Question

1. All Clear 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 of 0.050 inches. Let x̄ represent the mean thickness of 50 randomly selected windows. Please answer A-C

(a) Can we use a normal distribution to calculate probabilities of x̄? Explain

(b) Describe the mean (center) and standard error (spread) of the sampling distribution of x̄.

(c) Suppose a random sample of n = 50 windows yields a mean thickness of 0.392 inches. What is the likelihood of observing a sample with a mean thickness at least as thick as ours? (In other words, what is the probability that x̄ is greater than 0.392 inches?)

Answer 1

Solution :

Given that,

mean = = 0.375

standard deviation = = 0.050

n = 50

a ) = 1200

b ) The standard error

  ₌   / n = 0.050 50 = 0.0071

P ( > 0.392 )

= 1 - P ( < 0.392 )

= 1 - P ( - /  ) < ( 0.392 - 0.375 / 0.0071 )

= 1 - P ( z < 0.017 / 0.0071 )

= 1 - P ( z < 2.39 )

Using z table

= 1 - 0.9916

= 0.0084

Probability = 0.0084