7.25) The New York times reported that the mean time to downloadthe home page fo

Question

7.25) The New York times reported that the mean time to downloadthe home page for the Internal Revenue Service was 0.8 second.Suppose that the download time was normally distributed, with astandard deviation of 0.2 second. If you select a random sample of30 download times.

A)WHAT IS THE PROBABILITY THAT THE SAMPLE MEAN IS LESS THAN 0.75SECOND?

b)WHAT IS THE PROBABILITY THAT THE SAMPLE MEAN IS BETWEEN 0.70AND 0.90SECOND?              

C) the probability is 80% that the sample mean is between whattwo values, symmetrically distributed around the populationmean?

D) the probability is 90% that the sample mean is less than whatvalue?

Explanation / Answer

mean, = 0.8 second.
standard deviation, = 0.2 second. n = 30 standard error, e = / n = 0.2/30 =0.04 A)THE PROBABILITY THAT THE SAMPLE MEAN IS LESS THAN 0.75SECOND
P(X<0.75) P(Z<(X-)/e) = P(Z<(0.75-0.8)/0.04) =P(Z<-1.25) =0.1056 b) THE PROBABILITY THAT THE SAMPLE MEAN IS BETWEEN 0.70 AND0.90SECOND?               P(X<0.70) P(Z<(X-)/e) = P(Z<(0.70-0.8)/0.04) =P(Z<-2.5) =0.0062 P(X<0.90) P(Z<(X-)/e) = P(Z<(0.90-0.8)/0.04) = P(Z<2.5) = 0.9938 PROBABILITY THAT THE SAMPLE MEAN IS BETWEEN 0.70 AND 0.90SECOND = 0.9938-0.0062 =0.9876 P(Z<(X-)/e) = P(Z<(0.90-0.8)/0.04) = P(Z<2.5) = 0.9938 PROBABILITY THAT THE SAMPLE MEAN IS BETWEEN 0.70 AND 0.90SECOND = 0.9938-0.0062 =0.9876 C) the probability is 80% that the sample mean is between whattwo values, symmetrically distributed around the populationmean? probability =0.8 so z value corresponding to that = 0.85 confidence interval = ± z e = 0.8 ± (0.85 x0.04) = ( 0.766 , 0.834) D) the probability is 90% that the sample mean is less thanwhat value? for probability = 0.9, z = 1.29 hence z = (x-)/e now, 1.29 = (x-0.8)/0.04 hence x = 0.8516

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