The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Here, CI stands for class interval and Fr for frequency.
CI 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900-1000
Fr 2 5 x 12 17 20 y 9 7 4
Answers
Answer:
Step-by-step explanation:
The median class is 500-600 and the median will be 525
Median = l + n/2 -cf × h
where f = median frequency l = lower limit of median class = 500
h = class interval = 100 - 0 = 100
∑f_i = n = 100 cf = cumulative frequency before class median = 36 + x
= 525 = 500 + {50 - (36 + x)} * 5
= 525 - 500 = {50 - (36 + x)}*5
= 36 + x = 50 - 5
= x = 45 - 36
= x = 9
Now,
∑f_i = 76 + x + y
Substituting the values
100 = 76 + x + y
9 + y + 76 = 100
y = 100 - 85
y = 15
Answer:
Step-by-step explanation:
Step-by-step explanation:
The median class is 500-600 and the median will be 525
Median = l + n/2 -cf × h
where f = median frequency l = lower limit of median class = 500
h = class interval = 100 - 0 = 100
∑f_i = n = 100 cf = cumulative frequency before class median = 36 + x
= 525 = 500 + {50 - (36 + x)} * 5
= 525 - 500 = {50 - (36 + x)}*5
= 36 + x = 50 - 5
= x = 45 - 36
= x = 9
Now,
∑f_i = 76 + x + y
Substituting the values
100 = 76 + x + y
9 + y + 76 = 100
y = 100 - 85
y = 15