = 40, which is even.
So, the median terms will be the $\frac{n}{2} t h$ and the $(\frac{n}{2} + 1) t h$ observations, i.e. the 20th and the 21st observations.
Both the 20th and the 21st observations are 54.
So, the median term, M= 54.
Deviations from the median are tabulated below
$\begin{matrix} X & F & D = | X - M | & F D 22 & 1 & 32 & 32 34 & 2 & 20 & 40 39 & 2 & 15 & 30 45 & 8 & 9 & 72 54 & 8 & 0 & 0 56 & 9 & 2 & 18 68 & 4 & 14 & 56 78 & 4 & 24 & 96 84 & 2 & 30 & 60 \sum F = 40 & \sum F D = 404 \end{matrix}$
$M . D (M) = \frac{404}{40} = 10.1$
$C o e f f i c i e n t o f M . D (M) = \frac{M . D (M)}{M e d i a n} = \frac{10.1}{54} = 0.187$

Question 15.

The following are the scores of Adil and Kratu in 5 subjects. Adil's scores are out of 10 and Kratu's are out of 50.
Adil: 7, 9, 6, 8, 4
Kratu: 36, 44, 49, 41, 25
Compare the variability in scores using the coefficient of variation.

Adil's scores have higher variability
Kratu's scores have higher variability
Both set of scores have equal variability
Nothing can be concluded

Discuss Question

Answer: Option A. -> Adil's scores have higher variability
:
A

M e a n_{A} = \frac{7 + 9 + 6 + 8 + 4}{5} = 6.8

M e a n_{K} = \frac{36 + 44 + 49 + 41 + 25}{5} = 39

σ_{A} = \sqrt{{0.2}^{2} + {2.2}^{2} + {0.8}^{2} + {1.2}^{2} + {2.8}^{2}} 5 = \sqrt{2.96} = 1.72

σ_{K} = \sqrt{3^{2} + 5^{2} + 10^{2} + 2^{2} + 14^{2}} 5 = \sqrt{66.8} = 8.17

C . O . V_{A} = \frac{1.72}{6.8} \times 100 = 25.29 %

C . O . V_{K} = \frac{8.17}{39} \times 100 = 20.95 %

Hence, Adil's scores have higher variability

Question 16.

Number of goals scored by Ronaldo in champions league for past 10 years are as follows. {1, 3, 8, 4, 7, 6, 10, 12, 17, 10} Find the mean deviation about the mean.

Discuss Question

Answer: Option B. -> 3.6
:
B

$Mean = \frac{1 + 3 + 8 + 4 + 7 + 6 + 10 + 12 + 17 + 10}{10}$

$¯ X = \frac{78}{10} = 7.8$
$\sum | X - 7.8 | = 6.8 + 4.8 + 0.2 + 3.8 + 0.8 + 1.8 + 2.2 + 4.2 + 9.2 + 2.2 = 36$
$∴ Mean deviation = \frac{36}{10} = 3.6$

Question 17.

If the coefficient of variation and standard deviation of a distribution are 2 and 0.4 respectively, then mean of the distribution is

Discuss Question

Answer: Option B. -> 20
:
B

$C . V = 2; σ = 0.4 (g i v e n) C . V = 100 \times \frac{s . d}{m e a n} 2 = \frac{100 \times 0.4}{m e a n} M e a n = \frac{40}{2} = 20$

Question 18.

Inter-quartile range = ___ .

$\frac{Q_{3} - Q_{1}}{2}$
$\frac{Q_{1} - Q_{3}}{2}$
$Q_{3} - Q_{1}$
$Q_{1} - Q_{3}$

Discuss Question

Answer: Option C. ->

Q_{3} - Q_{1}

:
C

I n t e r - q u a r t i l e r a n g e = Q_{3} - Q_{1}

Question 19.

If the S.D of a set of observations is 4 and if each observation is divided by 4, the S.D of the new set of observations will be

Discuss Question

Answer: Option D. -> 1
:
D

We know that if we multiply all numbers of a set by a constant "k" then the mean of these numbers will also be a multiple of the old mean.
And the variance of these numbers will be multiple of " $k^{2}$ ".
Since standard deviation is nothing but the square root of variance, s.d will be multiple of "k" of the old standard deviation.
We are given the s.d was 4.
So after multiplying each observation by (1/4), the s.d will also be multiplied by (1/4), thus giving the value = 4 $\times \frac{1}{4}$ = 1