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