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) =

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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:

a) P(550 < x < 1300) =

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

X = 1300

has a pvalue of 0.9772

X = 550

has a pvalue of 0.0013

0.9772 - 0.0013 = 0.9759

P(550 < x < 1300) = 0.9759

b)

Now we have

P(900 < x < 1100) =

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

X = 1100

By the Central Limit Theorem

has a pvalue of 0.9826

X = 900

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

has a pvalue of 1

X = 850

has a pvalue of 0.0008

1 - 0.0008 = 0.9992

P(850 < x < 1300) = 0.9992

the answer is c.

for continuous functions, they continue in the same pattern and you have to find that pattern kinda like slope

my guess would be b and d, but i didn’t actually get out the pen and paper. lol