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babunello [35]
3 years ago
15

A mechanical engineer is creating a new composite carbon fiber material made from two different materials. The strength of the c

arbon fiber material depends on how much pressure it can withstand, and it follows a standard normal distribution. The new carbon fiber material will be slated for research and development purposes if it can withstand more than 2.15 standard deviations above the mean of pressure compared to the individual materials. Fiber material that withstand less than 2.15 standard deviations above the mean of pressure when compared to the individual materials are not of research interest. What is the probability that the carbon fiber composite material can withstand pressure less than 2.15 standard deviations above the mean when compared to the individual materials
Mathematics
1 answer:
Nataly [62]3 years ago
3 0

Answer:

0.9842 = 98.42% probability that the carbon fiber composite material can withstand pressure less than 2.15 standard deviations above the mean when compared to the individual materials

Step-by-step explanation:

When the distribution is normal, we use 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.

What is the probability that the carbon fiber composite material can withstand pressure less than 2.15 standard deviations above the mean when compared to the individual materials

This is the pvalue of Z = 2.15.

Z = 2.15 has a pvalue of 0.9842

0.9842 = 98.42% probability that the carbon fiber composite material can withstand pressure less than 2.15 standard deviations above the mean when compared to the individual materials

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