The table shows the distribution of the number of hours per day spent in studying by 50 students. Number of hours per day 4 5 6 7 8 9 10 11 Number of studen...
Question 1 Report
The table shows the distribution of the number of hours per day spent in studying by 50 students.
| Number of hours per day | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Number of students | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Calculate, correct to two decimal places,
the: (a) mean; (b) standard deviation.
Answer Details
Given distribution.
| Hours per day (x) | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| Students (f) | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Total number of students: \(\sum f = 50\).
(a) Mean. Compute \(\sum fx\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| f | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
| fx | 20 | 35 | 30 | 63 | 96 | 36 | 30 | 55 |
\[\sum fx = 20+35+30+63+96+36+30+55 = 365.\]
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{365}{50} = 7.30.\]
(b) Standard deviation. Compute \(\sum fx^{2}\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| fx\(^2\) | 80 | 175 | 180 | 441 | 768 | 324 | 300 | 605 |
\[\sum fx^{2} = 80+175+180+441+768+324+300+605 = 2873.\]
\[\text{Variance} = \frac{\sum fx^{2}}{\sum f} - \bar{x}^{2} = \frac{2873}{50} - (7.3)^{2} = 57.46 - 53.29 = 4.17.\]
\[\text{S.D.} = \sqrt{4.17} = 2.04 \text{ (2 d.p.).}\]
The mean study time is 7.30 hours and the standard deviation is 2.04 hours.
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