Question

A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

No. of defective tubes	Less than 5	5 – 9	10 – 14	15 – 9	20 – 24	25 – 29	30 and above
No. of tubes	45	51	84	39	20	8	4

Estimate the number of defective tubes in the central batch.

Answer 1

To find the number of defective tubes in the central batch, we have to find Q₂.
Since, the given data is not continuous, we have to convert it in the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be
Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

No. of defective tubes	No. of tubes (f)	Less than cumulative frequency (c.f.)
Less than 4.5	45	45
4.5 – 9.5	51	96
9.5 – 14.5	84	180 ← Q2
14.5 – 19.5	39	219
19.5 – 24.5	20	239
24.5 – 29.5	8	247
29.5 and above	4	251
Total	251

Here, N = 251
Q₂ class = class containing `((2"N")/4)^"th"` observation

∴ `(2"N")/4=(2xx251)/4` = 125.5

Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q₂ lies in the class 9.5 – 14.5
∴ L = 9.5, h = 5, f = 84, c.f. = 96

∴ Q₂ = `"L" "h"/"f"((2"N")/4-"c.f.")`

= `9.5 + 5/84 (125.5 - 96)`

= `9.5 + 5/84 xx 29.5`

= `9.5 + (147.5)/84`

= 9.5 + 1.76
= 11.26