1. A manufacturing process has a defect rate of 10%, meaning that 10% of the ite
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Question
1. A manufacturing process has a defect rate of 10%, meaning that 10% of the items produced are defective. If batches of 80 items are produced, the mean number of defects per batch is 8.0 and the standard deviation is 2.7.True or False: It would be unusual to get only five defects in a batch
2. Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and the standard deviation sigma. Then use the range rule of thumb to find the minimum usual value and the maximum usual value. Round each answer to one decimal place.
n = 60, p = 0.25
mu =
sigma =
Minimum Usual Value =
Maximum Usual Value =
Explanation / Answer
1)The range rule of thumb for identifying unusaul results.
Maximum usual value= µ+2s Minimum usual value= µ-2s When µ=8,s=2.7. Then, Maximum usual value = µ+2s =8+2(2.7) =8+5.4 =13.4 Minimum usual value = µ-2s =8-2(2.7) =8-5.4 =2.6. Since the number of defects is 5, which falls between 2.6 and 13.4. It is usual and it is not unusual. Hence it is true. Minimum usual value = µ-2s =8-2(2.7) =8-5.4 =2.6. Since the number of defects is 5, which falls between 2.6 and 13.4. It is usual and it is not unusual. Hence it is true. 2) We have binomial distribution, the mean of the binomial distribution when we have n=60,p=0.25. The mean is µ=np = 60(0.25) = 15. The standard deviation is s=vnpq =v60(0.25)(0.75) = v11.25 = 3.354102. ˜3.3541. Maximum usual value = µ+2s =15+2(3.3541) =15+6.708204 =21.7082 Minimum usual value = µ-2s =15-2(3.3541) =15-6.708204 =8.291796. Maximum usual value = µ+2s =15+2(3.3541) =15+6.708204 =21.7082 Minimum usual value = µ-2s =15-2(3.3541) =15-6.708204 =8.291796. Minimum usual value = µ-2s =15-2(3.3541) =15-6.708204 =8.291796. =15-2(3.3541) =15-6.708204 =8.291796.
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