A particular variety of watermelon weighs on average 21.9 pounds with a standard
ID: 3131500 • Letter: A
Question
A particular variety of watermelon weighs on average 21.9 pounds with a standard deviation of 1.49 pounds. Consider the sample mean weight of 94 watermelons of this variety. Assume the individual watermelon weights are independent. What is the expected value of the sample mean weight? Give an exact answer. What is the standard deviation of the sample mean weight? Give your answer to four decimal places. What is the approximate probability the sample mean weight will be less than 22.08? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question. What is the value c such that the approximate probability the sample mean will be less than c is 0.97? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.Explanation / Answer
a)
The expected sample mean weight is the population mean. Hence,
E(X) = 21.9 [ANSWER]
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b)
By central limit theorem,
sigma(X) = sigma/sqrt(n) = 1.49/sqrt(94) = 0.153681766 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 22.08
u = mean = 21.9
n = sample size = 94
s = standard deviation = 1.49
Thus,
z = (x - u) * sqrt(n) / s = 1.171251509
Thus, using a table/technology, the left tailed area of this is
P(z < 1.171251509 ) = 0.879251152 [ANSWER]
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d)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.97
Then, using table or technology,
z = 1.880793608
As x = u + z * s / sqrt(n)
where
u = mean = 21.9
z = the critical z score = 1.880793608
s = standard deviation = 1.49
n = sample size = 94
Then
x = critical value = 22.18904368 [ANSWER]
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