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Question

A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of $32 and a standard deviation of$20. Is it likely that the next 50 Sunday customers will spend an average of at least $40? Explain.

Solution

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Given:

μ=$32\mu=\$32

σ=$20\sigma=\$20

n=50n=50

The sampling distribution of the sample mean has mean μ\mu and standard deviation σn\dfrac{\sigma}{\sqrt{n}}.

The sampling distribution is also approximately Normal, by the central limit theorem, because the sample size of 50 is at least 30.

The z-score is the value decreased by the mean, divided by the standard deviation:

z=xμσ/n=403220/502.83z=\dfrac{x-\mu}{\sigma/\sqrt{n}}=\dfrac{40-32}{20/ \sqrt{50}}\approx 2.83

Determine the corresponding probability using table Z in appendix F.

P(x>$40)=P(Z>2.83)=1P(Z<2.83)=10.9977=0.0023=0.23%P(\overline{x}>\$40)=P(Z>2.83)=1-P(Z<2.83)=1-0.9977=0.0023=0.23\%

Since the probability is almost zero, it is unlikely that the customers will spend on average at least $40.

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