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

| Number of wickets | $20$-$60$

Question

The distribution given below shows that the number of wickets taken by bowler 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 | $7$ | $5$ | $16$ | $12$ | $2$ | $3$ |

Accepted Answer

Step 1: Taking assumed mean $a = 120$, $h = 40$.

| Class | $f_i$ | $x_i$ | $u_i$ | $f_i u_i$ | cf |
|-------|-------|-------|-------|-----------|----|
| $20$-$60$ | $7$ | $40$ | $-2$ | $-14$ | $7$ |
| $60$-$100$ | $5$ | $80$ | $-1$ | $-5$ | $12$ |
| $100$-$140$ | $16$ | $120$ | $0$ | $0$ | $28$ |
| $140$-$180$ | $12$ | $160$ | $1$ | $12$ | $40$ |
| $180$-$220$ | $2$ | $200$ | $2$ | $4$ | $42$ |
| $220$-$260$ | $3$ | $240$ | $3$ | $9$ | $45$ |
| **Sum** | $45$ | | | $6$ | |
Step 2: Mean $= a + \frac{\sum f_i u_i}{\sum f_i} 	imes h = 120 + \frac{6 	imes 40}{45} = 120 + 5.33 = 125.33$
Step 3: $\frac{N}{2} = \frac{45}{2} = 22.5$. Median class: $100$-$140$ ($cf = 28 \geq 22.5$). $l = 100$, $f = 16$, $cf = 12$, $h = 40$. Median $= l + \frac{\frac{N}{2} - cf}{f} 	imes h = 100 + \frac{22.5 - 12}{16} 	imes 40 = 100 + 26.25 = 126.25$

Question

Solution