Question

the width of each of five continuous classes in a frequency distribution is 6 and the lower class-limit of the lowest class is 12. The upper class limit of the highest classes is?​

Answer 1

Answer:

Let x and y be the upper and lower class limit of frequency distribution.

Given, width of the class =5

⇒x−y=5 ……….(i)

Also, given lower class (y)=10

On putting y=10 in Eq. (i), we get

x−10=5⇒x=15

So, the upper class limit of the lowest class is 15.

Hence, the upper limit of the highest class

= (Number of continuous classes x Class with + Lower class limit of the lowest class)

=5×5+10=25+10=35

Hence, the upper class limit of the highest class is 35.

Alternate method

After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15, 15-20, 20-25, 25-30, and 30-35.

Thus, the highest class is 30-35,

Hence, the upper limit of the class is 35.

Step-by-step explanation:

Answer 2

Answer:

48

Step-by-step explanation:

let x and y be the upper and lower class limit of frequency distribution.

Given -

• Width of the class - 6

==> x - y = 6 .... (i)

Also, given lower class y = 12 on putting y = 12 in Eq. (i) we get

x - 12 = 6 ===> x = 18

So, the upper class limit of the lowest class is 18. Hence, the upper limit of the highest class.

= (Number of continuous class × class width + lower class limit of the lowest class)

= 6 × 6 + 12 = 36 + 12 = 48

Hence, the upper class limit of the highest class is 48.