Question

Compute arithmetic mean and median for the following frequency distribution
Class 50-59 60-69 70-79 80-89 90-99 100-109 110-119
Frequency 9 11 10 17 21 6 4​

Answer 1

Answer:

In this question, total frequency(N) = 5+15+20+30+20+8=98

So,N2=49

If, we calculate cumulative frequency that is equal to or just greater than 49, it comes 70, making 90−99 our median class.

As, 70 is the less than type cumulative frequency corresponding to the class boundary 89.5

So, lower class boundary(l)=89.5

Cumulative frequency of the class preceding the median class(F)=40

Frequency of median class(f)=30

Interval(i)=10

We know, Median =(l+(N2)−Ff)∗i

Putting above values in formula,

Median = 89.5+(49−4030)∗10

=89.5+9∗1030

Median=89.5+3=92.5

Step-by-step explanation:

hope it helps

please mark as brainliest

please follow me