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

11th Grade > Statistics

MEASURES OF DISPERSION MCQs

Total Questions : 30 | Page 1 of 3 pages

Find the range of the following data
$\begin{matrix} 9 & 26 & 29 & - 10 & 6 22 & 7 & 34 & 24 & 1 23 & 13 & 30 & 2 & 34 13 & - 9 & 5 & - 6 & - 4 \end{matrix}$

Discuss Question

Answer: Option D. -> 44
:
D
Range is the difference between the largest and the smallest observation.

R a n g e = 34 - (- 10) = 44

Question 2.

Steven Gerrard, arguably one of the most consistent and skilled players in the English premier league was given the following monthly ratings by a football website for the past 40 months.

$\begin{matrix} R a t i n g & 6 & 8 & 8.5 & 9 & 7 & 10 N u m b e r o f m o n t h s w i t h t h e r a t i n g & 2 & 8 & 10 & 7 & 8 & 5 \end{matrix}$

Calculate how his rating is deviating from his average performance in terms of mean deviation about mean.

0.825
0.85
0.9
1

Discuss Question

Answer: Option A. -> 0.825
:
A

$M e a n = \frac{\sum F X}{\sum F}$
$M e a n, ¯ X = \frac{6 (2) + 8 (8) + (8.5) (10) + 9 (7) + 7 (8) + 10 (5)}{2 + 8 + 10 + 7 + 8 + 5} = \frac{330}{40} = 8.25$
The deviations from the mean are tabulated below.
$\begin{matrix} X & F & D = | X - ¯ X | & F D 6 & 2 & 2.25 & 4.5 8 & 8 & 0.25 & 2 8.5 & 10 & 0.25 & 2.5 9 & 7 & 0.75 & 5.25 7 & 8 & 1.25 & 10 10 & 5 & 1.75 & 8.75 \sum F = 40 & \sum F D = 33 \end{matrix}$

$M . D (¯ X) = \frac{\sum F D}{\sum F} = \frac{33}{40} = 0.825$

$∴$ Mean deviation about mean for his performance across 40 months about his average performance rating of 8.25 is 0.825 which shows he is quite consistent.

Question 3.

The heights of six students in a class are 169 cm, 172 cm, 178 cm, 170 cm, 166 cm and 174 cm. Find the standard deviation of heights using the assumed mean method.

2.85
3.46
3.82
4.15

Discuss Question

Answer: Option C. -> 3.82
:
C
Let 170 be the assumed mean (A).

\begin{matrix} X & D = X - A & D^{2} 169 & - 1 & 1 172 & 2 & 4 178 & 8 & 64 170 & 0 & 0 166 & - 4 & 16 174 & 4 & 16 \sum D = 9 & \sum D^{2} = 101 \end{matrix}

σ = \sqrt{\frac{\sum D^{2}}{N} - {(\frac{\sum D}{N})}^{2}} = \sqrt{\frac{101}{6} - {(\frac{9}{6})}^{2}} = \sqrt{14.58} = 3.82

Question 4.

The mark distribution of 40 students in an exam is shown below. Find the standard deviation of the distribution.
$\begin{matrix} Marks & 0 - 10 & 10 - 20 & 20 - 30 & 30 - 40 & 40 - 50 No. of Students & 7 & 10 & 12 & 6 & 5 \end{matrix}$

10.28
12.49
14.76
15.83

Discuss Question

Answer: Option B. -> 12.49
:
B

$\begin{matrix} X & F & F X & D = X - ¯ X & D^{2} & F D^{2} 5 & 7 & 35 & - 18 & 324 & 2268 15 & 10 & 150 & - 8 & 64 & 640 25 & 12 & 300 & 2 & 4 & 48 35 & 6 & 210 & 12 & 144 & 864 45 & 5 & 225 & 22 & 484 & 2420 \sum F = 40 & \sum F X = 920 & \sum F D^{2} = 6240 \end{matrix}$
$M e a n, ¯ X = \frac{\sum F X}{\sum F} = \frac{920}{40} = 23$
$σ = \sqrt{\frac{\sum F D^{2}}{\sum F}} = \sqrt{\frac{6240}{40}} = \sqrt{156} = 12.49$

Question 5.

