Posted by **Sarah** on Thursday, November 25, 2010 at 6:12am.

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

- Statistics -
**PsyDAG**, Thursday, November 25, 2010 at 10:28am
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.

- Statistics -
**Sarah**, Sunday, November 28, 2010 at 3:55am
i still dont get how to do it?

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