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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.)125-175175-225225-275275-325325-375375-425425-475475-525
Number of workers22219143461


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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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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\[{{x}_{i}}\] \[{{f}_{i}}\] \[{{f}_{i}}{{x}_{i}}\] \[\left( x-\overline{x} \right)\] \[{{\left( x-\overline{x} \right)}^{2}}\]
125-1751502300-124.6515537.6225
175-225200224400-74.655572.6225
225-275250194750-24.65607.6225
275-32530014420025.35642.6225
325-3753503105075.355677.6225
375-42540041600125.3515712.6225
425-47545062700175.3530747.6225
475-5255001500225.3550782.6225
\[\sum{{{f}_{i}}=71}\] \[\sum{{{f}_{i}}{{x}_{i}}=19500}\] \[\sum{{{\left( x-\overline{x} \right)}^{2}}}=125280.98\]

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.