The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the number of wickets taken. 

Number of wickets : 20-60 60-100 100-140 140-180 180-220 220-260
Number of bowlers  7 5 16 12 2 3

 

 

 

 
 
 
 
 

Answers (1)
S safeer

Assume the mean a = 120 and h = 40

\begin{array}{|c|c|c|c|c|c|c|c|} \hline \text { No. of wickets } & 20-60 & 60-100 & 100-140 & 140-180 & 180-220 & 220-260 & \\ \hline \text { No. of bowlers(f) } & 7 & 5 & 16 & 12 & 2 & 3 & \sum f_i = 45 \\ \hline \mathrm{x}_{\text {i }} & 40 & 80 & 120 & 160 & 200 & 240 & \\ \hline \mathrm{u}_{\text {i }}=(x_i-a)/h & -2 & -1 & 0 & 1 & 2 & 3 & \\ \hline \mathrm{f}_{i} \mathrm{u}_{\text {i }} & -14 & -5 & 0 & 12 & 4 & 9 & \sum f_i u_i = 6 \\ \hline \text { cf } & 7 & 12 & 28 & 40 & 42 & 45 & \\ \hline \end{array}

\\ \text { Mean }=a+\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h\\ =120+\frac{6 \times 40}{45} \\ =125.33 \\

N= 45 \implies \frac{N}{2} = 22.5
\therefore Median class = 100-140; Cumulative Frequency = 28; Lower limit, l = 100;

c.f. = 12; f = 16,  h = 40
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 100 + \left (\frac{22.5-12}{16} \right ) \times 40 \\ \\ = 126.25

Thus, the median of the data is 126.25

 

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