Question

In a continuous frequency distribution table the mid value of a class is 43 and the length of the class is 6. Find the lower class limit and the upper class limit.​

Answer 1

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(b) Let x and y be the upper and lower class limit in a frequency distribution.

Now, mid value of a class (x + y )/2=10 [given]

⇒ x + y = 20 …(i)

Also, given that, width of class x- y = 6 …(ii)

On adding Eqs. (i) and (ii), we get

2x =20+ 6

⇒ 2x =26 ⇒ x = 13

On putting x = 13 in Eq. (i), we get

13+y = 20 ⇒ y = 7

Hence, the lower limit of the class is 7.