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In a frequency distribution, the mid-value of a class is 10 and the width of the class is 6. The lower limit of the class is :

(A) \; 6

(B) \; 7

(C) \; 8

(D) \; 12
 

Answers (1)

Answer : B

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

\text{The mid-value of given class}=\frac{x+y}{2}

Here, the mid-value of the class is given as 10.

\frac{x+y}{2}=10

x+y=20\; \; \; \; \; \; \; \; \; \; \; \; ....(1)

Also, given that width of the class is 6

x-y=6\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(2)

Now, adding equation (1) and (2),

x+y+x-y=20+6

x=\frac{26}{2}=13

Putting x=13 in equation (1)

x+y=20

13+y=20

y=20-13=7

Hence, the lower limit of the class is 7

Therefore option (B) is correct.

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