please answer ASAP
Answers
the answer is the greatest numbers
Solution :-
Class interval Frequency Cumulative frequency
0 - 10 10 10
10 - 20 x (10 + x)
20 - 30 25 (35 + x)
30 - 40 30 (65 + x)
40 - 50 y (65 + x + y)
50 - 60 10 (75 + x + y)
Total (n) 100
so,
→ 75 + x + y = 100
→ x + y = 100 - 75
→ x + y = 25 --------------- Eqn.(1)
now, given that, Median = 32 which lies in 30 - 40 .
so,
- Median class = 30 - 40 .
- Lower limit (l) of median class = 30 .
- Cumulative frequency (CF) of class preceding the median class = (35 + x) .
- Frequency (f) of median class = 30 .
- Class size (h) = 10 .
we know that,
- Median = l + [ {(n/2) - CF} / f ] * h
putting all values we get,
→ 32 = 30 + [(50 - 35 - x)/30] * 10
→ 32 - 30 = (15 - x) / 3
→ 2 * 3 = 15 - x
→ x = 15 - 6
→ x = 9 .
putting value of x in Eqn.(1) now,
→ 9 + y = 25
→ y = 25 - 9
→ y = 16 .
therefore, value of x is 9 and y is 16 .
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