| 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.
Explanation
(a)
| Marks(x) | Frequency (f) | fx |
| 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 |
(sum f = 16 + x + y = 30)
(implies x + y = 14 … (1))
(bar{x} = frac{sum fx}{sum f})
(52 = frac{900 + 30x + 50y}{30})
(1560 = 900 + 30x + 50y implies 30x + 50y = 660)
(3x + 5y = 66 …. (2))
Solving equation (1) and (2),
From (1), x = 14 – y
(3(14 – y) + 5y = 42 – 3y + 5y = 66)
(2y = 24 implies y = 12)
(x = 14 – y = 14 – 12 = 2)
(x, y) = (2, 12).
(b)
| Class interval | Frequency | Upper class boundary |
| 1 – 10 | 1 | 10.5 |
| 11 – 20 | 1 | 20.5 |
| 21 – 30 | 2 | 30.5 |
| 31 – 40 | 5 | 40.5 |
| 41 – 50 | 12 | 50.5 |
| 51 – 60 | 1 | 60.5 |
| 61 – 70 | 4 | 70.5 |
| 71 – 80 | 3 | 80.5 |
| 81 – 90 | 1 | 90.5 |
(c)
(d) From the histogram, Mode = 44.