Asked by michelle
A manufacturer of automobile batteries claims that the distribution of battery life is
54 months with a standard deviation of 6 months. We take a random sample of 50 batteries.
a. Find the probability that their mean life is less than 52 months.
b. Find the probability that their mean life is more than 53 months.
54 months with a standard deviation of 6 months. We take a random sample of 50 batteries.
a. Find the probability that their mean life is less than 52 months.
b. Find the probability that their mean life is more than 53 months.
Answers
Answered by
PsyDAG
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
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