It is known that, across the adult population, the average number of words read per minute is 261 and the

population standard deviation is 20. You work for a company with many employees across the country, and you would like to know what the average words per minute is for company employees.

Your colleague proposes that he thinks the average will be the same as the average in the general population: 261. You decide to survey 100 employees, and find that in this sample the average words per minute is 257.47.

a)Based on the assumption that the population mean is 261 and that the population standard deviation is 20, calculate the z-score of the sample mean in your survey. Give your answer as a decimal to 2 decimal places.

z = 

b)Determine the proportion of the standard normal distribution that lies to the left of this z-score. That is, determine the area to the left of this z-score in the standard normal distribution. You may find this standard normal table useful. Give your answer as a percentage to 2 decimal places.

Area =  %

c)Denote by x% the percentage proportion you calculated in part b). Consider the following five potential conclusions:

A: There is a chance of x% that your friend is correct, that the true population mean is 261.

B: If your colleague is correct and the true population mean is 261, then x% of all samples will produce a sample mean of 257.47 or lower.

C: If your colleague is correct and the true population mean is 261, then x% of all samples will produce a sample mean of 257.47 or higher.

D: There is a chance of x% that the true population mean is 261 or lower.

E: There is a chance of x% that the true population mean is 261 or higher.

Select the statement that can be inferred from your findings:

A
B
C
D
E

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