Question 1 of 5 A sampling distribution of the mean has a mean u = 50 and a standard error o x = 5 based on a random sample of n = 10. a. What is the population mean? 0 E b. What is the population standard deviation? 0.00 EQuestion 2 of 5 Waiters at a restaurant chain earn an average of $250 per shift (regular pay + tips} with a standard deviation of $37. For random samples of 39 shifts at the restaurant, within what range of dollar values will their sample mean earnings fall, with 80% probability? I Standard Normal Distribution Table I Range: $0.00 Eto $0.00 5 Round to the nearest cent Question 3 of 5 In a growing industry, the mean number of hours of productivity lost by employees per week due to online social media engagement is 8 hours, with a standard deviation of 1.8 hours. Note: Assume the population data is normally distributed. Standard Normal Distribution Table a. What is the probability that an employee will lose more than 10 hours of productivity due to online social media engagement? P(X > 10) = 0.0000a. What is the probability that an employee will lose more than 10 hours of productivity due to online social media engagement? P(X :> 10) : 0.0000 5 Round to four decimal places if necessary b. What is the probability that 10 employees will lose more than 9 hours of productivity due to online social media engagement? P(}_( >9): 0.0000 5 Round to four decimal places if necessary Question 4 of 5 The amount of time travellers at an airport spend with customs officers has a mean of u = 32 seconds and a standard deviation of o = 11 seconds. For a random sample of 30 travellers, what is the probability that their mean time spent with customs officers will be: Standard Normal Distribution Table a. Over 30 seconds? 0.0000 E Round to four decimal places if necessaryb. Under 35 seconds? 0.0000 Round to four decimal places if necessary c. Under 30 seconds or over 35 seconds? 0.0000 Round to four decimal places if necessaryQuestion 5 of 5 If it is appropriate to do so, use the normal approximation to the 15 -distribution to calculate the indicated probability: Standard Normal Distribution Table n = 80,19 : 0.715 P(1")>0.75)= 0 5 Enter 0 if it is not appropriate to do so