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

11th Grade > Statistics

MEASURES OF CENTRAL TENDENCY MCQs

Total Questions : 30 | Page 3 of 3 pages

Find the modal class of the following distribution.
$\begin{matrix} C l a s s & F r e q u e n c y 0 - 100 & 3 100 - 200 & 13 200 - 300 & 24 300 - 400 & 8 400 - 500 & 16 500 - 600 & 22 600 - 700 & 18 700 - 800 & 9 \end{matrix}$

200-300
400-500
500-600
600-700

Discuss Question

Answer: Option C. -> 500-600
:
C
The grouping and analysis table for the distribution is given below.

Find The Modal Class Of The Following Distribution.ClassFreq...

From the analysis table, the class 500-600 has the maximum score. Hence, 500 - 600 is the modal class.

Question 22.

For which of the following sets of data is the mode and the median the same?

1, 7, 5, 2, 3, 7, 8, 1, 7, 10
1, 6, 5, 3, 9, 2, 2, 3, 4, 2
2, 1, 9, 2, 3, 0, 7, 7, 2, 1
7, 5, 3, 6, 8, 2, 7, 5, 8, 7

Discuss Question

Answer: Option C. -> 2, 1, 9, 2, 3, 0, 7, 7, 2, 1
:
C
Arrange the data in the ascending order.
A) 1, 1, 2, 3, 5, 7, 7, 7, 8, 10
B) 1, 2, 2, 2, 3, 3, 4, 5, 6, 9
C) 0, 1, 1, 2, 2, 2, 3, 7, 7, 9
D) 2, 3, 5, 5, 6, 7, 7, 7, 8, 8
It can be seen that for option C, the median, as well as the mode, is 2.

Question 23.

The mean of the following distribution is
$\begin{matrix} x_{i} & 10 & 13 & 16 & 19 f_{i} & 2 & 5 & 7 & 6 \end{matrix}$

15.2
15.35
15.55
16

Discuss Question

Answer: Option C. -> 15.55
:
C

$\begin{matrix} x_{i} & f_{i} & f_{i} x_{i} 10 & 2 & 20 13 & 5 & 65 16 & 7 & 112 19 & 6 & 114 \sum f_{i} = 20 & \sum f_{i} x_{i} = 311 \end{matrix}$

Mean = $\frac{\sum f_{i} x_{i}}{\sum f_{i}}$ = $\frac{311}{20}$ = $15.55$

Question 24.

If the median of the following data is 50, then find p.
$\begin{matrix} Class & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 Frequency & 10 & 7 & 16 & p & 8 \end{matrix}$

Discuss Question

Answer: Option D. -> 9
:
D

$\begin{matrix} Class & Frequency & C u m u l a t i v e F r e q u e n c y 0-20 & 10 & 10 20-40 & 7 & 17 40-60 & 16 & 33 60-80 & p & 33+p 80-100 & 8 & 41+p \end{matrix}$

Since median is 50, median class is 40-60.
$\frac{N}{2} = \frac{(41 + p)}{2}$
$M e d i a n = 40 + [\frac{\frac{(41 + p)}{2} - 17}{16} \times 20] = 50$
$[\frac{\frac{(41 + p)}{2} - 17}{16} \times 20] = 10 \Rightarrow \frac{(41 + p)}{2} = 25$
p = 9

Question 25.

In the ___ method of calculating mean, we divide the deviations by a common factor to simplify calculations.

Discuss Question

Answer: Option D. -> 9
:

The method in question is the step-deviation method.

Question 26.

The below table shows the profit made by a group of shops in mall. Then the median is

$\begin{matrix} P r o f i t p e r s h o p l e s s t h a n (%) & 10 & 20 & 30 & 40 & 50 & 60 N o . o f s h o p s & 12 & 30 & 57 & 77 & 94 & 100 \end{matrix}$

27.4
31.2
26.8
23.7

Discuss Question

Answer: Option A. -> 27.4
:
A

Cumulative frequencies are given by

$\begin{matrix} C . I & F r e q u e n c y & C F 0 - 10 & 12 & 12 10 - 20 & 30 & 18 20 - 30 & 57 & 27 30 - 40 & 77 & 20 40 - 50 & 94 & 17 50 - 60 & 100 & 6 T o t a l & 100 \end{matrix}$

$\frac{N}{2} = \frac{100}{2} = 50$

CF nearest to 50 and greater than 50 is 57

$∴ M e d i a n = l + (\frac{\frac{N}{2} - C F}{f}) \times h M e d i a n = 20 + (\frac{50 - 30}{27}) \times 10$

Median = 27.4

Question 27.

Calculate the mean of the following data.
$\begin{matrix} C l a s s & F r e q u e n c y 0 - 100 & 16 100 - 200 & 11 200 - 300 & 7 300 - 400 & 10 400 - 500 & 6 \end{matrix}$

Discuss Question

Answer: Option B. -> 208
:
B

\begin{matrix} C l a s s & X & F & F X 0 - 100 & 50 & 16 & 800 100 - 20 & 150 & 11 & 1650 200 - 300 & 250 & 7 & 1750 300 - 400 & 350 & 10 & 3500 400 - 500 & 450 & 6 & 2700 T o t a l & 50 & 10400 \end{matrix}

M e a n = \frac{\sum F X}{\sum F} = \frac{10400}{50} = 208

Question 28.

The class mark for a class interval while calculating the mean is its upper limit.

True
False
$l + (\frac{D_{1} + D_{2}}{D_{1}}) \times h$
$l + (\frac{D_{1} + D_{2}}{D_{2}}) \times h$

Discuss Question

Answer: Option B. -> False
:
B

The class mark is the midpoint of the lower and upper limits of a class. It is calculated by taking the average of the upper and lower limits i.e.
Class mark = $\frac{U p p e r l i m i t + L o w e r l i m i t}{2}$