Respuesta :
Answer:
97.59% probability that a single randomly selected specimen will have a permeability index between 550 and 1300.
Step-by-step explanation:
Problems of normally distributed samples can be solved using 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.
In this problem, we have that:
[tex]\mu = 1000, \sigma = 150[/tex].
How likely is it that a single randomly selected specimen will have a permeability index between 550 and 1300?
This is the pvalue of Z when X = 1300 subtracted by the pvalue of Z when X = 550.
Z = 1300
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1300 - 1000}{150}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
Z = 550
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{550 - 1000}{150}[/tex]
[tex]Z = -3[/tex]
[tex]Z = -3[/tex] has a pvalue of 0.0013
So there is a 0.9772 - 0.0013 = 0.9759 = 97.59% probability that a single randomly selected specimen will have a permeability index between 550 and 1300.
