Question

The arithmetic mean of the following frequency distribution is \[50\]. Find the value of \[p\].

Class	Frequency
\[0 - 20\]	\[17\]
\[20 - 40\]	\[p\]
\[40 - 60\]	\[32\]
\[60 - 80\]	\[24\]
\[80 - 100\]	\[19\]

Answer 1

Hint: In this question, we will apply the direct method of arithmetic mean of grouped data or classified data. To solve grouped data, we need to calculate the mid value of each class interval. Also, we need to calculate the total of given frequencies.

Formula used: The arithmetic mean for grouped data
\[\overline x = \dfrac{1}{n}\sum\limits_{i = 1}^n {{f_i}{m_i}} \]
Where,
\[\overline x \]\[ = \] Arithmetic mean of given data.
\[{f_i}\]\[ = \] Frequency of \[{i^{th}}\] class interval
\[{m_i}\]\[ = \] Mid value of \[{i^{th}}\] class interval
\[n\]\[ = \]\[\sum {{f_i}} \], Sum of all frequencies.

Complete step-by-step solution:
Here, in the given question, a grouped data has been given.
So, we have to use the formula to calculate the arithmetic mean for grouped data or classified data.
The given table is as follows:

Class	Frequency
\[0 - 20\]	\[17\]
\[20 - 40\]	\[p\]
\[40 - 60\]	\[32\]
\[60 - 80\]	\[24\]
\[80 - 100\]	\[19\]

To calculate the mid value, we have to add the upper limit and lower limit of the class interval and then divide it by \[2\].
Like, \[{m_1} = \dfrac{{0 + 20}}{2} = \dfrac{{20}}{2} = 10.\]
Rewrite the above table in the following form:

Class	Frequency\[({f_i})\]	Mid-value\[({m_i})\]	\[({f_i} \times {m_i})\]
\[0 - 20\]	\[17\]	\[10\]	\[170\]
\[20 - 40\]	\[p\]	\[30\]	\[30p\]
\[40 - 60\]	\[32\]	\[50\]	\[1600\]
\[60 - 80\]	\[24\]	\[70\]	\[1680\]
\[80 - 100\]	\[19\]	\[90\]	\[1710\]
Total\[ = \]	\[92 + p\]		\[5160 + 30p\]

So, from the above table we found that,
\[n = \sum {{f_i} = (92 + p} ).\]
\[\sum {{f_i} \times {m_i}} = (5160 + 30p).\]
So, the arithmetic mean of given distribution is \[\dfrac{{\sum {{f_i} \times {m_i}} }}{n}\]
\[ \Rightarrow \dfrac{{5160 + 30p}}{{92 + p}}\] .
But it is given that the arithmetic mean of the above frequency distribution is \[ = \]\[50\].
So, we can write it as,
\[ \Rightarrow \dfrac{{5160 + 30p}}{{92 + p}} = 50.\]
By using cross multiplication, we get:
\[ \Rightarrow (5160 + 30p) = 50 \times (92 + p).\]
On multiply the bracket terms and we get,
\[ \Rightarrow (5160 + 30p) = (4600 + 50p).\]
Taking the variable part into the R.H.S and constant part into the L.H.S:
\[ \Rightarrow (5160 - 4600) = (50p - 30p).\]
Let us subtract the term and we get
\[ \Rightarrow 560 = 20p.\]
Divide both sides by 20,
\[ \Rightarrow p = \dfrac{{560}}{{20}}.\]
On divide the term and we get,
\[ \Rightarrow p = 28.\]

\[\therefore \] The value of \[p\] is \[28\].

Note: This question is based on grouped (classified) data. So, to find the arithmetic mean of grouped data, we need to apply either the direct method or step deviation method to find the required arithmetic mean of the given distribution.
For, calculating the arithmetic mean of grouped data set, the following assumptions shall be made:
The class interval must be closed.
The width of each class interval shall be equal.
The value of the observation in each class interval must be uniformly distributed between upper limit and lower limit of the class interval.
Mid value of each class represents the average of the lower limit and upper limit of the given distribution.