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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 45 hours and a standard deviation equal to 6.75 hours. To keep the assembly flow moving on schedule, this component needs to have a build time between 37.5 and 54 hours. Find the propability that the bulid time will be such that assembly will stay on schedule.

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sqdancefan

Answer:

  p(on schedule) ≈ 0.7755

Step-by-step explanation:

A suitable probability calculator can show you this answer.

_____

The z-values corresponding to the build time limits are ...

  z = (37.5 -45)/6.75 ≈ -1.1111

  z = (54 -45)/6.75 ≈ 1.3333

You can look these up in a suitable CDF table and find the difference between the values you find. That will be about ...

  0.90879 -0.13326 = 0.77553

The probability assembly will stay on schedule is about 78%.

Ver imagen sqdancefan

Using the normal distribution, it is found that there is a 0.7747 = 77.47% probability that the build time will be such that assembly will stay on schedule.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It 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 , which is the percentile of measure X.

In this problem:

  • Mean of 45 hours, thus [tex]\mu = 45[/tex].
  • Standard deviation of 6.75 hours, thus [tex]\sigma = 6.75[/tex].
  • The probability of the time being between 37.5 and 54 hours is the p-value of Z when X = 54 subtracted by the p-value of Z when X = 37.5, then:

X = 54

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{54 - 45}{6.75}[/tex]

[tex]Z = 1.33[/tex]

[tex]Z = 1.33[/tex] has a p-value of 0.9082.

X = 37.5

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{37.5 - 45}{6.75}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a p-value of 0.1335.

0.9082 - 0.1335 = 0.7747.

0.7747 = 77.47% probability that the build time will be such that assembly will stay on schedule.

A similar problem is given at https://brainly.com/question/24663213

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