Question

You are given the following incomplete frequency distribution table it is known that the total frequency is 1000 and the median is 412.50 find the missing frequencies class interval 300-325, 325-350,350-375,375-400,400-425,425-450,450-475,475-500 frequency 5,17,80,x,326,y,88,9

Answer 1

Answer:

x = - 215, y = 690

Step-by-step explanation:

Given the total frequency is 1000 and the median is 412.50 find the missing frequencies class interval 300-325, 325-350,350-375,375-400,400-425,425-450,450-475,475-500 frequency 5,17,80,x,326,y,88,9

So calculate cf or cumulative frequency = 5   22  102  102 + x  428 + x  428 + x + y  516 + x + y  525 + x + y

Total frequency is 1000 = N

We know that median = 412.50, l = lower limit = 400 and h = 25(size)

 M = l + [N/2 - cf / f] h

412.5 = 400 + [100/2 - (102 + x)] 25 / 326

412.5 - 400 = (50 - 102 - x)/326 x 25

- 25 x = 5375

 x = - 215

525 + x + y = 1000

525 - 215 +  y = 1000

 y = 690

So x = - 215 and y = 690

Answer 2

Answer:

Step-by-step explanation:

Given the total frequency is 1000 and the median is 412.50 find the missing frequencies class interval 300-325, 325-350,350-375,375-400,400-425,425-450,450-475,475-500 frequency 5,17,80,x,326,y,88,9

So calculate cf or cumulative frequency = 5   22  102  102 + x  428 + x  428 + x + y  516 + x + y  525 + x + y

Total frequency is 1000 = N

We know that median = 412.50, l = lower limit = 400 and h = 25(size)

 M = l + [N/2 - cf / f] h

412.5 = 400 + [1000/2 - (102 + x)] 25 / 326

412.5 - 400 = (500 - 102 - x)/326 x 25

12.50 =(398-x)/326×25

 12.50×326=( 398-x)×25

4075= (398-x)×25

4075/25 = (398-x)

163 =398-x

X=398-163=235

525 + x + y = 1000

525 + 235 +  y = 1000

 y = 240

So x = 235 and y = 240