The monthly income of a group of 320 employees in a company is given below :

Monthly income	No. of employees
6 - 7	20
7 - 8	45
8 - 9	65
9 - 10	95
10 - 11	60
11 - 12	30
12 - 13	5

Draw an ogive of the given distribution on a graph sheet taking 2 cm = ₹ 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine :

(i) the median wage.

(ii) the number of employees whose income is below ₹ 8500.

(i) Cumulative frequency distribution table :

Here, n = 320, which is even.

By formula,

Median = $\dfrac{n}{2}$ th term

= $\dfrac{320}{2}$ = 160th term.

Steps of construction :

1. Since, the scale on x-axis starts at 6, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 6.

2. Take 2 cm along x-axis = 1 thousand rupees.

3. Take 1 cm along y-axis = 40 employees.

4. Plot the point (6, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (7, 20), (8, 65), (9, 130), (10, 225), (11, 285), (12, 315) and (13, 320).

6. Join the points by a free hand curve.

7. Draw a line parallel to x-axis from point M (no. of employees) = 160, touching the graph at point N. From point N draw a line parallel to y-axis touching x-axis at point O.

From graph, O = 9.2 (thousands)

Hence, median = ₹ 9200.

(ii) Draw a line parallel to y-axis from point P (income) = ₹ 8.5 (thousands), touching the graph at point Q. From point Q draw a line parallel to x-axis touching y-axis at point R.

From graph, R = 95.

Hence, 95 employees have income less than ₹ 8500.