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

Answer:

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

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

P(850 < x < 1300) = 0.9992

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

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 $s = \frac{\sigma}{\sqrt{n}}$.

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1300 - 1000}{150}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 550

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{550 - 1000}{150}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0013

0.9772 - 0.0013 = 0.9759

P(550 < x < 1300) = 0.9759

b)

Now we have

$n = 10, s = \frac{150}{\sqrt{10}} = 47.43$

P(900 < x < 1100) =

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

X = 1100

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1100 - 1000}{47.43}$

$Z = 2.11$

$Z = 2.11$ has a pvalue of 0.9826

X = 900

$Z = \frac{X - \mu}{s}$

$Z = \frac{900 - 1000}{47.43}$

$Z = -2.11$

$Z = -2.11$ 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

$Z = \frac{X - \mu}{s}$

$Z = \frac{1300 - 1000}{47.43}$

$Z = 6.32$

$Z = 6.32$ has a pvalue of 1

X = 850

$Z = \frac{X - \mu}{s}$

$Z = \frac{850 - 1000}{47.43}$

$Z = -3.16$

$Z = -3.16$ has a pvalue of 0.0008

1 - 0.0008 = 0.9992

P(850 < x < 1300) = 0.9992

Answer 2

Answer:

a. 0.8185

b. 0.9817

Step-by-step explanation:

Please see attachment