A sample of size 50 is to be taken from a population whose mean and standard deviation are 52 and 20, respectively. The probability that the sample mean will be larger than 49 is:

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The problem asks for the probability that the sample mean from n = 50 drawn from a population with mean μ = 52 and standard deviation σ = 20 exceeds 49.
First, identify the sampling distribution: the sample mean X̄ for a sample of size n is approximately normal with mean μ_X̄ = μ = 52 and standard deviation (the standard error) SE = σ/√n = 20/√50.
Compute the standard error: √50 ≈ 7.0711, so SE ≈ 20 / 7.0711 ≈ 2.828.
Next, standardize the target value: z = (49 − μ_X̄) / SE = (49 − 52) / 2.828 = −3 / 2.828 ≈ −1.061.
The probability we want is P(X̄ > 49) = P(Z > −1.061) for Z ~ N(0,1).
Using the symmetry of the normal distribution, P(Z > −1.061) = P(Z < 1.061) ≈ Φ(1.061).
Consulting or estimating the standard normal CDF at 1.061 yields Φ(1.061) ≈ 0.855 (rounded to four decimals, about 0.8554).
A few practical notes: the normal approximation is reasonable here because n = 50 is large enough for the Central Limit Theorem to apply, and the population is assumed to have finite variance; rounding of intermediate steps can affect the last digit but not the overall conclusion significantly.
If there were multiple choices, you would compare each option to this computed value, considering typical rounding (e.g., 0.855, 0.8554, 0.8556) and choose the option that matches most closely. Since the provided information includes no list of answer choices, this calculation yields the probability value you would select.

COMM111_S2 Quiz 5: Sampling Distributions

A sample of size 50 is to be taken from a population whose mean and standard deviation are 52 and 20, respectively. The probability that the sample mean will be larger than 49 is:

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