Question

The arithmetic mean of the following frequency distribution is 50.

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequency	16	p	30	32	14

Answer 1

1. Identify class marks

Calculate the midpoint (x_i) for each class interval using the formula:

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

For 0 – 10: `x_1 = (0 + 10)/2 = 5`

For 10 – 20: `x_2 = (10 + 20)/2 = 15`

For 20 – 30: `x_3 = (20 + 30)/2 = 25`

For 30 – 40: `x_4 = (30 + 40)/2 = 35`

For 40 – 50: `x_5 = (40 + 50)/2 = 45`

2. Set up a frequency table

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

Class Interval	Frequency (fi)	Class Mark (xi)	fixi
0 – 10	16	5	16 × 5 = 80
10 – 20	p	15	15p
20 – 30	30	25	30 × 25 = 750
30 – 40	32	35	32 × 35 = 1120
40 – 50	14	45	14 × 45 = 630
Total	Σfi = 92 + p		Σfixi = 2580 + 15p

3. Apply mean formula

Substitute the values into the grouped mean formula:

Mean = `(sumf_ix_i)/(sumf_i)`

`25 = (2580 + 15p)/(92 + p)`

4. Solve for p

Cross-multiply and isolate the variable p to find its value:

25(92 + p) = 2580 + 15p

2300 + 25p = 2580 + 15p

Subtract 15p from both sides:

2300 + 10p = 2580

Subtract 2300 from both sides:

10p = 2580 – 2300

10p = 280

p = `280/10`

p = 28

The value of the missing frequency p is 28.