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

2. What is the probability that a random sample of 49 different shopping hours will yield a sample mean between 441 and 446 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.

2. What is the probability that a random sample of 49 different shopping hours will yield a sample mean between 441 and 446 shoppers?

BlackTom AI · Accepted Answer

We start by identifying the given information and the target probability. The population mean number of shoppers per hour is mu = 448, with standard deviation sigma = 21. The sample consists of n = 49 hours, so the sampling distribution of the sample mean X̄ is Normal with mean mu_X̄ = mu = 448 and standard error SE = sigma / sqrt(n) = 21 / sqrt(49) = 21 / 7 = 3.

We want P(441 ≤ X̄ ≤ 446). Standardize both endpoints to Z using Z = (X̄ − mu_X̄) / SE. This gives:
- For 441: Z1 = (441 − 448) / 3 = −7 / 3 ≈ −2.333...
- For 446: Z2 = (446 − 448) / 3 = −2 / 3 ≈ −0.666...

Thus, the probability is P(−2.333 ≤ Z ≤ −0.666).

Using the standard normal distribution values (or a z-table/normal CDF):
- Φ(−0.666...) ≈ 0.251,
- Φ(−2.333...) ≈ 0.0098.

Therefore, P(−2.333 ≤ Z ≤ −0.666) = Φ(−0.666...) − Φ(−2.333...) ≈ 0.251 − 0.0098 ≈ 0.2412.

In approximate terms, the probability is about 0.24 (roughly 24%).

If you instead use more precise Φ values, you’d get a result in the vicinity of 0.24 to 0.243, which matches the expected range for this calculation given the normal approximation and the chosen endpoints.

Suppose that during any hour in a large department store, the average number of shoppers is 448, with a standard deviation of 21 shoppers. 2. What is the probability that a random sample of 49 different shopping hours will yield a sample mean between 441 and 446 shoppers? Numerical

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