Question

question
the distribution of the ages of the 65 attendees at an event is given in the table above. which of the following could be the median age, in years, of the attendees?
19
28
32
38
age (years) frequency
0–10 12
11–20 15
21–30 6
31–40 32

Explanation:

Step1: Calculate cumulative frequencies

For age 0 - 10, cumulative frequency $CF_1=12$. For 11 - 20, $CF_2 = 12 + 15=27$. For 21 - 30, $CF_3=27 + 6=33$. For 31 - 40, $CF_4=33+32 = 65$.

Step2: Determine median class

There are $n = 65$ data - points. The median is the $\frac{n + 1}{2}=\frac{65+1}{2}=33$ - rd value. Since the 33 - rd value lies in the 31 - 40 age group (as cumulative frequency up to 21 - 30 is 33 and we start counting from the lower limit of the next class for the median in grouped data), the median class is 31 - 40.

Step3: Use median formula for grouped data

The formula for the median of grouped data is $M = L+\frac{\frac{n}{2}-CF}{f}\times w$, where $L$ is the lower limit of the median class ($L = 31$), $n$ is the total number of data - points ($n = 65$), $CF$ is the cumulative frequency of the class before the median class ($CF = 33$), $f$ is the frequency of the median class ($f = 32$), and $w$ is the class width ($w=10$). But since the 33 - rd value is the first value of the 31 - 40 class, the median is close to the lower limit of the 31 - 40 class. In this case, we can say the median is within the 31 - 40 class.

Answer:

32