Question

The following table is given the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.

Wages(in Rs.)	125-175	175-225	225-275	275-325	325-375	375-425	425-475	475-525
Number of workers	2	22	19	14	3	4	6	1

Answer 1

Hint: First of all, modify the given table by adding four more columns in the given table namely \[{{x}_{i}}\] , \[{{f}_{i}}{{x}_{i}}\] , \[\left( x-\overline{x} \right)\] , and \[{{\left( x-\overline{x} \right)}^{2}}\] . In the given table, frequency \[\left( {{f}_{i}} \right)\] is the row of “Number of workers''. Calculate \[{{x}_{i}}\] for each class interval by using the formula, \[\text{Class}\,\text{Mark=}\dfrac{\text{actual}\,\text{upper}\,\text{limit+actual lower}\,\text{limit}}{\text{2}}\] . Use the formula, \[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] and calculate the mean \[\left( \overline{x} \right)\] . Now, for every class interval calculate \[\left( x-\overline{x} \right)\] , and \[{{\left( x-\overline{x} \right)}^{2}}\] . For the calculation of standard deviation, use the formula, \[\sigma =\sqrt{\dfrac{\sum{{{\left( x-\overline{x} \right)}^{2}}}}{\sum{{{f}_{i}}}}}\] . Similarly, for the calculation of the coefficient of variation, use the formula, \[CV=\dfrac{\sigma }{\overline{x}}\times 100\] . Now, solve it further and get the value of the standard deviation and the coefficient of variation.

Complete step-by-step solution:
  According to the question, we have a table that is showing the daily wages of workers in a factory.
In the given table, frequency \[\left( {{f}_{i}} \right)\] is the row of “Number of workers”.
First of all, we need to modify the given table.
Let us add four more columns in the given table namely \[{{x}_{i}}\] , \[{{f}_{i}}{{x}_{i}}\] , \[\left( x-\overline{x} \right)\] , and \[{{\left( x-\overline{x} \right)}^{2}}\] .
Here, \[{{x}_{i}}\] is the classmark which can be calculated by using the formula, \[\text{Class}\,\text{Mark=}\dfrac{\text{actual}\,\text{upper}\,\text{limit+actual lower}\,\text{limit}}{\text{2}}\] for each class intervals ……………………………………(1)
Now, using equation (1), we get
The class mark \[\left( {{x}_{i}} \right)\] for the class interval “125-175” = \[\dfrac{125+175}{2}=150\] ,
 Similarly, the class mark \[\left( {{x}_{i}} \right)\] for the class intervals “175-225”, “225-275”, “275-325”, “325-375”, “375-425”, “425-475”, and “475-525” are 200, 250, 300, 350, 400, 450, and 500 respectively.
Now, using the data of \[{{x}_{i}}\] and \[{{f}_{i}}\] columns from the table, calculating the mean \[\overline{x}\] with the help of the formula, \[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\] .
 \[Mean\left( \overline{x} \right)=\dfrac{300+4400+4750+4200+1050+1600+2700+500}{2+22+19+14+3+4+6+1}=\dfrac{19500}{71}=274.65\] ………………………………….(2)
Now, using equation (2) and calculating \[\left( x-\overline{x} \right)\] and \[{{\left( x-\overline{x} \right)}^{2}}\] for each class intervals to modify the given table.
The modified table is given below,

Wages	\[{{x}_{i}}\]	\[{{f}_{i}}\]	\[{{f}_{i}}{{x}_{i}}\]	\[\left( x-\overline{x} \right)\]	\[{{\left( x-\overline{x} \right)}^{2}}\]
125-175	150	2	300	-124.65	15537.6225
175-225	200	22	4400	-74.65	5572.6225
225-275	250	19	4750	-24.65	607.6225
275-325	300	14	4200	25.35	642.6225
325-375	350	3	1050	75.35	5677.6225
375-425	400	4	1600	125.35	15712.6225
425-475	450	6	2700	175.35	30747.6225
475-525	500	1	500	225.35	50782.6225
		\[\sum{{{f}_{i}}=71}\]	\[\sum{{{f}_{i}}{{x}_{i}}=19500}\]		\[\sum{{{\left( x-\overline{x} \right)}^{2}}}=125280.98\]

We also know the formula for the standard deviation, \[\sigma =\sqrt{\dfrac{\sum{{{\left( x-\overline{x} \right)}^{2}}}}{\sum{{{f}_{i}}}}}\] …………………………………….(3)
Now, on using the data from the table and equation (3), we get
Standard deviation, \[\sigma =\sqrt{\dfrac{125280.98}{71}}=42.0061\approx 42\] ……………………………………(4)
For the calculation of the coefficient of variation, we have a formula, \[CV=\dfrac{\sigma }{\overline{x}}\times 100\] ……………………………………….(5)
Mow, from equation (2), equation (4), and equation (5), we get
Coefficient of variation = \[\dfrac{42}{274.65}\times 100=15.3\] ……………………………………………….(6)
From equation (4) and equation (6), we have the standard deviation and the coefficient of variation respectively.
Therefore, the standard deviation and the coefficient of variation of the given table are 42 and 15.3 respectively.

Note: We can see that it is very complex to solve this question without using the formula. Therefore, to solve this question, always keep in mind the formula for the standard deviation and the coefficient of variation that are \[\sigma =\sqrt{\dfrac{\sum{{{\left( x-\overline{x} \right)}^{2}}}}{\sum{{{f}_{i}}}}}\] and \[CV=\dfrac{\sigma }{\overline{x}}\times 100\] respectively.