The table shows the distribution of marks scored by students in a test. Mark (%) 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 90 - 99 Fre...
Question 1 Report
The table shows the distribution of marks scored by students in a test.
| Mark (%) |
10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 4 | 7 | 12 | 18 | 20 | 14 | 9 | 4 | 2 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw a cumulative frequency curve for the distribution.
(c) Use the curve to estimate the:
(i) median;
(ii) probability that a student selected at random obtained distinction, if the lowest mark for distinction is 75%.
Answer Details
(a) Cumulative frequency table
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 10–19 | 9.5–19.5 | 4 | 4 |
| 20–29 | 19.5–29.5 | 7 | 11 |
| 30–39 | 29.5–39.5 | 12 | 23 |
| 40–49 | 39.5–49.5 | 18 | 41 |
| 50–59 | 49.5–59.5 | 20 | 61 |
| 60–69 | 59.5–69.5 | 14 | 75 |
| 70–79 | 69.5–79.5 | 9 | 84 |
| 80–89 | 79.5–89.5 | 4 | 88 |
| 90–99 | 89.5–99.5 | 2 | 90 |
Total frequency, \(N=90\).
(b) Cumulative frequency curve
The less-than ogive is plotted using the upper class boundaries and their cumulative frequencies.
(c)(i) Median
\[\frac{N}{2}=\frac{90}{2}=45.\]
From the curve, the mark corresponding to cumulative frequency 45 is approximately \(51.5\%\).
\[\boxed{\text{Median}=51.5\%}\]
(c)(ii) Probability of obtaining distinction
At \(75\%\), the cumulative frequency read from the curve is approximately \(80\). Hence, the number scoring at least \(75\%\) is
\[90-80=10.\]
Therefore,
\[P(\text{distinction})=\frac{10}{90}=\frac{1}{9}\approx0.111.\]
Download The App On Google Playstore
Everything you need to excel in your exams
