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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

Respuesta :

Answer:

P (average within 10) = 0.9984

Step-by-step explanation:

Given:

Number of consecutive shoppers observed, n = 40

Standard deviation, σ = $21.51

[tex]\bar{x} - \mu = 10[/tex]

Now,

Standard error (SE) = [tex]\frac{\sigma}{\sqrt{n}}[/tex]

or

Standard error (SE) = [tex]\frac{21.51}{\sqrt{40}}[/tex]

or

Standard error (SE) = 3.40

also,

z = [tex]\frac{\bar{x} - \mu}{SE}[/tex]

on substituting the values, we have

z = [tex]\frac{10}{3.40}[/tex]

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

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