Long Answer Questions 1 (Graph-based) — Question 3

Cumulative frequency distribution table :

Here, n = 120, which is even.

Median = $\dfrac{n}{2} = \dfrac{120}{2}$ = 60th term.

Steps of construction :

1. Plot daily wages on x-axis.

2. Plot cumulative frequency on y-axis.

3. Mark points (40, 8), (50, 22), (60, 34), (70, 51), (80, 71), (90, 97), (100, 110) and (110, 120).

4. Draw a free hand curve passing through the points marked, strating from the lower limit of first class and terminating at upper limit of the last class.

5. Mark A = 60 on y-axis, draw a horizontal line which meets curve at B.

6. Through point B, draw a vertical line which meets x-axis at point C. The value of point C on x-axis is the median. ∴ Median wage is ₹74.

7. Mark D = 84 on x-axis, draw a vertical line which meets curve at E.

8. Through point E, draw a horizontal line which meets y-axis at point F. The value of point F on y-axis represents no. of employees earning less than or equal to ₹ 84 per day.

From graph,

F = 81.

No. of employees earning more than ₹ 84 per day = 120 - 81 = 39.

Percentage of employees earning more than ₹ 84 = $\dfrac{\text{No. of employees earning more than ₹ 84}}{\text{Total employees}} \times 100$

$= \dfrac{39}{120} \times 100 = \dfrac{3900}{120}$ = 32.5 %.

9. Mark G = 56 on x-axis, draw a vertical line which meets curve at H.

10. Through point H, draw a horizontal line which meets y-axis at point I. The value of point I on y-axis represents no. of employees earning less than or equal to ₹ 56 per day.

From graph,

I = 30.

No. of employees earning less than or equal to ₹ 56 per day = 30.