Answer
Verified
373.8k+ views
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,
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.
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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE