Question

Is it likely that the next 50 Sunday customers will spend an average of at least​ $40? Explain. Choose the correct answer below​ and, if​ necessary, fill in the answer box to complete your choice. A. No it is not likely. The probability that the next 50 Sunday customers will spend an average of at least​ $40 is nothing. ​(Type an integer or decimal rounded to four decimal places as​ needed.) B. Yes it is likely. The probability that the next 50 Sunday customers will spend an average of at least​ $40 is nothing. ​(Type an integer or decimal rounded to four decimal places as​ needed.) C. The probability cannot be determined.

Answer 1

Answer:

B. Yes it is likely. The probability that the next 50 Sunday customers will spend an average of at least​ $40 is P=0.0023.

Step-by-step explanation:

The question is incomplete:

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.

We have to assume certain conditions to calculate the probability that the next 50 Sunday customers will spend an average of at least​ $40.

First, this 50 purchases are representative of the total purchases made in the store (the same as saying it is a random sample).

Second, the 10% condition: the 50 sales represent less than 10% of all purchases.

Third, the sample of 50 sales is large enough to make an approximation to the normal distribution.

If all these conditions are met, we can approximate the probabiltity that the next 50 Sunday customers will spend an average of at least​ $40.

We have a sampling distribution, with mean 32 (equal to the population mean) and standard deviation:

$\sigma_M=\dfrac{20}{\sqrt{50}}=\dfrac{20}{7.07}=2.83$

Then, we calculate the z-score

$z=\dfrac{X-\mu_M}{\sigma_M}=\dfrac{40-32}{2.83}=\frac{8}{2.83} =2.83$

The probabilty can be calculated then as:

$P(X_{50}>40)=P(z>2.83)=0.0023$