Given a population with a mean of 75 and a standard deviation of 12, the probability that the mean of a sample of 36 observations taken at random from this population exceeds 78 is:

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BlackTom AI · Accepted Answer

The question describes sampling from a population with mean μ = 75 and standard deviation σ = 12, and asks for P(
  Xbar > 78
) where Xbar is the sample mean of n = 36 observations.

First, recognize that for a sample mean from a population with finite variance, the sampling distribution of Xbar is approximately normal with mean μ_Xbar = μ = 75 and standard deviation (σ_Xbar) = σ / sqrt(n) = 12 / sqrt(36) = 12 / 6 = 2.

We convert the probability to a standard normal: P(Xbar > 78) = P(Z > (78 - μ_Xbar) / σ_Xbar) = P(Z > (78 - 75) / 2) = P(Z > 3 / 2) = P(Z > 1.5).

Consulting the standard normal distribution (or using a Z-table or a calculator), P(Z > 1.5) ≈ 0.0668. This value represents the probability that the sample mean exceeds 78 for a sample of 36 from this population.

In summary, the calculation hinges on the sampling distribution of the mean, the standard error being 2, and the standardized threshold corresponding to Z = 1.5, which yields approximately 0.0668 as the probability. If one were evaluating multiple-choice options, this is the numerical result you would compare against the provided choices.

COMM111_S2 Quiz 5: Sampling Distributions

Given a population with a mean of 75 and a standard deviation of 12, the probability that the mean of a sample of 36 observations taken at random from this population exceeds 78 is:

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