Marks 10 20 30 40 50 60 70 80 90 Frequency 1 1 x 5 y 1 4 3 1 The frequency distribution shows the marks distribution of a class of 30 students in an examina...
Question 1 Report
| Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |
| Frequency | 1 | 1 | x | 5 | y | 1 | 4 | 3 | 1 |
The frequency distribution shows the marks distribution of a class of 30 students in an examination.
The mean mark of the distribution is 52.
(a) Find the values of x and y.
(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.
(c) Draw a histogram for the distribution
(d) Use the histogram to estimate the mode.
Answer Details
(a) Determination of \(x\) and \(y\)
| Mark, \(m\) | Frequency, \(f\) | \(fm\) |
|---|---|---|
| 10 | 1 | 10 |
| 20 | 1 | 20 |
| 30 | \(x\) | \(30x\) |
| 40 | 5 | 200 |
| 50 | \(y\) | \(50y\) |
| 60 | 1 | 60 |
| 70 | 4 | 280 |
| 80 | 3 | 240 |
| 90 | 1 | 90 |
| Total | \(16+x+y\) | \(900+30x+50y\) |
Since there are 30 students,
\[16+x+y=30 \quad\Rightarrow\quad x+y=14\tag{1}\]
Also,
\[52=\frac{900+30x+50y}{30}\]
\[1560=900+30x+50y\]
\[3x+5y=66.\tag{2}\]
From (1), \(x=14-y\). Hence,
\[3(14-y)+5y=66\]
\[42+2y=66\Rightarrow y=12.\]
Therefore, \(x=14-12=2\).
\(\boxed{x=2,\ y=12}\)
(b) Grouped frequency distribution
| Class interval | Class boundaries | Frequency |
|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 1 |
| 11 - 20 | 10.5 - 20.5 | 1 |
| 21 - 30 | 20.5 - 30.5 | 2 |
| 31 - 40 | 30.5 - 40.5 | 5 |
| 41 - 50 | 40.5 - 50.5 | 12 |
| 51 - 60 | 50.5 - 60.5 | 1 |
| 61 - 70 | 60.5 - 70.5 | 4 |
| 71 - 80 | 70.5 - 80.5 | 3 |
| 81 - 90 | 80.5 - 90.5 | 1 |
(c) Histogram
(d) Mode
The modal class is \(41-50\). Reading the peak position from the histogram by the usual intersecting-lines method gives a mode of approximately
\[\boxed{44\text{ marks}}\]
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