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

3.What is the probability that more than 427 shoppers enter the store during a random hour?

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.

3.What is the probability that more than 427 shoppers enter the store during a random hour?

BlackTom AI · Accepted Answer

The problem gives the average number of shoppers per hour as 448 with a standard deviation of 21. We model the hourly count X as a normal random variable with these parameters: X ~ N(448, 21^2).

First, transform the value 427 into a Z-score: Z = (X - μ) / σ. Here, Z for 427 is (427 - 448) / 21 = -21 / 21 = -1.

We want P(X > 427). In terms of Z, this is P(Z > -1).

Using the symmetry of the standard normal distribution, P(Z > -1) = P(Z < 1) because the standard normal is symmetric about 0. The cumulative probability up to 1 is Φ(1) ≈ 0.8413.

Therefore, P(X > 427) ≈ 0.8413. This matches the intuition that being one standard deviation below the mean corresponds to roughly 84% probability of exceeding that threshold in a normal distribution.

Note: Since a normal approximation is being used with the given mean and standard deviation, the result relies on the assumption that shopper counts are well modeled by a normal distribution at this scale. If a discrete count model (like Poisson or binomial) were intended, the calculation would differ, but with the provided parameters, the above steps yield the standard normal probability corresponding to Z > -1.

BSTA011 Week 6 Problem solving _ sampling distribution of the sample mean

Suppose that during any hour in a large department store, the average number of shoppers is 448, with a standard deviation of 21 shoppers. 3.What is the probability that more than 427 shoppers enter the store during a random hour?

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