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Surface Areas And Volumes
EXERCISE 13.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Number of mangoes 50 - 52 53 - 55 56 - 58 59 - 61 62 - 64 Number of boxes 15 110 135 115 25 Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Step-by-Step Solution

Key Concept: For grouped data, the mean is obtained by using the class‑mark (mid‑point) of each interval. The formula is \(\bar{x}=\frac{\sum f_i x_i}{\sum f_i}\), where \(f_i\) is the frequency of the i‑th class and \(x_i\) is its class‑mark.
1. Form the class‑mark (mid‑point) for each interval
\[\begin{aligned}
50-52 &: x_1 = \frac{50+52}{2}=51 \\
53-55 &: x_2 = \frac{53+55}{2}=54 \\
56-58 &: x_3 = \frac{56+58}{2}=57 \\
59-61 &: x_4 = \frac{59+61}{2}=60 \\
62-64 &: x_5 = \frac{62+64}{2}=63
\end{aligned}\]
2. Write the frequencies
\[f_1=15,\; f_2=110,\; f_3=135,\; f_4=115,\; f_5=25\]
3. Compute \(f_i x_i\) for each class
\[\begin{aligned}
f_1x_1 &= 15\times51 = 765 \\
f_2x_2 &= 110\times54 = 5940 \\
f_3x_3 &= 135\times57 = 7695 \\
f_4x_4 &= 115\times60 = 6900 \\
f_5x_5 &= 25\times63 = 1575
\end{aligned}\]
4. Find the totals
\[\sum f_i = 15+110+135+115+25 = 400\]
\[\sum f_i x_i = 765+5940+7695+6900+1575 = 22875\]
5. Apply the mean formula
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{22875}{400}=57.1875\]
6. State the answer (to two decimal places)
\[\boxed{\bar{x}\approx 57.19\text{ mangoes per box}}\]
7. Method chosen: The "class‑mark (mid‑point) method" for grouped data was used, which is the standard technique taught in NCERT for finding the mean of a frequency distribution.

Correct Answer: 57.19 (approximately) mangoes per box
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