Plot a cumulative distribution curve for the following data:
| Weight in kg | $40 - 45$ | $45 - 50$ | $50 - 55$ | $55 - 60$ | $60 - 65$ | $65 - 70$ | $70 - 75$ | $75 - 80$ |
|---|---|---|---|---|---|---|---|---|
| Pupils | $7$ | $15$ | $28$ | $20$ | $12$ | $10$ | $6$ | $2$ |
Estimate the median and the quartiles.
If $50 \; kg$ is standard weight, find the number of pupils that are overweight.
| Weight in kg (Class Interval) | Pupils (Frequency) | Cumulative Frequency |
|---|---|---|
| $40 - 45$ | $7$ | $7$ |
| $45 - 50$ | $15$ | $22$ |
| $50 - 55$ | $28$ | $50$ |
| $55 - 60$ | $20$ | $70$ |
| $60 - 65$ | $12$ | $82$ |
| $65 - 70$ | $10$ | $92$ |
| $70 - 75$ | $6$ | $98$ |
| $75 - 80$ | $2$ | $100$ |
Number of students $= N = \Sigma f_i = 100$
Taking weight (in kg) (class intervals) along X-axis and cumulative frequency along Y-axis, draw an ogive.
Median $= \left(\dfrac{N}{2}\right)^{th} \text{value} = \left(\dfrac{100}{2}\right)^{th} \text{value} = 50^{th} \; \text{value} = 55$
$\therefore \;$ Median number of pupils $= 55$
Lower quartile $= Q_1 = \left(\dfrac{N}{4}\right)^{th} \text{value} = \left(\dfrac{100}{4}\right)^{th} \text{value} = 25^{th} \; \text{value} = 51$
$\therefore \;$ Lower quartile $= 51$
Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th} \text{value} = \left(\dfrac{3 \times 100}{4}\right)^{th} \text{value} = 75^{th} \; \text{value} = 62$
$\therefore \;$ Upper quartile $= 62$
Given: Standard weight $= 50 \; kg$
From the ogive, number of pupils who weigh $50 \; kg = 22$
$\therefore \;$ Number of pupils who are overweight $= 100 - 22 = 78$