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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.

Respuesta :

Answer:

0.998 is the probability that the average money spent by  a sample of 40 shoppers is within $10 of the actual population mean.

Step-by-step explanation:

We are given the following information in the question:

Standard Deviation, σ = $21.51

We are given that the distribution of average money spend is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

We have to find:

P( average money spent is within $10 of the actual population mean.)

[tex]= P( z \leq \displaystyle\frac{10\times \sqrt{40}}{21.51}}) = P(z \leq 2.94)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(z \leq 2.94) = 0.998[/tex]

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