Question

The arithmetic mean of the following frequency distribution is 53. Find the value of x.

Class	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Frequency	12	15	32	x	13

Answer 1

1. Calculate class marks

Find the mid-value (x_i) for each class interval using the formula:

`x_i = ("Lower Limit" + "Upper Limit")/2`

For 0 – 20: `x_1 = (0 + 20)/2 = 10`

For 20 – 40: `x_2 = (20 + 40)/2 = 30`

For 40 – 60: `x_3 = (40 + 60)/2 = 50`

For 60 – 80: `x_4 = (60 + 80)/2 = 70`

For 80 – 100: `x_5 = (80 + 100)/2 = 90`

2. Compute frequency products

Multiply each class mark (x_i) by its corresponding frequency (f_i) to get f_ix_i:

Class Interval	Frequency (fi)	Class Mark (xi)	Product (fixi)
0 – 20	12	10	120
20 – 40	15	30	450
40 – 60	32	50	1600
60 – 80	x	70	70x
80 – 100	13	90	1170
Total	Σfi = 72 + x		Σfixi = 3340 + 70x

3. Set up equation

Use the formula for the arithmetic mean of grouped data:

Mean = `(sumf_ix_i)/(sumf_i)`

Substitute the given mean value (53) and our calculated sums into the equation:

`53 = (3340 + 70x)/(72 + x)`

4. Solve for variable x

Cross-multiply to clear the fraction and solve for x:

53(72 + x) = 3340 + 70x

3816 + 53x = 3340 + 70x

Rearrange the terms to isolate x on one side:

3816 – 3340 = 70x – 53x

476 = 17x

`x = 476/17`

x = 28