The mean expenditure at a tyre store is $85.00, with a population standard deviation of $9.00. From a random sample of 60 customers, what is the probability that the sample average sale per customer will be between $83 and $86?

Question

BlackTom AI · Accepted Answer

The problem asks for the probability that the sample mean of 60 independent tyre purchases falls between $83 and $86, given a population mean of $85 and population standard deviation of $9.

First, recognize that the sampling distribution of the sample mean is normal (or approximately so) with:
- mean μ_x̄ = μ = 85
- standard error SE = σ / √n = 9 / √60

Compute the standard error: √60 ≈ 7.746, so SE ≈ 9 / 7.746 ≈ 1.162.

Next, convert the bounds to z-scores using z = (X̄ − μ_x̄) / SE:
- For X̄ = 83: z1 = (83 − 85) / 1.162 ≈ −2 / 1.162 ≈ −1.72
- For X̄ = 86: z2 = (86 − 85) / 1.162 ≈ 1 / 1.162 ≈ 0.86

Now find the probabilities from the standard normal distribution:
- P(Z ≤ 0.86) ≈ 0.805
- P(Z ≤ −1.72) ≈ 0.043 (since P(Z ≤ −a) = 1 − P(Z ≤ a), and P(Z ≤ 1.72) ≈ 0.957, so 1 − 0.957 ≈ 0.043)

The probability that the sample mean lies between 83 and 86 is:
P(83 ≤ X̄ ≤ 86) = P(Z ≤ 0.86) − P(Z ≤ −1.72) ≈ 0.805 − 0.043 ≈ 0.762

Rounding to three decimals gives a value around 0.763. This aligns with the typical approximation using the normal model for the sampling distribution of the mean.

BRED001 Week 6 Problem solving _ sampling distribution of the sample mean

The mean expenditure at a tyre store is $85.00, with a population standard deviation of $9.00. From a random sample of 60 customers, what is the probability that the sample average sale per customer will be between $83 and $86?

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