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

where did you get them? if you picked from the grass, is is probably poisonous.

step-by-step explanation:

the population of other specie of birds will decline or other specie of birds will find some alternate source of food to survive.

explanation:

according to the competitive exclusion principle, which is also known as gause's law, two specie living in the same niche and feeding on the same resource cannot co-exist stably.

as we have well studied the theory of natural selection which states that only those organisms survive which have better tendency to compete for food and adapt to the environmental conditions.

coming towards the question, what wi ll happen when one species outcompetes the other for the small fish , the other specie will not get enough food and will start to die or decline in population. eventually either other specie needs to find another alternative food source for itself or migrate at some other location where there is less competition otherwise it will just not exist anymore at that niche.

hope it !