The mean weight of pears grown in a certain orchard is 6 ounces with a standard deviation of .5 ounces. A random sample of 30 pears is selected. What is the approximate probability that the sample mean weight is less than 5.5 ounces?

Answer 1

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since 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 proportion corresponding to that Z score.

Answer 2

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