A population, with a normal distribution, has a mean of 80 and a standard deviation of 7. For a sample of 49, the probability that the sample mean will be larger than 82 is

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BlackTom AI · Accepted Answer

We start by identifying the sampling distribution of the sample mean. Since the population is normal with mean μ = 80 and standard deviation σ = 7, and the sample size is n = 49, the sampling distribution of the sample mean has:
- mean: μ_xbar = μ = 80
- standard error: SE = σ / sqrt(n) = 7 / sqrt(49) = 7 / 7 = 1

Now we want P( X̄ > 82 ). Standardize this to a Z value:
Z = (X̄ − μ_xbar) / SE = (82 − 80) / 1 = 2

Thus P( X̄ > 82 ) = P( Z > 2 ) = 1 − Φ(2) ≈ 1 − 0.9772 = 0.0228.

Option-by-option analysis:
- Option 0.0228: This matches the calculation above, since the tail probability beyond Z = 2 in a standard normal distribution is about 0.0228. This is consistent with the correct tail area for z = 2.
- Option 0.5228: This would imply a probability just over 0.5, which would occur if the threshold were near the mean or if Z were near 0; but here Z = 2, so the tail probability is not around 0.5. Therefore this overestimates the probability and is incorrect.
- Option 0.9772: This equals Φ(2), the probability that Z ≤ 2, i.e., the cumulative area up to 2. It represents P( X̄ ≤ 82 ) rather than P( X̄ > 82 ); since we want the > case, this option pertains to the complementary region and is not the requested probability.
- Option 0.4772: This value does not correspond to a standard normal tail or cumulative probability at Z = 2; it resembles a miscalculation or a mix-up of areas. It is not consistent with the proper Z-score interpretation here.

In summary, the correct reasoning yields a small tail probability for X̄ exceeding 82, while the other options misstate either the tail or the cumulative area related to Z = 2, leading to incorrect conclusions.

OSC 2030-009 Fall 2025 Exam 2

A population, with a normal distribution, has a mean of 80 and a standard deviation of 7. For a sample of 49, the probability that the sample mean will be larger than 82 is

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