# Free Measures of Central Tendency 03 Practice Test - 11th Grade - Commerce

### Question 1

Class mark of a class is __________

upper limit + lower limit

12×(upper limit + lower limit)

12×(upper limit - lower limit)

upper limit - lower limit

#### SOLUTION

Solution :B

Class mark is the midpoint of a class interval. Therefore, its formula is given by

upper limit + lower limit2.

### Question 2

If the mean of the following data is 17 then the value of p is

X10p182125F1015799

#### SOLUTION

Solution :XFFX1010100p1515p187126219189259225

∑F=50

∑FX=640+15p

Given mean=17

17=640+15p50⇒15p=50×17−640=210⇒p=21015=14

### Question 3

A survey was conducted on 20 families in a locality by a group of students .What will be the mode of the data?

Age of family member0−2020−4040−6060−8080−100Number of students78221

Given that modal class is 20-40

21.85

22.86

23.87

24.87

#### SOLUTION

Solution :B

Given that the modal class is 20- 40

Frequency of modal class= 8

Frequency of the preceding class = 7

Frequency of the succeeding class = 2

l=20; D1=8−7=1

D2=8−2=6; h=20

Mode=l+(D1D1+D2)×h=20+17×20=22.86

### Question 4

The median for grouped data is found by using the formula:

l+(N2−CFf)×h

l−(N2−CFf)×h

l+(N2+CFf)×h

l−(N2+CFf)×h

#### SOLUTION

Solution :A

The median for grouped data is formed by l+(N2−CFf)×h.

Where l is the lower class limit of the median class, N is the total number of observations, CF is the cumulative frequency of the class preceding the median class, f is the frequency of the median class and h is the class size.

### Question 5

The Mean of the first 10 prime numbers is

#### SOLUTION

Solution :The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29

The sum of numbers = 129

The Mean = 129/10 = 12.9

### Question 6

If the preceding and succeeding classes of the modal class have the same frequency, then the mode will be at the midpoint of the modal class.

True

False

#### SOLUTION

Solution :A

The formula for the mode is as follows.

Mode=l+(f1−f02f1−f0−f2)×h

l= lower boundary of the modal class

h= size of the modal class interval

f1= frequency of the modal class.

f0= frequency of the class preceding the modal class

f2= frequency of the class succeeding the modal class

If the preceding and succeeding classes have the same frequency, then f0=f2=f(say).Then the equation reduces to

Mode=l+(f1−f2f1−f−f)×h

Mode=l+(f1−f2(f1−f))×h

Mode=l+12×h, which is the midpoint of the modal class.∴ If the preceding and succeeding classes of the modal class have the same frequency, then the mode will be at the midpoint of the modal class.

### Question 7

The algebraic sum of the deviations of a frequency distribution from its mean is :

#### SOLUTION

Solution :C

The algebraic sum of the deviations of a frequency distribution from its mean is 0

### Question 8

In the assumed mean method, if A is the assumed mean, then deviation D is :

#### SOLUTION

Solution :A

The deviation is D=X−A

### Question 9

The number of hours of classes attended by students in a semester and the corresponding number of students is tabulated below. Find the median number of hours of classes attended by the students.

Number of hours100−120120−140140−160160−180180−200Number of students192830167

128

133

138

142

#### SOLUTION

Solution :D

Class IntervalFrequencyCumulative Frequency100−1201919120−1402847140−1603077160−1801693180−2007100

N2=50

Median class is 140-160

M=L+N2−c.ff×h=140+50−4730×20=142

### Question 10

The time (in seconds) taken by 150 athelets to run a 110 m hurdle race are tabulated below.

Class13.8−1414−14.214.2−14.414.4−14.614.6−14.814.8−15Frequency245714820

The number of atheletes who completed the race in less than 14.6 s is

11

71

82

130

#### SOLUTION

Solution :C

The number of athletes who completed the race in less than 14.6 is given by the cumulative frequency of the class 14.4 - 14.6

= 2 + 4 + 5 + 71

= 82