Question

A marketing consultant observed 40 consecutive shoppers to estimate the average money spent by shoppers in a supermarket store. Assume that the money spent by the population of shoppers follow a Normal distribution with a standard deviation of $21.51. What is the probability that the average money spent by a sample of 40 shoppers is within $10 of the actual population mean. 0.1389

Answer 1

P (average within 10) = 0.9984

Explanation:

Given:

Number of consecutive shoppers observed, n = 40

Standard deviation, σ = $21.51

`\bar{x} - \mu = 10`

Now,

Standard error (SE) =
`(\sigma)/(√(n))`

or

Standard error (SE) =
`(21.51)/(√(40))`

or

Standard error (SE) = 3.40

also,

z =
`\frac{\bar{x} - \mu}{SE}`

on substituting the values, we have

z =
`(10)/(3.40)`

or

z = 2.94

Now, from the Table of Area Under Standard Noral Curve

for z = 2.94 ; area = 0.9984

we have P (average within 10) = 0.9984