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An SBI health insurance agent found the following data for distribution of ages of $100$ policy holders. The health insurance policies are given to persons of age $15$ years and onwards, but less than $60$ years.
Age (in yrs)
$15-20$
$20-25$
$25-30$
$30-35$
$35-40$
$40-45$
$45-50$
$50-55$
$55-60$
Number of policy holders
$2$
$4$
$18$
$21$
$33$
$11$
$3$
$6$
$2$
Find the modal age and median age of the policy holders.
Age (in yrs)
$15-20$
$20-25$
$25-30$
$30-35$
$35-40$
$40-45$
$45-50$
$50-55$
$55-60$
Number of policy holders
$2$
$4$
$18$
$21$
$33$
$11$
$3$
$6$
$2$
Find the modal age and median age of the policy holders.
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For correct table
Age (in years)
f
cf
$15-20$
$2$
$2$
$20-25$
$4$
$6$
$25-30$
$18$
$24$
$30-35$
$21$
$45$
$35-40$
$33$
$78$
$40-45$
$11$
$89$
$45-50$
$3$
$92$
$50-55$
$6$
$98$
$55-60$
$2$
$100$
Total
$100$ (1 Mark)
Modal class = $35-40$ (1/2 Mark)
Mode $= 35 + \frac{33-21}{2(33)-21-11} \times 5$ (1 Mark)
$= \frac{625}{17} = 36.7$(approx.) (1/2 Mark)
$\therefore$ Modal age = $36.7$ years (approx.)
$\frac{n}{2} = 50$, Median class = $35- 40$ (1/2 Mark)
Median $= 35 + \frac{50-45}{33} \times 5$ (1 Mark)
$= \frac{1180}{33} = 35.7$(approx.) (1/2 Mark)
$\therefore$ Median age = $35.7$ years (approx.)
Age (in years)
f
cf
$15-20$
$2$
$2$
$20-25$
$4$
$6$
$25-30$
$18$
$24$
$30-35$
$21$
$45$
$35-40$
$33$
$78$
$40-45$
$11$
$89$
$45-50$
$3$
$92$
$50-55$
$6$
$98$
$55-60$
$2$
$100$
Total
$100$ (1 Mark)
Modal class = $35-40$ (1/2 Mark)
Mode $= 35 + \frac{33-21}{2(33)-21-11} \times 5$ (1 Mark)
$= \frac{625}{17} = 36.7$(approx.) (1/2 Mark)
$\therefore$ Modal age = $36.7$ years (approx.)
$\frac{n}{2} = 50$, Median class = $35- 40$ (1/2 Mark)
Median $= 35 + \frac{50-45}{33} \times 5$ (1 Mark)
$= \frac{1180}{33} = 35.7$(approx.) (1/2 Mark)
$\therefore$ Median age = $35.7$ years (approx.)