Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 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