Question

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches. What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long

Answer 1

Answer:

62.93%

Step-by-step explanation:

Z score is a score used in statistics to measure by how many standard deviations that the raw score is above or below the mean. A positive z score means the raw score is above the mean and a negative z score means the raw score is below the mean. The z score is given as:

$z=\frac{x-\mu}{\sigma}$

Given that:

Mean (μ) = 30.05 inches, standard deviation (σ) = 0.3 inches

For x > 29.95

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{29.95-30.05}{0.3} =-0.33$

From the normal distribution table, P(x > 29.95) = P(z > -0.33) = 1 - P(z < 0.33) = 1 -  0.3707 = 0.6293 = 62.93%