Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
The designed life should be of 21,840 vehicle miles.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.
This means that [tex]\mu = 35000, \sigma = 7000[/tex]
Find its designed life if a .97 reliability is desired.
The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.88 = \frac{X - 35000}{7000}[/tex]
[tex]X - 35000 = -1.88*7000[/tex]
[tex]X = 21840[/tex]
The designed life should be of 21,840 vehicle miles.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
