The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Daily pocket 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25 allowance (in `) Number of children 7 6 9 13 f 5 4 182
Step-by-Step Solution
Key Concept: For grouped data, the mean \(\bar{x}\) is given by \(\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}\), where \(x_i\) are the class mid‑points. By expressing \(\sum f_i x_i\) and \(\sum f_i\) in terms of the unknown frequency \(f\) and using the given mean, we obtain a linear equation in \(f\). Solving it yields the required frequency.
1. Find the class mid‑points (average of lower and upper limits):
\[
\begin{aligned}
11-13 &: \; x_1 = \frac{11+13}{2}=12 \\
13-15 &: \; x_2 = \frac{13+15}{2}=14 \\
15-17 &: \; x_3 = \frac{15+17}{2}=16 \\
17-19 &: \; x_4 = \frac{17+19}{2}=18 \\
19-21 &: \; x_5 = \frac{19+21}{2}=20 \\
21-23 &: \; x_6 = \frac{21+23}{2}=22 \\
23-25 &: \; x_7 = \frac{23+25}{2}=24
\end{aligned}
\]
2. Write the expressions for \(\sum f_i\) and \(\sum f_i x_i\)
\[
\sum f_i = 7+6+9+13+f+5+4 = 44+f
\]
\[
\sum f_i x_i = 12\times7 + 14\times6 + 16\times9 + 18\times13 + 20\times f + 22\times5 + 24\times4
\]
Calculating the known products:
\[
\begin{aligned}
12\times7 &= 84 \\
14\times6 &= 84 \\
16\times9 &= 144 \\
18\times13 &= 234 \\
22\times5 &= 110 \\
24\times4 &= 96
\end{aligned}
\]
Hence
\[
\sum f_i x_i = 84+84+144+234+20f+110+96 = 752 + 20f
\]
3. Use the given mean (\(\bar{x}=18\))
\[
\bar{x}=\frac{\sum f_i x_i}{\sum f_i}\;\Rightarrow\; 18 = \frac{752+20f}{44+f}
\]
4. Solve the linear equation for \(f\)
\[
752 + 20f = 18(44+f) = 792 + 18f
\]
\[
20f - 18f = 792 - 752 \quad\Rightarrow\quad 2f = 40
\]
\[
f = \frac{40}{2} = 20
\]
5. Check (optional)
Total frequency = 44 + 20 = 64.
\(\displaystyle \frac{752+20\times20}{64}=\frac{1152}{64}=18\), confirming the mean.
Therefore, the missing frequency is \(f = 20\).
Correct Answer: 20