Monthly income (in ₹ thousand)	No. of workers
6-7	40
7-8	68
8-9	86
9-10	120
10-11	90
11-12	40
12-13	26

Draw an ogive for the given data and from it, determine :

(i) the median income

(ii) the number of employees whose income exceeds ₹ 11,800.

(i) Cumulative frequency distribution table :

Here, n = 470, which is even.

By formula,

Median = $\dfrac{n}{2} = \dfrac{470}{2}$ = 235th term.

Steps of construction :

1. Take 2 cm along x-axis = 1000 rupees.

2. Take 1 cm along y-axis = 50 workers.

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

4. Plot the points (7, 40), (8, 108), (9, 194), (10, 314), (11, 404), (12, 444) and (13, 470).

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

6. Draw a line parallel to x-axis from point I (no. of workers) = 235, touching the graph at point J. From point J draw a line parallel to y-axis touching x-axis at point K.

From graph, K = 9.35

Since, the money is in thousands.

∴ 9.35 × 1000 = ₹ 9350.

Hence, median = ₹ 9350.

(ii) Draw a line parallel to y-axis from point L (money) = 11.8, touching the graph at point M. From point M draw a line parallel to x-axis touching y-axis at point N.

From graph, N = 435.

∴ 435 employees earn less than or equal to ₹ 11800.

Employees earning more than ₹ 11800 = 35 (470 - 435).

Hence, 35 employees earn more than ₹ 11800.