Question

If the median of the following frequency distribution
is 46, then find the value of and respectively.



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Answer 1

Answer

m = 34

n = 46

Explanation

Total frequency (N) = 230

→ N/2 = 115

Given that median of given distribution = 46

→ median class = 40 - 50 (since, 46 lies between 40 - 50)

→ lower limit of median class (l) = 40

→ class size (h) = 10 (difference of 50 and 40)

We will find cumulative frequency of class preceding median class (cf)

→ cf = 12 + 30 + m = 42 + m

Now, we know that median of a group distribution is given as

$\sf median = l + \frac{\frac{N}{2} - cf}{f}\times h$

Where, N is the total frequency, f is the frequency of median class and rest are denoting the same as said above.

We have,

N/2 = 115

l = 40

h = 10

cf = 42 + m

median = 46

and f = 65

→ $\sf 46 =40+ \frac{(115 - 42 - m)}{65}\times 10$

→ $\sf 46 - 40= \frac{(73 - m)}{65}\times 10$

→ $\sf 6 \times \frac{65}{10} = 73 - m$

→ $\sf 39 = 73 - m$

→ m = 73 - 39

→ m = 34.

Now, total frequency = 230

→ 12 + 30 + m + 65 + n + 25 + 18 = 230

→ 12 + 30 + 34 + 65 + n + 25 + 18 = 230

→ n + 184 = 230

→ n = 230 - 184

→ n = 46.