Question

A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:

Life time (in hrs)	Number of tubes
300 - 399	14
400 - 499	46
500 - 599	58
600 - 699	76
700 - 799	68
800 - 899	62
900 - 999	48
1000 - 1099	22
1100 - 1199	6

Answer 1

Here, the class-intervals are formed by exclusive method. If we make the series an inclusive one the mid-values remain same. So, there is no need to convert the series.
Let the assumed mean be A = 749·5 and h = 100.

Calculation of Mean

Life time (in hrs)	Frequency `(f_i)`	Mid-values `(x_i)`	`d_i = x_i - "A" = x_i - 749·5`	`mu_i = (x_i - "A")/"h"` = `(x_i - 749·5)/(100)`	`f_imu_i`
300 - 399	14	349·5	-400	-4	-56
400 - 499	46	449·9	-300	-3	-138
500 - 599	58	549·5	-200	-2	-116
600 - 699	76	649·5	-100	-1	-76
700 - 799	68	A = 749·5	0	0	0
800 - 899	62	849·5	100	1	62
900 - 999	48	949·5	200	2	96
1000 - 1099	22	1049·5	300	3	66
1100 - 1199	6	1149·5	400	4	24
	N = `sumf_i = 400`				`sumf_imu_i = -138`

Here,
N = 400, A = 749·5, h = 100 and `sumf_imu_i = -138`
∴ `bar"X" = "A" + "h"/"N" sumf_imu_i`
⇒ `bar"X" = 749·5 + 100 xx ((-138)/400)`
= `749·5 - (138)/(4)`
= 749·5 - 34·5
= 715.
Hence, the average life time of a tube is 715 hours.