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

11th Grade > Statistics

MEASURES OF DISPERSION MCQs

Total Questions : 30 | Page 3 of 3 pages

The average ambient temperatures (in degree celsius) for a region for 15 years from 2001-2015 is given below.
26.3, 29.7, 31, 29.8, 26.8, 27.6, 27.4, 27.0, 27.6, 27.8, 30.0, 32.0, 28.9, 30.5, 29.5
Find the inter-quartile range of the temperatures.

Discuss Question

Answer: Option D. -> 2.6
:
D
Arrange the observations in ascending order.
26.3, 26.8, 27.0, 27.4, 27.6, 27.6, 27.8, 28.9, 29.5, 29.7, 29.8, 30, 30.5, 31, 32
Here, N=15

1 s t q u a r t i l e, Q_{1} = s i z e o f (\frac{N + 1}{4}) t h o b s e r v a t i o n = s i z e o f 4 t h o b s e r v a t i o n = 27.4

3 r d q u a r t i l e, Q_{3} = s i z e o f 3 (\frac{N + 1}{4}) t h o b s e r v a t i o n = s i z e o f 12 t h o b s e r v a t i o n = 30

I n t e r - q u a r t i l e r a n g e = Q_{3} - Q_{1} = 30 - 27.4 = 2.6

Question 22.

The distance travelled by a scooter during five months are respectively 890 km, 940 km, 640 km, 570 km and 760 km. Find the standard deviation of the distances.

141.3 km
160.5 km
180.7 km
192.8 km

Discuss Question

Answer: Option A. -> 141.3 km
:
A

$Mean = \frac{890 + 940 + 640 + 570 + 760}{5} = 760 k m$
$σ = \sqrt{\frac{\sum (X - ¯ X)^{2}}{N}} = \sqrt{\frac{1}{5} \times [130^{2} + 180^{2} + 120^{2} + 190^{2} + 0]} \sqrt{\frac{99800}{5}} = \sqrt{19960} = 141.3 k m$

Question 23.

Find the coefficient quartile deviation of the given data:
$\begin{matrix} W a g e s (i n R s .) & 200 - 300 & 300 - 400 & 400 - 500 & 500 - 600 & 600 - 700 N o . o f L a b o u r e r s & 3 & 5 & 20 & 10 & 6 \end{matrix}$

0.14
0.18
0.23
0.47

Discuss Question

Answer: Option A. -> 0.14
:
A
The cumulative frequency distribution for the data is :

\begin{matrix} W a g e s & F r e q u e n c y & C u m u l a t i v e f r e q u e n c y 200 - 300 & 3 & 3 300 - 400 & 5 & 8 400 - 500 & 20 & 28 500 - 600 & 10 & 38 600 - 700 & 6 & 44 \end{matrix}

In the above distribution the total number of observations (labourers) is 44. Hence N = 44.

\frac{N}{4} = 11 & \frac{3 N}{4} = 33

Class corrsponding to the first quartile is 400 - 500

Q_{1} = L + \frac{\frac{N}{4} - C F}{f} \times h = 400 + \frac{11 - 8}{20} \times 100 = 415

Class corresponding to the third quartile is 500 - 600

Q_{3} = L + \frac{\frac{3 N}{4} - C F}{f} \times h = 500 + \frac{33 - 28}{10} \times 100 = 550

C o e f f i c i e n t o f Q D = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} = \frac{135}{965} = 0.14

Question 24.

The scores of 8 students in a test are 16, 13, 21, 14, 9, 19, 24, 25. Find the standard deviation of their scores using direct method.

3.8
5.24
6.06
7.12

Discuss Question

Answer: Option B. -> 5.24
:
B

\begin{matrix} X & X^{2} 16 & 256 13 & 169 21 & 441 14 & 196 9 & 81 19 & 361 24 & 576 25 & 625 \sum X = 141 & \sum X^{2} = 2705 \end{matrix}

M e a n, ¯ X = \frac{\sum X}{N} = \frac{141}{8} = 17.62

σ = \sqrt{\frac{\sum X^{2}}{N} - {¯ X}^{2}} = \sqrt{\frac{2705}{8} - {17.62}^{2}} = \sqrt{338.12 - 310.46} = \sqrt{27.48} = 5.24

Question 25.

The runs scored by a batsman in the 5 matches of a one-day series are 51, 27, 44, 71 and 82. Calculate the mean deviation about the mean.

12.4
13.6
15.6
17.2

Discuss Question

Answer: Option D. -> 17.2
:
D

M e a n, ¯ X = \frac{51 + 27 + 44 + 71 + 82}{5} = \frac{245}{5} = 55

The deviations for the scores are tabulated below

\begin{matrix} O b s e r v a t i o n (X) & D = | X - ¯ X | 51 & 4 27 & 28 44 & 11 71 & 16 82 & 27 \end{matrix}

M . D (¯ X) = \frac{4 + 28 + 11 + 16 + 27}{5} = \frac{86}{5} = 17.2

Question 26.

Distance travelled by 100 cars in a month are tabulated below
$\begin{matrix} D i s t a n c e (k m) & 0 - 400 & 400 - 800 & 800 - 1200 & 1200 - 1600 & 1600 - 2000 N o . o f c a r s & 25 & 40 & 20 & 10 & 5 \end{matrix}$
Calculate the coefficient of variation.

30.5 %
45.5 %
61.1 %
72%

Discuss Question

Answer: Option C. -> 61.1 %
:
C
Let the assumed mean be i.e. A = 1000 and the dividing factor (c) be 400.

\begin{matrix} X & F & D = X - A & D^{'} = \frac{D}{c} & F D^{'} & F D^{' 2} 200 & 25 & - 800 & - 2 & - 50 & 100 600 & 40 & - 400 & - 1 & - 40 & 40 1000 & 20 & 0 & 0 & 0 & 0 1400 & 10 & 400 & 1 & 10 & 10 1800 & 5 & 800 & 2 & 10 & 20 \sum F = 100 & \sum F D^{'} = - 70 & \sum F D^{' 2} = 170 \end{matrix}

¯ X = A + \frac{\sum F D}{\sum F} \times c = 1000 - \frac{70}{100} \times 400 = 720

σ = \sqrt{\frac{\sum F D^{' 2}}{\sum F} - {(\frac{\sum F D^{'}}{\sum F})}^{2}} \times c = \sqrt{\frac{170}{100} - {(\frac{- 70}{100})}^{2}} \times 400 = \sqrt{1.7 - 0.49} \times 400 = \sqrt{1.21} \times 400 = 1.1 \times 400 = 440

C . O . V = \frac{440}{720} \times 100

Question 27.

The Lorenz curve given below shows the distribution of land among the population of a country.

What percentage of the land is owned by the top 40% of the people?

Discuss Question

Answer: Option D. -> 80%
:
D
From the Lorenz curve, it can be observed that the bottom 60% of the people own 20% of the land. Hence, the top 40% of the people own 80% of the land.

Question 28.

Number of goals scored by Messi in champion's league for past 10 years is a follows: {1, 1, 6, 9, 8, 12, 14, 8, 8, 10}. Find the mean deviation about median.

Discuss Question

Answer: Option B. -> 2.9
:
B

First, we have to calculate the median of the goals.
Arranging the given values in ascending order, we get,
1, 1, 6, 8, 8, 8, 9, 10, 12, 14,

5^th and 6^th values are 8 and 8.

$∴ M e d i a n = \frac{8 + 8}{2} = 8$
$\sum | X - M | = 7 + 7 + 2 + 1 + 0 + 4 + 6 + 0 + 0 + 2 = 29$