SeeClear Windows makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch. Suppose a random sample of n = 50 windows yields a sample mean of 0.392 inches. What is the probability if the windows meet the standards?
A) 0.0612
B) 0.0082
C) 0.0015
D) 0.0009

Question

Quizplus · Accepted Answer

To find the probability, we calculate the z-score using the formula: z = (X̄ - μ) / (σ/√n), where X̄ = 0.392, μ = 0.375, σ = 0.050, and n = 50. The z-score is approximately 2.40. Using a standard normal distribution table, the probability of a z-score greater than 2.40 is approximately 0.0082.

SeeClear Windows Makes Windows for Use in Homes and Commercial