Question
The table lists the number of defective 60-watt lightbulbs found in samples of 100 bulbs selected...
The table lists the number of defective 60-watt lightbulbs found in samples of 100 bulbs selected over 25 days from a manufacturing process. Assume that during this time the manufacturing process was not producing an excessively large fraction of defectives.
| Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Defectives | 5 | 2 | 6 | 8 | 4 | 4 | 5 | 6 | 7 | 2 |
| Day | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| Defectives | 2 | 5 | 3 | 4 | 1 | 3 | 4 | 2 | 5 | 0 |
| Day | 21 | 22 | 23 | 24 | 25 | |||||
| Defectives | 2 | 2 | 4 | 6 | 4 | |||||
A hardware store chain purchases large shipments of lightbulbs from the manufacturer described above and specifies that each shipment must contain no more than 5% defectives. When the manufacturing process is in control, what is the probability that the hardware store's specifications are met? (Round your answer to four decimal places.)
Homework Answers
Answer #1
Answer:
The mean proportion of defective bulbs over 25 days is 3.84 per 100 bulbs. Let, p = 3.8%
The standard deviation of proportion of defective bulbs over 25 days is,
= 0.0382
We know that the sampling distribution of proportions follow a normal distribution. So the proportion of defective bulbs follow normal distribution with mean = 0.038 and standard deviation = 0.0382
Probability that hardware store sepcificantion met = Probability that proportion of defective bulbs is less than or equal to 5% = P(p < 0.05)
= P[Z < (0.05 - 0.038)/0.0382] = P(Z < 0.3141)
= 0.6217
So, the probability that hardware store sepcificantion met is 0.6217
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