## Related questions with answers

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

VerifiedGiven:

$\mu=\$32$

$\sigma=\$20$

$n=50$

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

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:

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

Determine the corresponding probability using table Z in appendix F.

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

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

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