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: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 = $21.51We 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]