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Find missing frequency, {{40-59,60-79,80-99,100-119,120-139},{50,?,500,?,50}},median=87.50,N=1000

Solution:
Your problem -> missing frequency, {{40-59,60-79,80-99,100-119,120-139},{50,?,500,?,50}},median=87.50,N=1000

 Class(1) Frequency (f)(2) cf(3) 40-59 50 50 50=0+50(3)=Previous (3)+(2) 60-79 a 50 + a 50 + a=50+a(3)=Previous (3)+(2) 80-99 500 550 + a 550 + a=50 + a+500(3)=Previous (3)+(2) 100-119 b 550 + a + b 550 + a + b=550 + a+b(3)=Previous (3)+(2) 120-139 50 600 + a + b 600 + a + b=550 + a + b+50(3)=Previous (3)+(2) --- --- --- -- n=1000n=a + b + 600

n=1000

a+b+600=1000

a+b=400 ->(1)

To find median class
Here, median is 87.5.

:. The median class is 79.5 - 99.5.

Now,
:. L = lower boundary point of median class =79.5

:. n = Total frequency =1000

:. cf = Cumulative frequency of the class preceding the median class =50 + a

:. f = Frequency of the median class =500

:. c = class length of median class =20

Median M = L + (( n)/2 - cf)/f * c

87.5=79.5 + (( 1000)/2 - (50 + a))/500 * 20

87.5 - 79.5=(500 - (50 + a))/500 * 20

8 = (-a+450)/500 * 20

8*500=(-a+450)*20

4000=-20a+9000

20a=5000

a=5000/20

a=250

Substituting in (1)

250 + b = 400

b = 400 - 250

b = 150

Thus, the missing frequencies are 250 and 150 respectively.

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