Question

If the arithmetic average of the data given below be 165 rupees, find the missing area.

Monthly Wages in Rs.	100	150	200	____	300	500
Number of Labourers	30	20	15	10	4	1

Answer 1

Hint: To solve this question, we will first calculate the \[\sum{{{f}_{i}}{{x}_{i}}}\] by calculating the table of \[\sum{{{f}_{i}}}\] and \[\sum{{{x}_{i}}}\] and then by using the formula of mean given by \[\text{Arithmetic mean}=\overline{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}.\] Finally, we will assume a variable for the missing term and get its value by the mean formula.

Complete step by step answer:
We are given that the arithmetic average of the data given below is 165 rupees. The data is given as

Monthly Wages in Rs.	100	150	200	____	300	500
Number of Labourers	30	20	15	10	4	1

Let us make the frequency table of the above data where the monthly wages is \[{{x}_{i}},i=1,2,3,4,5\] and the number of laborers is \[{{f}_{i}}\] which is the frequency. Also, calculate \[{{x}_{i}}{{f}_{i}}\] from the same. Let the value to be determined be ‘x’.

\[{{x}_{i}}\]	\[{{f}_{i}}\]	\[{{x}_{i}}{{f}_{i}}\]
100	30	3000
150	20	3000
200	15	3000
x	10	10x
300	4	1200
500	1	500
	\[\sum{{{f}_{i}}=80}\]	\[\sum{{{f}_{i}}{{x}_{i}}=10700+10x}\]

Then the total number of laborers can be calculated by using \[\sum{{{f}_{i}}}\] here.
\[\sum{{{f}_{i}}=30+15+20+10+4+1=80}\]
And the summation of \[{{f}_{i}}{{x}_{i}}\] is calculated by
\[\sum{{{f}_{i}}{{x}_{i}}}=3000+3000+3000+10x+1200+500\]
\[\Rightarrow \sum{{{f}_{i}}{{x}_{i}}}=10700+10x\]
We are given the question that the arithmetic mean is given by 165. Now, the arithmetic mean is calculated by using the below stated formula.
\[\text{Arithmetic mean}=\overline{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
The case when the frequency table is given then the arithmetic mean is given by the above formula,
\[\Rightarrow \overline{x}=165\]
\[\Rightarrow 165=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
Substituting the value of \[\sum{{{f}_{i}}{{x}_{i}}}\] as 10700 + 10x and the value of \[\sum{{{f}_{i}}}\] as 80 as stated above
\[\Rightarrow 165=\dfrac{10700+10x}{80}\]
\[\Rightarrow 165\times 80=10700+10x\]
\[\Rightarrow 165\times 80-10700=10x\]
\[\Rightarrow 1320-1070=x\]
\[\Rightarrow x=250\]
Therefore, the value of x is 250.

Hence, the missing term is given by x = 250.

Note: Remember that if all the values of the given \[{{x}_{i}}\] are positive then the mean would be definitely positive and if all the values of \[{{x}_{i}}\] are negative. Then the mean would be definitely negative. Also, if some values of \[{{x}_{i}}\] are positive and some are negative, then the mean can be any value from positive or negative in this case.