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Standard deviation and quality control

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A particular type of electronic component for use in PCs is mass produced and subject to quality control checks since it is known that 2% of all components produced in this way are defective. The quality of a day's output is monitored as follows. A sample of 10 components is drawn from the day's output (which may be assumed to be large) and inspected for defective components. If this sample contains 0 or 1 defectives the day's output is accepted, otherwise it is rejected. If it contains more than 2 defectives the output is rejected. If the sample contains 2 defective a second sample of 10 is taken. If this sample contains 0 defectives the output is accepted, otherwise it is rejected.

Use the binomial distribution to calculate the probability of

(i) 0
(ii) 1
(iii) 2

defectives in a sample of 10.

Hence calculate the probability that the day's output is accepted.

Suppose that it is estimated that it costs £100 to inspect a sample of size 10. What is the expected cost of a day's sampling?

A large food manufacturing company processes and packs leaf tea. One particular section packs tea in boxes. The production process is set so that the combined weight of tea and box is normally distributed with mean 500 g and standard deviation 15 g.

(i) What is the probabilty that weight of a box of tea is less than 520 g ?
(ii) What is the probabilty that weight of a box of tea is lies between 530 g and 490 g ?

Quality control regulations require that no more than 1 box in 40 weighs less than 463 g.

One day it is discovered that 5% of the boxes produced have weights less than 463 g. Assuming that the standard deviation has not changed, what has happened to the mean weight of box and tea produced that day?

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Solution Summary

This uses probability and standard deviation in a quality control context.

Solution provided by:
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
  • "Thank you"
  • "Thank you very much for your valuable time and assistance!"
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