During the recent week Lloyd Department surveyed 120 custome

During the recent week, Lloyd Department surveyed 120 customers to estimate amount spend per shopping trip. The standard deviation for all store is known to be $90 per trip. Estimate the standard error of the mean or average expenditure per trip. Estimate the probability that the estimated mean from this sample or the average amount spend per trip will be between $85 and $90. Estimate the 99% confidence level interval estimate on average amount spent per trip.

Solution

Standard error is equals to sigma x/?n which is 90/?120= 8.2

Probability of estimated mean from this sample equlas to area under normal distribution curve a bell shape symmetry curve and calculate the z-score as X-u/sigma/?n

90-85/8.21 =0.60 and z value from table for 0.60 is 0.2257 so probability in percent is 22.57percent