Free Measures of Central Tendency 01 Practice Test - 11th Grade - Commerce 

Question 1

Find the modal class of the following distribution.

ClassFrequency01003100200132003002430040084005001650060022600700187008009

A. 200-300
B. 400-500
C. 500-600
D. 600-700

SOLUTION

Solution : C

The grouping and analysis table for the distribution is given below.





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

Question 2

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

A. 1, 7, 5, 2, 3, 7, 8, 1, 7, 10
B. 1, 6, 5, 3, 9, 2, 2, 3, 4, 2
C. 2, 1, 9, 2, 3, 0, 7, 7, 2, 1
D. 7, 5, 3, 6, 8, 2, 7, 5, 8, 7

SOLUTION

Solution : 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 3

The mean of the following distribution is
xi10131619fi2576

A.

15.2

B.

15.35

C.

15.55

D.

16

SOLUTION

Solution : C

xififixi1022013565167112196114fi=20 fixi=311

Mean = fixifi = 31120 =15.55

Question 4

If the median of the following data is 50, then find p.
Class0-2020-4040-6060-8080-100Frequency10716p8

A.

6

B.

7

C.

8

D.

9

SOLUTION

Solution : D

 ClassFrequencyCumulative Frequency0-20101020-4071740-60163360-80p33+p80-100841+p

Since median is 50, median class is 40-60.
N2=(41+p)2
Median=40+[(41+p)21716×20]=50
[(41+p)21716×20]=10(41+p)2=25

p = 9

Question 5

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

SOLUTION

Solution :

The method in question is the step-deviation method.

Question 6

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

Profit per shop less than (%)102030405060No. of shops1230577794100

A.

27.4

B.

31.2

C.

26.8

D.

23.7

SOLUTION

Solution : A

Cumulative frequencies are given by

C.IFrequencyCF01012121020301820305727304077204050941750601006Total100

N2=1002=50

CF nearest to 50 and greater than 50 is 57

 Median=l+(N2CFf)×hMedian=20+(503027)×10

Median = 27.4

Question 7

Calculate the mean of the following data.

ClassFrequency010016100200112003007300400104005006

A. 198
B. 208
C. 226
D. 235

SOLUTION

Solution : B

ClassXFFX01005016800100201501116502003002507175030040035010350040050045062700Total5010400

Mean=FXF=1040050=208

Question 8

For a distribution,
FX=5x+2, F=12

If the mean of the distribution is 6, what is the value of x?

A.

12

B.

14

C.

15

D.

18

SOLUTION

Solution : B

Mean=FXF5x+212=6x=14

Question 9

Which of the following expressions is used to determine the mode?

A.

l+(D1D1+D2)×h 

B.

l+(D2D1+D2)×h ​​​​​​​

C.

l+(D1+D2D1)×h ​​​​​​​

D.

l+(D1+D2D2)×h ​​​​​​​

SOLUTION

Solution : A

The formula for mode is as follows.

Mode=l+(D1D1+D2)×h 

Question 10

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

A.

True

B.

False

SOLUTION

Solution : 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 = Upper limit+Lower limit2