# Free Measures of Central Tendency 02 Practice Test - 11th Grade - Commerce

### Question 1

Find the mode of the following data.

21417111572081710820183

#### SOLUTION

Solution :A

Mode is the most common observation. In the given data, 8 occurs 3 times, more than any other number. Hence, the mode is 8.

### Question 2

Mode is the

Most frequent value

Least frequent value

Middle most value

Average value

#### SOLUTION

Solution :A

Mode is the most frequently observed value.

### Question 3

The median class for the given data is

Height (in cm) Less than 140 Less than 145 Less than 150Less than 155Less than 160Less than 165No. of girls41129404651

145 - 150

150 - 155

155 - 160

160 - 165

#### SOLUTION

Solution :A

The given table can be written as:

Height (in cm) Less than 140 Less than 145 Less than 150Less than 155Less than 160Less than 165No. of girls41129404651

Height (in cm) Less than 140140−145145−150150−155155−160160−165Cumulative Frequency41129404651

n = 51

∴n2=512

= 25.5

Since there are a total of 51 observations, the median class is the one whose cumulative frequency is closest to and greater than 25.5. The cumulative frequency for the class '145 - 150' is 29 which is greater than 25.5. The median class is '140 - 145'. Thus, the class interval '145 - 150' is the required answer.

### Question 4

The mean ¯x of a discrete series of data is given by:

∑FX∑F

∑F∑FX

∑X∑F

∑FX∑X

#### SOLUTION

Solution :A

The mean of the data is the sum of the values of all the observations divided by the total number of observations.

### Question 5

The median for the data is

IQ60−7070−8080−9090−100100−110110−120120−130No. of pupils2351614137

102.86

98.46

97.47

103.26

#### SOLUTION

Solution :A

Class IntervalFrequencyCumulative Frequency60−702270−803580−9051090−1001626100−1101440110−1201353120−130760

n2=30

∴ the median class is 100-110

Median =l+(n2−cff)×h

=100+30−2614×10

=102.86

### Question 6

For calculating the mean of grouped data, the mid point of each class interval is chosen to represent all the observations from that class.

True

False

#### SOLUTION

Solution :A

For calculating the mean of grouped data, the midpoint of each class interval is chosen to represent all the observations from that class. The midpoint is called the class midpoint or class mark.

### Question 7

The following data shows monthly savings of 100 families, the difference between modal and mean monthly savings lies between

Monthly savings(Rs) Number of families 1000−2000142000−3000153000−4000214000−5000275000−600025

600-700

700-800

800-900

900-1000

#### SOLUTION

Solution :C

Modal class is 4000-5000

l=4000; D1=27−21=6

D2=27−25=2;h=1000Mode=l+(D1D1+D2)×h4000+68×1000=750

Mean=∑FX∑F=391000100=3910

Difference between mode and mean

= 4750 – 3910 = 840

840 lies between 800 and 900.

### Question 8

Find the range of the following data

92629−1062273424123133023413−95−6−4

25

34

43

44

#### SOLUTION

Solution :D

Range is the difference between the largest and the smallest observation.

Range = 34-(-10)=44

### Question 9

The median class for the following data is

MarksBelow 10Below 20Below 30Below 40Below 50 No. of students 510233140

10 - 20

20 - 30

30 - 40

40 - 50

#### SOLUTION

Solution :B

The given table can be written as:

Marks0−1010−2020−3030−4040−50 No. of students551389 Cumulative Frequency510233140

No. of students can be calculated as follows:

5

10 - 5 = 5

23 - 10 = 13

31 - 23 = 8

40 - 31 = 9

∴n2=402=20

Since there are a total of 40 observations, the median class is the one whose cumulative frequency is closest to and greater than 20.

Here the cumulative frequency for the class interval '10 - 20' is 23 which is greater than 20.

The class interval 20 - 30 is the required answer.

### Question 10

In the formula ¯x=A+∑FD∑D , the A stands for assumed mean.

True

False

#### SOLUTION

Solution :A

¯x=A+∑FD∑F

In the formula, the A stands for the assumed mean.