Final Answer:
The probability that Xˉ is less than 93, given a normal distribution with μ=101 and σ=25, and a sample size of n=25, is approximately 0.4452 or 44.52%.
Step-by-step explanation:
Given parameters:
- Population mean (μ) = 101
- Population standard deviation (σ) = 25
- Sample size (n) = 25
- Desired value for sample mean (Xˉ) = 93
Calculate the standard error of the sample mean (σ):
The formula for the standard error of the sample mean is σₓ = σ / √n.
σˣ = 25 / √25 = 25 / 5 = 5.
Calculate the Z-score:
The formula for the Z-score for sample means is Z = (Xˉ - μ) / σ.
Given Xˉ= 93, μ = 101, and σₓ = 5:
Z = (93 - 101) / 5 = -8 / 5 = -1.6.
Find the probability:
Using a standard normal distribution table or calculator, find the probability corresponding to a Z-score of -1.6.
The probability P(Z < -1.6) ≈ 0.0548.
However, the question asks for the probability that Xˉ is less than 93, so we need to find P(Z < -1.6) and then subtract this probability from 0.5 (as the total area under the curve is 1 for a standard normal distribution):
P(Xˉ < 93) = 0.5 - P(Z < -1.6) = 0.5 - 0.0548 ≈ 0.4452.
Therefore, the probability that the sample mean Xˉ is less than 93, given the provided parameters, is approximately 0.4452 or 44.52%.