Question

A recent survey revealed that an American's Christmas spending averaged $830. Use this as the population mean American's Christmas spending. Suppose Americans' Christmas spending is normally distributed with a standard deviation of $220. A random sample of size 100 is selected from the population of American consumers. 

What is the probability that the sample mean spending is no more than $840? 

Answer 1

solution:-

given that population mean μ = 830

standard deviation σ = 220

sample n = 100

and also given that x = 840

formula

z = (x - μ)/(σ/sqrt(n))

z = (840-830)/(220/sqrt(100))

z = 0.45

=> P(Z ≤ 0.45) = 0.6736