Question

An analysis of particular information is given in the following table.

Age Group	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequency	2	5	6	5	2

For this data, mode = median = 25. Calculate the mean. Observing the given frequency distribution and values of the central tendency interpret your observation.

Answer 1

Given: Mode = Median = 25

We know, 3 Median = Mode + 2 Mean

⇒ 3 (25) = 25 + 2 Mean

⇒ 75 – 25 = 2 Mean

⇒ Mean = 25

Now,

Class (Age Group)	Frequency `(f_i)`	Class mark `x_i`	`f_ix_i`	`c.f.`
0 – 10	2	5	10	2
10 – 20	5	15	75	2 + 5 = 7
20 – 30	6	25	150	7 + 6 = 13
30 – 40	5	35	175	13 + 5 = 18
40 – 50	2	45	90	18 + 2 = 20
	`sumf_i` = 20		`sumf_ix_i` = 500

Mean, `barX = (sumf_ix_i)/(sumf_i) = 500/20 = 25`

20 Thus, the mean of the data is 25.

Now, maximum frequency = 6.

Class corresponds to maximum frequency = 20 – 30

∴ Modal class = 20 – 30

So, L = Lower limit of the modal class = 20

h = Class interval of the modal class = 10

f₁ = Frequency of modal class = 6

f₀ = Frequency of class preceding modal class = 5

f₂ = Frequency of class succeeding modal class = 5

Now, Mode = `"L" + ((f_1 - f_0)/(2f_1 - f_0 - f_2)) xx h`

= `20 + ((6 - 5)/(2 xx 6 - 5 - 5)) xx 10`

= `20 + 1/2 xx 10`

= 20 + 5

= 25

Thus, the mode of given data is 25.

From the table, N = 20

∴ `"N"/2 = 20/2` = 10

Cumulative frequency just greater than 10 is 13, which belongs to the class interval 20 – 30.

∴ The middle class = 20 – 30

So, L = Lower limit of median class = 20

h = Class interval of the median class = 10

f = Frequency of median class = 6

c.f. = Cumulative frequency of class preceding median class = 7

Now, Median = `"L" + (("N"/2 - c.f.)/f) xx "h"`

= `20 + ((10 - 7)/6) xx 10`

= `20 + (3/6) xx 10`

= 20 + 5

= 25

As a result, the given data's median is 25.

As a result, the data's median, mode, and mean are 25, 25, and 25, respectively.