3. Another application of the sampling distribution of the mean There are 431 fu

Question

3. Another application of the sampling distribution of the mean There are 431 full-service restaurants in Alaska. The mean number of seats per restaurant is 77.9. [Data source: U.S. Census Bureau. Data based on the 2002 Economic Census.] Assume that the population standard deviation is 20. Suppose that the population mean of 77.9 is unknown. The Alaska tourism board selects a simple random sample of 50 full-service restaurants located within the state to estimate . The standard deviation of the distribution of sample means (the standard error) is Use the Distributions tool to answer the question that follows. Remember to set the parameters on the tool to the appropriate values for the distribution of sample means. Normaí Distribution Mean = 79.5 Standard Deviation-2.90 72 73 74 75 76 777879 80 81 82 83 84 There is a 0.25 probability that the sample mean is less than

Explanation / Answer

a)

b)

for 25th percentile z =-0.6745

therefore probability is 0.25 that sample mean is less than mean +Z*std deviaiton = 77.9-0.6745*2.8284

=75.99

( please revert for any clariifcation or any other subparts)

std deviation   == 20.0000 sample size       =n= 50 std error=x=/n= 2.8284

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