Question

Water permeability of concrete can be measured by letting water flow across the surface and determining the amount lost (in inches per hour). Suppose that the permeability index x for a randomly selected concrete specimen of a particular type is normally distributed with mean value 1000 and standard deviation 150. How likely is it that a single randomly selected specimen will have a permeability index between 550 and 1300?

Answer 1

97.59% probability that a single randomly selected specimen will have a permeability index between 550 and 1300.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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.

In this problem, we have that:

`\mu = 1000, \sigma = 150`.

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

`Z = (X - \mu)/(\sigma)`

`Z = (1300 - 1000)/(150)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

Z = 550

`Z = (X - \mu)/(\sigma)`

`Z = (550 - 1000)/(150)`

`Z = -3`

`Z = -3` 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.