Use graph paper to answer this question:

During a medical checkup of 60 students in a school, weights were recorded as follows:

Weight (in kg)	Number of students
28 - 30	2
30 - 32	4
32 - 34	10
34 - 36	13
36 - 38	15
38 - 40	9
40 - 42	5
42 - 44	2

Taking 2 cm = 2 kg along one axis and 2 cm = 10 students along the other axis, draw an ogive. Use your graph to find the:

(i) Median

(ii) Upper quartile

(iii) Number of students whose weight is above 37 kg

Cumulative frequency distribution table :

Here, n = 60, which is even.

(a) Median = $\dfrac{\text{n}}{2} \text{ th term} = \dfrac{60}{2}$ = 30th term.

Steps of construction :

1. Take 2 cm = 2 kg on x-axis.

2. Take 2 cm = 10 students on y-axis.

3. Since, x axis starts at 28 hence, a kink is drawn at the starting of x-axis. Plot the point (28, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58) and (44, 60).

5. Join the points by a free-hand curve.

6. Draw a line parallel to x-axis from point A (no. of students) = 30, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 36.2

Hence, median = 36.2 kg

(ii) Here, n = 60, which is even.

By formula,

Upper quartile = $\dfrac{3\text{n}}{4} = \dfrac{3 \times 60}{4} = \dfrac{180}{4}$ = 45 th term.

Draw a line parallel to x-axis from point D (no. of students) = 45, touching the graph at point E. From point E draw a line parallel to y-axis touching x-axis at point F.

From graph, F = 38.2 kg

Hence, upper quartile = 38.2 kg.

(iii) Draw a line parallel to y-axis from point G (weight) = 37 kg, touching the graph at point H. From point H draw a line parallel to x-axis touching y-axis at point I.

From graph, I = 36.

∴ 36 students have weight less than or equal to 36 kg.

No. of students whose weight is more than 36 kg = 60 - 36 = 24.

Hence, no. of students whose weight is more than 36 kg = 24.