Question

In a production process, the diameter measures of manufactured o-ring gaskets are known to be normally distributed with a mean diameter of 80 mm and a standard deviation of 3 mm. Any o-ring measuring 75 mm or less in diameter is defective and cannot be used. Using Excel, determine the percent or proportion of defective o-rings that will be produced.

Answer 1

Answer:

DIRECT WAY EXCEL

"=NORM.DIST(75,80,3,TRUE)"

And we got: $P(X\leq 75)= 0.04779$

OTHER WAY

$P(X\leq 75)=P(\frac{X-\mu}{\sigma}\leq \frac{75-\mu}{\sigma})=P(Z\leq \frac{75-80}{3})=P(Z$

And we can find this probability using the normal standard table or excel:

$P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the diameter of a population, and for this case we know the distribution for X is given by:

$X \sim N(80,3)$  

Where $\mu=80$ and $\sigma=3$

And we know that if the diameter is 75 or less the ring would be considered defective , so then in order to find the proportion of defective we need to find the following probability:

$P(X\leq 75)$

One way to do this in excel is with the following formula:

"=NORM.DIST(75,80,3,TRUE)"

And we got: $P(X\leq 75)= 0.04779$

And the other way is use the z score formula given by:

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

If we apply this formula to our probability we got this:

$P(X\leq 75)=P(\frac{X-\mu}{\sigma}\leq \frac{75-\mu}{\sigma})=P(Z\leq \frac{75-80}{3})=P(Z$

And we can find this probability using the normal standard table or excel:

$P(Z$