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

### Question 1

Find the modal class of the following distribution.

ClassFrequency0−1003100−20013200−30024300−4008400−50016500−60022600−70018700−8009

#### 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?

#### 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

15.2

15.35

15.55

16

#### SOLUTION

Solution :C

xififixi1022013565167112196114∑fi=20 ∑fixi=311

Mean = ∑fixi∑fi = 31120 =15.55

### Question 4

If the median of the following data is 50, then find p.

Class0-2020-4040-6060-8080-100Frequency10716p8

6

7

8

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)2−1716×20]=50

[(41+p)2−1716×20]=10⇒(41+p)2=25

p = 9

### Question 5

In the

#### 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

27.4

31.2

26.8

23.7

#### SOLUTION

Solution :A

Cumulative frequencies are given by

C.IFrequencyCF0−10121210−20301820−30572730−40772040−50941750−601006Total100

N2=1002=50

CF nearest to 50 and greater than 50 is 57

∴ Median=l+(N2−CFf)×hMedian=20+(50−3027)×10

Median = 27.4

### Question 7

Calculate the mean of the following data.

ClassFrequency0−10016100−20011200−3007300−40010400−5006

#### SOLUTION

Solution :B

ClassXFFX0−1005016800100−20150111650200−30025071750300−400350103500400−50045062700Total5010400

Mean=∑FX∑F=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?

12

14

15

18

#### SOLUTION

Solution :B

Mean=∑FX∑F⇒5x+212=6⇒x=14

### Question 9

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

l+(D1D1+D2)×h

l+(D2D1+D2)×h

l+(D1+D2D1)×h

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.

True

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