Question

The thickness of glass sheets produced by a certain process are normally distributed with a mean of μ = 3.00 millimeters and a standard deviation of σ = 0.12 millimeters. What is the value of c for which there is a 99% probability that a glass sheet has a thickness within the interval ? Round your answer to four decimal places.

Answer 1

Answer:

The values of c for which there is a 99% probability that a glass sheet has a thickness within the interval is between 2.604mm and 3.3948mm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The thickness of glass sheets produced by a certain process are normally distributed with a mean of μ = 3.00 millimeters and a standard deviation of σ = 0.12 millimeters. This means that $\mu = 3, \sigma = 0.12$.

What is the value of c for which there is a 99% probability that a glass sheet has a thickness within the interval ?

The lower limit of this interval is the value of X when Z has a pvalue of 0.005

Z = -3.3 has a pvalue of 0.005. So

$Z = \frac{X - \mu}{\sigma}$

$-3.3 = \frac{X - 3}{0.12}$

$X - 3 = -0.396$

$X = 2.604$

The upper limit of this interval is the value of X when Z has a pvalue of 0.995

Z = 3.29 has a pvalue of 0.995. So:

$Z = \frac{X - \mu}{\sigma}$

$3.29 = \frac{X - 3}{0.12}$

$X - 3 = 0.3948$

$X = 3.3948$

The values of c for which there is a 99% probability that a glass sheet has a thickness within the interval is between 2.604mm and 3.3948mm.