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Lockheed Martin, the defense contractor designs and build communication satellite systems to be used by the U.S. military. Because of the very high cost the company performs numerous test on every component. These test tend to extend the component assembly time. Suppose the time required to construct and test (called build time) a particular component is thought to be normally distributed, with a mean equal to 60 hours and a standard deviation equal to 9.4 hours. To keep the assembly flow moving on schedule, this component needs to have a build time between 52 and 70 hours. Find the probability that the build time will be such that assembly will stay on schedule.

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Answer:

[tex]P(52<X<70)=P(\frac{52-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{52-60}{9.4}<Z<\frac{70-60}{9.4})=P(-0.851<z<1.064)[/tex]

And we can find this probability with this difference:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)[/tex]

And if we use the normal standard distribution or excel we got:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)=0.856-0.197=0.659[/tex]

Step-by-step explanation:

Let X the random variable that represent the time required to construct and test a particular component of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(60,9.4)[/tex]  

Where [tex]\mu=60[/tex] and [tex]\sigma=9.4[/tex]

We want to find this probability:

[tex]P(52<X<70)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Using this formula we got:

[tex]P(52<X<70)=P(\frac{52-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{52-60}{9.4}<Z<\frac{70-60}{9.4})=P(-0.851<z<1.064)[/tex]

And we can find this probability with this difference:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)[/tex]

And if we use the normal standard distribution or excel we got:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)=0.856-0.197=0.659[/tex]

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