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Question
An electric company producing electric bulb, has packed 100 bulbs in each box. Some bulbs from 16 such boxes are tested defective. The information of number of defective bulbs in 16 boxes is given below.
No. of defective bulbs | No. of boxes |
0 – 2 | 3 |
2 – 4 | 4 |
4 – 6 | 5 |
6 – 8 | 3 |
8 – 10 | 1 |
- How many boxes contains maximum number of defective bulbs ?
- Find the mean of defective bulbs.
- The box is selected at random. What is the probability that it will contain average 2 to 4 defective bulbs?
- Find the probability of getting highest number of defective bulbs.
Sum
Solution
i. 1 box contains the maximum number of defective bulbs.
ii.
No. of defective bulbs | No. of boxes `(f_i)` |
Class mark `x_i` |
`f_ix_i` |
0 – 2 | 3 | 1 | 3 |
2 – 4 | 4 | 3 | 12 |
4 – 6 | 5 | 5 | 25 |
6 – 8 | 3 | 7 | 21 |
8 – 10 | 1 | 9 | 9 |
`sumf_i` = 16 | `sumf_ix_i` = 70 |
Mean, `barX = (sumf_ix_i)/(sumf_i)`
= `70/16`
= 4.375
As a result, the average of defective bulbs is 4.375.
iii. Total number of boxes = 16
Boxes with 2 to 4 faulty bulbs on average = 4
Thus, P(E) = `4/16 = 1/4`
iv. Total number of boxes = 16
Boxes with the greatest number of faulty bulbs = 1.
Thus, P(E) = `1/16`
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Method of Finding Mean for Grouped Data: Direct Method
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