Question

1) The U.S. Bureau of Labor statistics conducte periodic surveys to collect information on the labor market. According to a recent survey, hourly workers in the private sector in a particular metropolitan area earn an average of $11.75 per hour, with a standard deviation of $2.10 per hour The pattern of these sample results suggests that "hourly private sector wages follows a normal distribution for this metro area. Based on this information: a) what percentage of workers in this area earn between $11.50 and $12.50 per hour ? b) a random sample of 49 such workers is obtained from this metropolitan area. What is the probability the mean earnings for the workers in this sample is between $11.50 and $12.50 per hour? 2) Suppose the true population mean spending for fast food $$ per week by Normandale students is $14, and the population standard deviation is $16, (reflecting that the population distribution is skewed notably to the right). if a random sample of 64 students is selected from this population, what is the probability the sample mean will be within $1 of the true population mean (i.e. that the sample mean will be between $13 and $15) ? b) How large would the sample need to be, so as to give an 80% probability that the sample mean would be within $1 of the true population mean?

Answer 1

my dear student as per Chegg rules I have solved the first one question

M - $ 11-75 o = $ 2.10 P ($11.50 CXL$ 1. 2-10 0.1882 =P(.5.50 - CZE 12.30-- ) - PC -50 - 1145 CZ.C 12.50 – 1.75) = P(+0.12. CZC 0:36) = P(Z <0.36) - P(Z 2-0-12) = 0.6404 – 0.4522 = 10.1882) -16.5 M 105 -0-12° 0:36 (0) n = 849 P ( 11-50 LX212-50) pľ 1150 -11.750 (ZL 12:50 – 11675 ) 2010/√49 - P(-0.83 LZL 2.50) - PIZZ 2.50) - P(22-0-83) a 2.10149 20.9938 – 0.2033 -10-7905