The number of years of education acquired by a sample of 100 adults in a locality is as follows.
$\begin{matrix} Years of education & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 & 20 - 24 No. of people & 10 & 22 & 32 & 24 & 8 & 4 \end{matrix}$

2.85
3.75
4.95
6.05

Discuss Question

Answer: Option C. -> 4.95
:
C

Let the assumed mean be 10 i.e. A = 10
$\begin{matrix} X & F & D = X - A & F D & F D^{2} 2 & 10 & - 8 & - 80 & 640 6 & 22 & - 4 & - 88 & 352 10 & 32 & 0 & 0 & 0 14 & 24 & 4 & 96 & 384 18 & 8 & 8 & 64 & 512 22 & 4 & 12 & 48 & 576 \sum F = 100 & \sum F D = 40 & \sum F D^{2} = 2464 \end{matrix}$
$σ = \sqrt{\frac{\sum F D^{2}}{\sum F} - {(\frac{\sum F D}{\sum F})}^{2}} = \sqrt{\frac{2464}{100} - {(\frac{40}{100})}^{2}} = \sqrt{24.64 - 0.16} = \sqrt{24.48} = 4.95$

Question 6.

The average ambient temperatures (in degrees celsius) for a region for 15 years from 2001-2015 is given below.
26.3, 29.7, 31.0, 29.8, 26.8, 27.6, 27.4, 27.0, 27.6, 27.8, 30.0, 32.0, 28.9, 30.5, 29.5
Calculate the coefficient of range for the temperatures.

0.036
0.049
0.078
0.098

Discuss Question

Answer: Option D. -> 0.098
:
D
The lowest temperature is 26.3 degrees and the highest temperature is 32.0 degrees.

C o e f f i c i e n t o f r a n g e = \frac{L - S}{L + S} = \frac{32^{\circ} C - {26.3}^{\circ} C}{32^{\circ} C + {26.3}^{\circ} C} = \frac{{5.7}^{\circ} C}{{58.3}^{\circ} C} = 0.098

Question 7.

The coefficient of variation of data pertaining to the goals scored by two teams A and B in a football season are 0.35 and 0.72 respectively. The true statement is

Team B is more consistent than team A
Two teams are equally consistent
Team A is more consistent than team B
Nothing can be concluded

Discuss Question

Answer: Option C. -> Team A is more consistent than team B
:
C
The coefficient of variation is a measure of consistency. Lower, the C.V, more consistent the data is. In this case, team A has a lower C.V and hence, is more consistent.

Question 8.

Given below is the Lorenz curve representing the percentage revenues of car manufacturing companies worldwide.

The top 20 percentage of the companies generate what percentage of revenue?

Discuss Question

Answer: Option D. -> 60%
:
D
From the Lorenz curve, the bottom 80% of the companies generate 40% of the revenue. Hence, the top 20% of the companies generate 60% of the revenue.

Question 9.

For a data set, the coefficient of range is 0.5. The largest observation is 12. Find the smallest observation.

Discuss Question

Answer: Option C. -> 4
:
C
Given that the coefficient of range is 0.5.

\frac{L - S}{L + S} = 0.5 \Rightarrow L = 3 S

Given that L=12.

∴ 3 S = 12 S = 4

Question 10.

Data A has a mean of 100 and standard deviation of 10. Data B has a mean of 1000 and standard deviation of 50. Which data shows more variability?

Data A shows more variability
Data B shows more variability
Both show equal variability
Nothing can be concluded

Discuss Question

Answer: Option A. -> Data A shows more variability
:
A
To compare dissimilar data, relative measures should be used. Here, the coefficient of variation of set A and set B can be compared.