Asked by Sarah
The weights of the fish in a certain lake are normally distributed with a mean of 11 lb and a
standard deviation of 6. If 4 fish are randomly selected, what is the probability that the mean
weight will be between 8.6 and 14.6 lb?
A) 0.0968 B) 0.4032 C) 0.3270 D) 0.6730
standard deviation of 6. If 4 fish are randomly selected, what is the probability that the mean
weight will be between 8.6 and 14.6 lb?
A) 0.0968 B) 0.4032 C) 0.3270 D) 0.6730
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.
Answered by
Sarah
i still don't get how to do it?
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