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Maharashtra State BoardSSC (English Medium) 9th Standard

In the table given below, class-mark and frequencies are given. Construct the frequency table taking inclusive and exclusive classes. Class width 5 15 25 35  Frequency 3 9 15 13

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Question

In the table given below, class-mark and frequencies are given. Construct the frequency table taking inclusive and exclusive classes.

Class width  Frequency
5 3
15 9
25 15
35 13
Sum
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Solution

Suppose, the lower limit of the  5 classes is x1 and the upper limit is x2.

Class mark = `("Lower class limit" + "Upper class limit")/2`

5 = `(x_1 + x_2)/2`

∴ x1 + x2 = 10   ...(1)

The difference between the classes of two adjacent classes is equal to the class width of each class.

∴ class width = 15 − 5 = 25 − 15 = 10

∴ x2 + x1 = 10

∴ − x1 + x2 = 10   ...(2)

Let's sum the equations (1) and (2).

      x1 + x2 = 10
+ − x1 + x2 = 10
___________________
           2x2 = 20

∴ x2 = 10

The value of x2 = 10 is put into equation (1)

x1 + 10 = 10

∴ x1 = 10 − 10

∴ x1 = 0

∴ First-class = 0 − 10. From this the next class are 10 − 20, 20 − 30 and 30 − 40.

So, the exclusive frequency table is given by:

Exclusive class  Classmark  Frequency
0 − 10 5 3
10 − 20 15 9
20 − 30  25 15
30 − 40 35 13

class width = Upper limit − Lower limit

= 10 − 0 = 10

Also, the inclusive frequency table is given by:

Inclusive class  Classmark  Frequency
0.5 − 9.5 5 3
10.5 − 19.5 15 9
20.5 − 29.5 25 15
30.5 − 39.5 35 13
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Chapter 7: Statistics - Practice set 7.3 [Page 119]

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Balbharati Mathematics 1 [English] Standard 9 Maharashtra State Board
Chapter 7 Statistics
Practice set 7.3 | Q (7) (i) | Page 119

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