Question

You are in charge of monitoring the pressure of a compressed air tank in a processing plant. After repeating the pressure measurement a large number of times, you found that the pressure values have Gaussian distribution with a mean value of 69 psi and a standard deviation of 10 psi. If it is important that the pressure stays below 86 psi, what is the probability (in percent) that the pressure will exceed this value? (Provide your answer as a percentage using two decimal places. Do not include the % symbol in your answer)

Answer 1

Answer:

4.46% probability that the pressure will exceed this value.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Gaussian distribution with a mean value of 69 psi and a standard deviation of 10 psi.

Gaussian distribution = normal.

This means that $\mu = 69, \sigma = 10$

If it is important that the pressure stays below 86 psi, what is the probability (in percent) that the pressure will exceed this value?

As a proportion, this probability is 1 subtracted by the pvalue of Z when X = 86. So

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

$Z = \frac{86 - 69}{10}$

$Z = 1.7$

$Z = 1.7$ has a pvalue of 0.9554

1 - 0.9554 = 0.0446

0.0446*100 = 4.46%

4.46% probability that the pressure will exceed this value.