A particular variety of watermelon weighs on average 25 pounds witha standard de

Question

A particular variety of watermelon weighs on average 25 pounds witha standard deviation of 1.55 pounds. Consider the sample meanweight of 65 watermelons of this variety. Assume the individualwatermelon weights are independent.

1. What is the standard deviation of the sample mean weight? Giveyour answer to four decimal places.

2. What is the approximate probability the sample mean weight willbe greater than 24.71? Give your answer to four decimal places. Usethe standard deviation as you entered it above to answer thisquestion.

Explanation / Answer

1. )    Variance of the sample mean is2/n so standard deviation is /n So, standard deviation of the sample mean weight =1.55/65 = 0.1923 2)   Assuming that the sample mean has a normaldistribution , then it has mean = 25 and standard deviation =0.1923     P[ Sample mean > 24.71]    = P[ Z > (24.71 -25)/0.1923 ]   = P[ Z > - 1.5081 ] = 0.9342   ... usingtables

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