Question #282144

The IQ test results of a group of students in a certain college were taken and are presented in a frequency distribution below.


Classes f

70-75 2

76-81 8

82-87 19

88-93 21

94-99 28

100-105 38

106-111 15

112-117 9


1. Use the frequency distribution above, determine the following:

a. Size of the class interval

b. Lower boundary of the median class

C. Median class

d. Number of students with an IQ greater than 99.5


2. Compute the value of:

A. Mean

B. Median

C. Mode


1
Expert's answer
2021-12-23T12:11:13-0500

1.

a. The class width for a class in a frequency distribution is found by subtracting the lower (or upper) class limit of one class from the lower (or upper) class limit of the next class


7670=676-70=6

b.

Calculate the total frequency


n=2+8+19+21+28+38+15+9=140n=2+8+19+21+28+38+15+9=140

Calculate n/2n/2


n/2=140/2=70n/2=140/2=70

Calculate cumulative frequency. Select Cumulative frequency just greater than or equal to n/2.n/2.


2+8+19+21=50,2+8+19+21=50,




2+8+19+21+28=78,2+8+19+21+28=78,

We find that the 70th observation lies in the class 9499.94-99.

Lower boundary of the median class is l=93.5.l=93.5.


c. Median class is 93.599.5.93.5-99.5.


d.


38+15+9=6238+15+9=62

There are 62 students with an IQ greater than 99.5.


2.

A.


xˉ=1ni=1nfixi\bar{x}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nf_ix_i

=1140(2(72.5)+8(78.5)+19(84.5)+21(90.5)=\dfrac{1}{140}(2(72.5)+8(78.5)+19(84.5)+21(90.5)

+28(96.5)+38(102.5)+15(108.5)+9(114.5))+28(96.5)+38(102.5)+15(108.5)+9(114.5))

=1353414096.6714=\dfrac{13534}{140}\approx96.6714

B.

l=l= lower boundary point of median class =93.5=93.5

n=n= Total frequency =140=140

cf=cf= Cumulative frequency of the class preceding the median class =50=50

f=f= Frequency of the median class =28=28

c=c= class length of median class =6=6


Median M=l+n/2cffcMedian\ M=l+\dfrac{n/2-cf}{f}\cdot c

=93.5+705028697.7857=93.5+\dfrac{70-50}{28}\cdot 6\approx97.7857

C.

To find Mode Class

Here, maximum frequency is 38.

The mode class is 99.5105.5.99.5-105.5.

L=L= lower boundary point of mode class =99.5=99.5

f1=f_1=  frequency of the mode class =38=38

f0=f_0= frequency of the preceding class =28=28

f2=f_2=  frequency of the succedding class =15=15

c=c= class length of mode class =6=6


Z=L+(f1f02f1f0f2)cZ=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\cdot c

=99.5+(38282(38)2815)6c=99.5+(\dfrac{38-28}{2(38)-28-15})\cdot 6c

101.3182\approx101.3182


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