The frequency distribution of marks obtained by 100 students in a certain subject is given below :Marks 0-10 10-20 20-30 30-40 No. of students 10 20 f 40
The problem asks for the standard deviation of marks obtained by 100 students, given a frequency distribution.
Frequency Distribution Analysis
The total number of students is 100. The frequencies for the given class intervals are 10 (0-10 marks), 20 (10-20 marks), f (20-30 marks), and 40 (30-40 marks). We can find the missing frequency 'f' using the total number of students:
\(10 + 20 + f + 40 = 100\)
\(70 + f = 100\)
\(f = 100 - 70 = 30\)
Now, we can construct the complete frequency distribution table including the class marks (\(x_i\)).
| Marks | Class Mark (\(x_i\)) | No. of Students (\(f_i\)) | \(f_i x_i\) | \(x_i^2\) | \(f_i x_i^2\) |
|---|---|---|---|---|---|
| 0-10 | 5 | 10 | \(10 \times 5 = 50\) | 25 | \(10 \times 25 = 250\) |
| 10-20 | 15 | 20 | \(20 \times 15 = 300\) | 225 | \(20 \times 225 = 4500\) |
| 20-30 | 25 | 30 | \(30 \times 25 = 750\) | 625 | \(30 \times 625 = 18750\) |
| 30-40 | 35 | 40 | \(40 \times 35 = 1400\) | 1225 | \(40 \times 1225 = 49000\) |
| Total | \(N = 100\) | \(\sum f_i x_i = 2500\) | \(\sum f_i x_i^2 = 72500\) |
Mean Calculation
The mean (\(\bar{x}\)) of the distribution is calculated using the formula:
\(\bar{x} = \frac{\sum f_i x_i}{N}\)
Substituting the values from the table:
\(\bar{x} = \frac{2500}{100} = 25\)
Standard Deviation Calculation
The standard deviation (\(\sigma\)) is calculated using the formula:
\(\sigma = \sqrt{\frac{\sum f_i x_i^2}{N} - (\bar{x})^2}\)
Substituting the calculated values:
\(\sigma = \sqrt{\frac{72500}{100} - (25)^2}\)
\(\sigma = \sqrt{725 - 625}\)
\(\sigma = \sqrt{100}\)
\(\sigma = 10\)
The standard deviation is 10.
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