Question

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution.

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

Find the mean and median number of mangoes kept in a packing box.

Answer 1

Here I need to find the mean and median number of the mangoes kept. We will first find the class mark because the given class intervals are not continuous. Then we will use the formula of mean to find the mean number of mangoes kept in a packing box. We will then use the formula of median to find the median number of mangoes.

Formula Used:
We will use the following formulas:

1. The formula to calculate the mean is given by, $\overline X = A + h\left( {\dfrac{1}{N}\sum\limits_{i = 1}^n {{f_i}{u_i}} } \right)$ where, A is the assumed mean, $h$ is the uniform class width, $N$ is the total frequency, ${f_i}'s$ are the frequencies corresponding to ${i^{th}}$ class interval, and ${u_i}$ is the ratio of ${d_i} = {x_i} - A$ and $h$ ; ${x_i}$ is mid-value of ${i^{th}}$ class interval, which is known as class mark.
2. Formula to calculate class mark is given by ${x_i} = \dfrac{{{\text{upper class limit}} + {\text{lower class limit}}}}{2}$ .
3. The formula of median is $Median = l + h\left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right)$, where $l$ is the lower limit of the median class and $f$ is the frequency corresponding to median class.

Complete step by step solution:
Here, we know that the class width $h = 3$ and we shall assume the mean to be $A = 57$ .
We will use the formula of class marks to find the respective class mark.
Therefore, we will create the following table:

Number of mangoes	Mid value $\left( {{x_i}} \right)$	Number of boxes $\left( {{f_i}} \right)$	$\begin{array}{l}{d_i} = {x_i} - A\\\\{\text{ }} = {x_i} - 57\end{array}$	$\begin{array}{l}{u_i} = \dfrac{1}{h}\left( {{d_i}} \right)\\\\{\text{ }} = \dfrac{1}{3}\left( {{d_i}} \right)\end{array}$	${f_i}{u_i}$
50-52	$\dfrac{{50 + 52}}{2} = 51$	15	$51 - 57 = - 6$	$-2$	$15 \times - 2 = - 30$
53-55	$\dfrac{{53 + 55}}{2} = 54$	110	$54 - 57 = - 3$	$-1$	$54 \times - 1 = - 54$
56-58	$\dfrac{{56 + 58}}{2} = 57$	135	$57 - 57 = 0$	0	$57 \times 0 = 0$
59-61	$\dfrac{{59 + 61}}{2} = 60$	115	$60 - 57 = 3$	1	$60 \times 1 = 60$
62-64	$\dfrac{{62 + 64}}{2} = 63$	25	$63 - 57 = 6$	2	$63 \times 2 = 126$
	$\sum {{f_i} = 400}$			$\sum {{f_i}{u_i}} = 25$

Now, we will substitute all the calculated values in the formula of mean, $\overline X = A + h\left( {\dfrac{1}{N}\sum\limits_{i = 1}^n {{f_i}{u_i}} } \right)$ to find mean of the given distribution. Therefore, we get
$\overline X = 57 + 3\left( {\dfrac{1}{{400}} \times 25} \right)$
On multiplying the terms inside the bracket, we get
$\Rightarrow \overline X = 57 + 3 \times \dfrac{1}{{16}}$
On further multiplication, we get
$\Rightarrow \overline X = 57 + 0.187$
On adding the terms, we get
$\Rightarrow \overline X = 57.187$
Therefore, the required mean is equal to $57.187$
Now, we will calculate the median for the given table.
We will first calculate the cumulative frequency and the median class and then we will apply the formula of median.

Number of mangoes	Mid value $\left( {{x_i}} \right)$	Number of boxes $\left( {{f_i}} \right)$	Cumulative frequency $\left( {cf} \right)$
50-52	$\dfrac{{50 + 52}}{2} = 51$	15	15
53-55	$\dfrac{{53 + 55}}{2} = 54$	110	$15 + 110 = 125$
56-58	$\dfrac{{56 + 58}}{2} = 57$	$135 = f$	$125 + 135 = 260$
59-61	$\dfrac{{59 + 61}}{2} = 60$	115	$260 + 115 = 375$
62-64	$\dfrac{{62 + 64}}{2} = 63$	25	$375 + 25 = 400$
	$\sum {{f_i} = 400}$

Median class is 56-58 as $\dfrac{N}{2} = \dfrac{{400}}{2} = 200$. So $cf$ is equal to 125.
From the table we get the following values:
$\begin{array}{l}l = 56\\\N = 400\\\h = 3\\\f = 135\\\cf = 125\end{array}$
Substituting all the values in the formula of median $Median = l + h\left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right)$ , we get
${\text{Median}} = 56 + 3\left( {\dfrac{{\dfrac{{400}}{2} - 125}}{{135}}} \right)$
On simplifying the terms, we get
$\begin{array}{l} \Rightarrow {\text{Median}} = 56 + 3\left( {\dfrac{{200 - 125}}{{135}}} \right)\\\ \Rightarrow {\text{Median}} = 56 + 3 \times \dfrac{{75}}{{135}}\end{array}$
On multiplying the terms, we get
$\Rightarrow {\text{Median}} = 56 + 1.666 = 57.6667$
Therefore, the required median is equal to $57.6667$

Therefore, the mean and the median number of mangoes kept in a packing box are equal to $57.187$ and $57.6667$ respectively.

Note:
Since, we have calculated the mean and median of the data. There are some differences between mean and median.

1. Mean represents the center of gravity of a data but median represents the center of gravity of the midpoint of the data.
2. Mean takes into account every value of data set but median does not take into account every value of data set.