Question

Find the mean, median and mode of the given data:

Class	65 – 85	85 – 105	105 – 125	125 – 145	145 – 165	165 – 185	185 –205
Frequency	8	7	22	17	13	5	3

Answer 1

Class `(f_i)`	Frequency `(x_i)`	Class mark	`d_i = x_i - A` `h = 20`	`u_i = d_i/h`	`f_iu_i`
65 – 85	8	75	– 60	– 3	– 24
85 – 105	7	95	– 40	– 2	– 14
105 –125	22	115	– 20	– 1	– 22
125 – 145	17	135 = A	0	0	0
145 –165	13	155	20	1	13
165 – 185	5	175	40	2	10
185 –205	3	195	60	3	9
	`sumf_i = 75`				`sumf_i u_i = - 28`

We know, Mean `bar"X" = "A" + ((sumf_i u_i)/(sumf_i)) xx h`

= `135 + ((-28)/75) xx 20`

= `135 + ((-28)/15) xx 4`

= 135 – 7.47

= 127.53

Thus, the mean of the data is 127.53.

Now, the maximum frequency is 22 belonging to class interval 105 – 125.

∴ Modal class = 105 – 125

So, L = 105, f₁ = 22, f₀ = 7, f₂ = 17, h = 20.

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

= `105 + ((22 - 7)/(2(22) - 7 - 17)) xx 20`

= `105 + (15/(44 - 24)) xx 20`

= `105 + (15/20) xx 20`

= 105 + 15

= 120

Thus, the mode is 120.

Class	Frequency `(f_i)`	Cumulative frequency
65 – 85	8	8
85 – 105	7	8 + 7 = 15
105 –125	22	15 + 22 = 37
125 – 145	17	37 + 17 = 54
145 –165	13	54 + 13 = 67
165 – 185	5	67 + 5 = 72
185 –205	3	72 + 3 = 75

Here, N = 75

∴ `(N/2)^("th") "term" = 75/3 = (37.5)^("th") "term"`

Since, 37.5^th term lies in the class interval 125 – 145.

∴ Median class =125 – 145

`"So", L = 125, f = 17, N/2 = 37.5, c.f. = 37, h = 20.`

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

= `125 + ((37.5 - 37)/17) xx 20`

= `125 + (0.5/17) xx 20`

= `125 + (5/17) xx 2`

= 125 + 0.588

= 125.588

As a result, the median is 125.588.

As a result, the data's mean, median, and mode are 127.53, 125.588, and 120, respectively.