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.

(a) How likely is it that a single randomly selected specimen will have a permeability index between 550 and 1300? (Round your answer to four decimal places.)
P(550 < x < 1300) =

(b) If the permeability index is to be determined for each specimen in a random sample of size 10, how likely is it that the sample average permeability index will be between 900 and 1100? (Round your answers to four decimal places.)
P(900 < x < 1100) =

Between 850 and 1300?
P(850 < x < 1300) =

Answer 1

a. 0.8185

b. 0.9817

Explanation:

Please see attachment

Answer 2

a) P(550 < x < 1300) = 0.9759

b) P(900 < x < 1100) = 0.9652

P(850 < x < 1300) = 0.9992

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 1000, \sigma = 150`

a) P(550 < x < 1300) =

pvalue of Z when X = 1300 subtracted by the pvalue of Z when X = 550.

X = 1300

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

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

`Z = 2`

`Z = 2` has a pvalue of 0.9772

X = 550

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

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

`Z = -3`

`Z = -3` has a pvalue of 0.0013

0.9772 - 0.0013 = 0.9759

P(550 < x < 1300) = 0.9759

b)

Now we have

`n = 10, s = (150)/(√(10)) = 47.43`

P(900 < x < 1100) =

pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900.

X = 1100

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

By the Central Limit Theorem

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

`Z = (1100 - 1000)/(47.43)`

`Z = 2.11`

`Z = 2.11` has a pvalue of 0.9826

X = 900

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

`Z = (900 - 1000)/(47.43)`

`Z = -2.11`

`Z = -2.11` has a pvalue of 0.0174

0.9826 - 0.0174 = 0.9652

P(900 < x < 1100) = 0.9652

P(850 < x < 1300) =

pvalue of Z when X = 1300 subtracted by the pvalue of Z when X = 850. So

X = 1300

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

`Z = (1300 - 1000)/(47.43)`

`Z = 6.32`

`Z = 6.32` has a pvalue of 1

X = 850

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

`Z = (850 - 1000)/(47.43)`

`Z = -3.16`

`Z = -3.16` has a pvalue of 0.0008

1 - 0.0008 = 0.9992

P(850 < x < 1300) = 0.9992