Question

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying the 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 mean did you choose?

Answer 1

Hint: To solve this question, we will, first of all, calculate \[{{x}_{i}}\] by calculating the midpoint of class interval and write the frequency in the given terms \[{{f}_{i}}\] where i = 1, 2, 3, 4, 5. Finally, we will make the table and calculate \[\sum{{{f}_{i}}{{x}_{i}}},\] where i = 1, 2, 3, 4, 5 and calculate the mean using the formula, \[\text{mean}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}.\]

Complete step by step answer:
We are given the data as

Number of Mangoes	50 – 52	53 – 55	56 – 58	59 – 61	62 – 64
Number of boxes	15	110	135	115	25

We are given the class interval as 50 – 52, 53 – 55, 56 – 58, 59 – 61, 62 – 64. Observing this, we will calculate the value of \[{{x}_{i}}\] by using the formula,
\[{{x}_{i}}=\dfrac{\left( \text{upper limit + lower limit} \right)}{2}\]
\[\Rightarrow {{x}_{1}}=\dfrac{50+52}{2}\]
\[\Rightarrow {{x}_{1}}=\dfrac{102}{2}\]
\[\Rightarrow {{x}_{i}}=51\]
\[\Rightarrow {{x}_{2}}=\dfrac{53+55}{2}\]
\[\Rightarrow {{x}_{2}}=\dfrac{108}{2}\]
\[\Rightarrow {{x}_{2}}=54\]
Similarly, all else. After this, we will make a table of \[{{f}_{i}}{{x}_{i}}\] and calculate \[\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}.\] Finally, we will get the mean using the formula.
\[\text{Mean}=\overline{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
We have obtained the table as

Class Interval	\[{{x}_{i}}\]	\[{{f}_{i}}\]	\[{{f}_{i}}{{x}_{i}}\]
50 – 52	51	15	765
53 – 55	54	110	5940
56 – 58	57	135	7695
59 – 61	60	115	6900
62 – 64	63	25	1575
		\[\sum{{{f}_{i}}=400}\]	\[\sum{{{f}_{i}}{{x}_{i}}=22875}\]

Finally, we will use the formula of the mean as
\[\text{Mean}=\overline{x}=\dfrac{\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
where \[\sum{{{f}_{i}}{{x}_{i}}=22875}\] and \[\sum{{{f}_{i}}=400.}\]
\[\Rightarrow \text{Mean}=\overline{x}=\dfrac{22875}{400}=57.1875\]

Hence, the value of the mean of the given data is 57.1875. We have used the standard process of finding the mean.

Note: Another method to solve this question can be using the shortest mean formula. This is done by assuming a mean A at the middle term or the near to middle term in the given series. If \[{{x}_{1}},{{x}_{2}},....{{x}_{n}}\] are the observation given with the respective frequencies \[{{f}_{1}},{{f}_{2}},....{{f}_{n}}.\] Let the deviation A take at any point, we have \[{{d}_{i}}={{x}_{i}}-A\] where i = 1, 2,….n.
\[\Rightarrow {{f}_{i}}{{d}_{i}}={{f}_{i}}\left( {{x}_{i}}-A \right);i=1,2,....n\]
Then the mean \[\overline{x}\] is given by the formula, \[\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}.\]
This is the shortest mean method explained. And the answer by both methods would be approximately the same.