The number of people reached by an online advertising campaign per day follows a normal distribution with a mean of 10,000 and a standard deviation of 1,500. If a random sample of 16 daily campaigns is selected, what is the probability that their average daily reach exceeds 10,500 people? (Round your final answer to 4 decimal places.)

Question

BlackTom AI · Accepted Answer

We start by identifying the sampling distribution of the sample mean for n = 16 daily campaigns. The underlying reach per day is Normal with mean μ = 10,000 and standard deviation σ = 1,500.

1) Distribution of the sample mean: If we take the mean of n independent observations from a Normal distribution, the sample mean is also Normal with:
- Mean μ_x̄ = μ = 10,000
- Standard deviation σ_x̄ = σ / sqrt(n) = 1,500 / sqrt(16) = 1,500 / 4 = 375.

2) Standardize the target value: We want P(x̄ > 10,500). Convert to a Z-score:
Z = (x̄ − μ_x̄) / σ_x̄ = (10,500 − 10,000) / 375 = 500 / 375 ≈ 1.3333.

3) Look up or approximate the tail probability: P(Z > 1.3333) = 1 − Φ(1.3333).
Using a standard normal CDF table or a calculator, Φ(1.3333) ≈ 0.9085 to 0.9088. Thus the tail probability is approximately
P(Z > 1.3333) ≈ 1 − 0.9088 ≈ 0.0912.

4) Final value: Rounding to four decimal places gives 0.0912.

Notes on the provided answer: If someone reports 0.0122 as the probability, that would correspond to a much larger Z-score (around 2.24 or 2.30). The correct computation with the given mean, standard deviation, and sample size yields a probability closer to 0.0912, so 0.0122 appears to be inconsistent with these inputs.

Summary steps in brief: compute σ_x̄ = 1500/4 = 375, compute Z = 500/375 ≈ 1.333, find P(Z > 1.333) ≈ 0.0912, final answer 0.0912.

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