Question

For the data given in the following table, find:

Class-mark	Frequency
15	10
25	12
35	14
45	16
55	8

1. mean
2. mode (using graph)
3. median (using grap)

Answer 1

Since the difference between two consecutive class-marks is 10, subtract `10/2` = 5 from each class-mark to get the lower limit and add 5 to get the upper limit.

Class-mark (x)	Class interval	Frequency (f)	fx	Cumulative frequency
15	10 - 20	10	150	10
25	20 - 30	12	300	22
35	30 - 40	14	490	36
45	40 - 50	16	720	52
55	50 - 60	8	440	60
Total	Σf = 60	Σfx = 2100

i. By formula,

Mean = `(Σfx)/(Σf) = 2100/.60`​ = 35.

Hence, mean = 35.

ii. Calculating mode:

Steps of construction:

1. Draw a histogram of the given distribution.
2. Inside the highest rectangle, which represents the maximum frequency (or modal class) draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.
3. Through the point K (the point of intersection of diagonals AC and BD), draw KM perpendicular to the horizontal axis.
4. The value of point M on the horizontal axis represents the value of mode.

∴ Mode = 42 (approx).

Hence, mode = 42 (approx).

iii. n = Σf = 60

Median = `(n/2)^(th)` term = `60/2`​ = 30th term.

Steps of construction of ogive:

1. Take 1 cm = 10 units along x-axis.
2. Take 1 cm = 10 units along y-axis.
3. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (10, 0).
4. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (20, 10), (30, 22), (40, 36), (50, 52) and (60, 60).
5. Join the points marked by a free hand curve.
6. Mark 30 on y-axis, draw a horizontal line which meets the curve at point A.
7. Through point A, on the curve, draw a vertical line which meets x-axis at point B.

From graph,

B = 36 (approx)

Hence, median = 36 (approx).