Suppose that during any hour in a large department store, the average number of shoppers is 448, with a standard deviation of 21 shoppers.

Assuming the population is normally distributed, what is the probability that a random sample of 9 different shopping hours will yield a sample mean less than 437 shoppers?

Question

Suppose that during any hour in a large department store, the average number of shoppers is 448, with a standard deviation of 21 shoppers.

Assuming the population is normally distributed, what is the probability that a random sample of 9 different shopping hours will yield a sample mean less than 437 shoppers?

BlackTom AI · Accepted Answer

We start by identifying the given quantities: the population mean for shoppers per hour is 448, the population standard deviation is 21, and we are sampling 9 hours (n = 9). For a sample mean from a normal population, the sampling distribution of the sample mean is itself normal with mean μ_Xbar = μ and standard deviation σ_Xbar = σ / sqrt(n).

Compute the standard deviation of the sample mean: σ_Xbar = 21 / sqrt(9) = 21 / 3 = 7.

We want P(Xbar < 437). Convert to a Z-score using Z = (Xbar - μ_Xbar) / σ_Xbar = (437 - 448) / 7 = -11 / 7 ≈ -1.5714.

Next, find the probability that Z is less than -1.5714 for a standard normal distribution. consulting standard normal tables or a calculator yields P(Z < -1.5714) ≈ 0.058.

Thus, the probability that the sample mean of 9 hours is less than 437 shoppers is approximately 0.058. The calculation hinges on the sampling distribution of the mean and standardizing the target value to a Z-score, followed by using the standard normal distribution to obtain the probability. The key steps are the use of the central limit behavior for the sample mean (even with a normal population this is exact) and the standardization to a Z variable for which probabilities are tabulated.

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

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