PVC pipe is manufactured with a mean diameter of 1.01 inch and a standard deviation of 0.003 inch. The diameters are known to be normally distributed. Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch. (Express the final answer to four decimal places (e. g. 0.9876).)

0.8185

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples can be 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 random variable X, with mean and standard deviation , the sample means with size n can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

Find the probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.

This is the pvalue of Z when X = 1.012 subtracted by the pvalue of Z when X = 1.009. So

X = 1.012

By the Central limit theorem

has a pvalue of 0.9772

X = 1.009

has a pvalue of 0.1587

0.9772 - 0.1587 = 0.8185

0.8185 = 81.85% probability that a random sample of n = 9 sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

gimme brainliest > : 3