Question

Calculate the arithmetic mean of the given frequency distribution table.

Class interval	0−10	10-20	20-30	30-40	40−50	50-60
Frequency	7	14	28	26	16	9

Answer 1

Hint: Here need to find the mean of the given frequency distribution table. So we will use the step deviation method to calculate it. The given class intervals are continuous, so we will calculate the mid value. Then we will use the formula of mean to find the mean number of mangoes kept in a packing box and then we will use the formula of median to find the median number of mangoes.

Formula Used:
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.

Complete step-by-step answer:
First, we will calculate the mean of the given frequency distribution table.
We know Formula to calculate class marks is given by \[{x_i} = \] (upper class limit \[ + \] lower class limit) \[ \div 2\].
Here, the class width \[h = {\rm{10}}\] and we shall assume the mean to be \[A = {\rm{35}}\] .
Therefore, we create the following table:

Number of mangoes	Mid value \[\left( {{x_i}} \right)\]	Number of boxes \[\left( {{f_i}} \right)\]	\[\begin{array}{l}{d_i} = {x_i} - A\\{\rm{ }} = {x_i} - 57\end{array}\]	\[\begin{array}{l}{u_i} = \dfrac{1}{h}\left( {{d_i}} \right)\\{\rm{ }} = \dfrac{1}{{10}}\left( {{d_i}} \right)\end{array}\]	\[{f_i}{u_i}\]
0-10	\[\dfrac{{0 + 10}}{2} = 5\]	7	\[5 - 35 = - 30\]	-3	\[7 \times - 3 = - 21\]
10-20	\[\dfrac{{10 + 20}}{2} = 15\]	14	\[15 - 35 = - 20\]	-2	\[14 \times - 2 = - 28\]
20-30	\[\dfrac{{20 + 30}}{2} = 25\]	28	\[25 - 35 = - 10\]	-1	\[28 \times - 1 = - 28\]
30-40	\[\dfrac{{30 + 40}}{2} = 35\]	26	\[35 - 35 = 0\]	0	\[26 \times 0 = 0\]
40-50	\[\dfrac{{40 + 50}}{2} = 45\]	16	\[45 - 35 = 10\]	1	\[16 \times 1 = 16\]
50-60	\[\dfrac{{50 + 60}}{2} = 55\]	9	\[55 - 35 = 20\]	2	\[9 \times 2 = 18\]
		\[\sum {{f_i} = 100} \]			\[\sum {{f_i}{u_i}} = - 43\]

Now, we will substitute all the values in the formula of mean \[\overline X = A + h\left( {\dfrac{1}{N}\sum\limits_{i = 1}^n {{f_i}{u_i}} } \right)\]. Therefore, we get
\[\overline X = 35 + 10\left( {\dfrac{1}{{100}} \times - 43} \right)\]
On multiplying the numbers inside the bracket, we get
\[ \Rightarrow \overline X = 35 + 10 \times \dfrac{{ - 43}}{{100}}\]
On further multiplication, we get
\[ \Rightarrow \overline X = 35 - 4.3\]
On subtracting the numbers, we get
\[ \Rightarrow \overline X = 30.7\]
Therefore, the required mean is equal to \[30.7\].
Note: Here, we calculated the mean from the given frequency distribution. A frequency distribution is a tabular representation of data, that represents the number of observations within a given class interval. The class intervals should always be mutually exclusive. Mean is also known as average and it is defined as the ratio of the sum of all numbers to the total number of numbers.