Question

The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the median of these data.
\[\begin{array}{|c|c|c|c|c|c|c|} \hline Age (in years) & 5-15 & 15-25 & 25-35 & 35-45 & 45-55 & 55-65 \\ \hline \text{Number of patients} & \text{6} & \text{11} & \text{21} & \text{23} & \text{14} & \text{5} \\ \hline \end{array}\]

Hint:

For the mode in grouped data, use $l+\dfrac{(f_1-f_0)}{(2f_1-f_0-f_2)}h$. For the median, locate the class where the cumulative frequency first exceeds $N/2$, then apply $l+\dfrac{(N/2-c_f)}{f}h$.

Answer 1

Total $N=6+11+21+23+14+5=80$, class width $h=10$. Mode:
Modal class $=35$--$45$ (highest frequency $f_1=23$). With $l=35,\ f_0=21,\ f_2=14$,
\[ \text{Mode}=l+\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}\,h =35+\frac{23-21}{2\cdot23-21-14}\times10 =35+\frac{2}{11}\times10 =35+\frac{20}{11}\approx \boxed{36.82\ \text{years}}. \] Median:
Cumulative frequencies: $6,17,38,61,75,80$. Since $N/2=40$, the median class is $35$--$45$ ($l=35$) with $f=23$ and cumulative before it $c_f=38$.
\[ \text{Median}=l+\frac{\left(\frac{N}{2}-c_f\right)}{f}\,h =35+\frac{(40-38)}{23}\times10 =35+\frac{20}{23} \approx \boxed{35.87\ \text{years}}. \]