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. a) Explain why you cannot determine the probability that the next Sunday customer will spend at least $40. b) Can you estimate the probability that the next 10 Sunday customers will spend an average of at least $40? Explain. c) Is it likely that the next 50 Sunday customers will spend an average of at least $40? Explain.

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07-12-2019
Mathematics

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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. a) Explain why you cannot determine the probability that the next Sunday customer will spend at least $40. b) Can you estimate the probability that the next 10 Sunday customers will spend an average of at least $40? Explain. c) Is it likely that the next 50 Sunday customers will spend an average of at least $40? Explain.

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aaregbe aaregbe · Answer 1 · 2019-12-07T16:21:37+01:00

Answer:

a) This is because the Sunday customer purchases is skewed. b) For a highly skewed distribution, it is not possible but if it is slightly skewed, it is possible. c) approximately 0.0023

Step-by-step explanation:

a) As stated in the question, the Sunday customer purchases is skewed, thus we cannot use the normal model to estimate the probability that the next Sunday customer will spend at least $40.

b) Using a sample of 10 Sunday customers won't be large enough to apply the Normal model for sampling. However, if it is just slightly skewed, the sample may be large enough for the analysis while if it is highly skewed it would be impossible to calculate the probability.

c) Using randomization principle, it can be assumed that 50 Sunday customer purchases can be considered a representative sample for all Sunday customer purchases. Using a 10% condition for the computation:

μ = μ[tex]_{ybar}[/tex] = $32

σ[tex]_{ybar}[/tex] = 20/[tex]\sqrt{50}[/tex] = 2.828

Then, the z-score is:

[tex]z_{40} = (40-32)/2.828 = 2.83[/tex]

Thus:

P (z ≥ 2.83)[tex][/tex] = normalcdf (2.83, E99, 0.1) = 0.0023

Therefore, the probability that the average of the Sunday customer purchases is at least $40 is 0.0023