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A life insurance agent found the following data for the distribution of $100$ policy holders on the basis of their ages.
begin{tabular}{|c|c|}
hline Age (in years) & Number of policy holders
hline $15-20$ & $2$
hline $20-25$ & $4$
hline $25-30$ & $18$
hline $30-35$ & $21$
hline $35-40$ & $33$
hline $40-45$ & $11$
hline $45-50$ & $3$
hline $50-55$ & $6$
hline $55-60$ & $2$
hline
end{tabular}
Find the median age of the policy holders.
begin{tabular}{|c|c|}
hline Age (in years) & Number of policy holders
hline $15-20$ & $2$
hline $20-25$ & $4$
hline $25-30$ & $18$
hline $30-35$ & $21$
hline $35-40$ & $33$
hline $40-45$ & $11$
hline $45-50$ & $3$
hline $50-55$ & $6$
hline $55-60$ & $2$
hline
end{tabular}
Find the median age of the policy holders.
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Solution:
begin{tabular}{|c|c|c|}
hline $CI$ & $f_i$ & $Cf$
hline $15-20$ & $2$ & $2$
hline $20-25$ & $4$ & $6$
hline $25-30$ & $18$ & $24$
hline $30-35$ & $21$ & $45$
hline $35-40$ & $33$ & $78$
hline $40-45$ & $11$ & $89$
hline $45-50$ & $3$ & $92$
hline $50-55$ & $6$ & $98$
hline $55-60$ & $2$ & $100$
hline
end{tabular}
$\frac{N}{2} = 50 \therefore$ median class : $35-40$
Median $= 35 + \frac{50 - 45}{33} \times 5 = 35.76$
Thus, the median age of the policy holders is $35.76$ years.
begin{tabular}{|c|c|c|}
hline $CI$ & $f_i$ & $Cf$
hline $15-20$ & $2$ & $2$
hline $20-25$ & $4$ & $6$
hline $25-30$ & $18$ & $24$
hline $30-35$ & $21$ & $45$
hline $35-40$ & $33$ & $78$
hline $40-45$ & $11$ & $89$
hline $45-50$ & $3$ & $92$
hline $50-55$ & $6$ & $98$
hline $55-60$ & $2$ & $100$
hline
end{tabular}
$\frac{N}{2} = 50 \therefore$ median class : $35-40$
Median $= 35 + \frac{50 - 45}{33} \times 5 = 35.76$
Thus, the median age of the policy holders is $35.76$ years.