The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Number of 20 - 60 60 - 100 100 - 150 150 - 250 250 - 350 350 - 450 wickets Number of bowlers
Step-by-Step Solution
Key Concept: For grouped (class‑interval) data the mean is obtained by using the class‑mid‑point (or assumed mean) as a representative value for each class. The formula is \(\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}\), where \(f_i\) is the frequency of the i‑th class and \(x_i\) is its midpoint.
1. Identify the class intervals and their frequencies (the frequencies are given in the table; denote them as \(f_1,f_2,\dots,f_6\)).
2. Find the midpoint of each class:
\[\begin{aligned}
m_1 &= \frac{20+60}{2}=40,\\
m_2 &= \frac{60+100}{2}=80,\\
m_3 &= \frac{100+150}{2}=125,\\
m_4 &= \frac{150+250}{2}=200,\\
m_5 &= \frac{250+350}{2}=300,\\
m_6 &= \frac{350+450}{2}=400.
\end{aligned}\]
3. Multiply each midpoint by its corresponding frequency to obtain \(f_i m_i\).
4. Add all the products to get \(\sum f_i m_i\).
5. Add all the frequencies to get \(\sum f_i\).
6. Compute the mean using the formula:
\[\displaystyle \bar{x}=\frac{\sum f_i m_i}{\sum f_i}.\]
7. Interpretation: The obtained mean represents the average number of wickets taken by a bowler in a one‑day match across the whole group of bowlers considered. It gives a single representative value that summarises the overall performance of the bowlers.
*If the actual frequencies are, for example, \(f = \{2, 5, 8, 6, 3, 1\}\), the calculation would be:*
\[\begin{aligned}
\sum f_i m_i &= 2\times40 + 5\times80 + 8\times125 + 6\times200 + 3\times300 + 1\times400 \\
&= 80 + 400 + 1000 + 1200 + 900 + 400 = 3980,\\
\sum f_i &= 2+5+8+6+3+1 = 25,\\
\bar{x} &= \frac{3980}{25}=159.2 \text{ wickets (approx.)}.
\end{aligned}\]
The same procedure is followed with the actual frequencies given in the question.
Correct Answer: Mean = \(\displaystyle \frac{\sum f_i\,m_i}{\sum f_i}\) where \(m_i\) are the class mid‑points (40, 80, 125, 200, 300, 400). Substituting the given frequencies yields the numerical value of the mean (e.g., 159.2 wickets for the sample frequencies shown). The mean signifies the average number of wickets taken by a bowler in a one‑day match for the whole group of bowlers.