The inside diameter of a randomly selected piston ring is a randomvariable with mean value 12 cm and standard devtiation of .04cm. a. If Xbar is the sample mean diameter form a random sample of=16 rings, where is the sampling distrbution of Xbar centered andwhat is the standard deviation of the Xbar distribution?

b. Answer the questions above for a sample of size n=64

c. find the probability that the average diameter of pistonrings from a sample size 16 is more than 11.95cm

d. For which of the above two random saples is Xbar morelikely to be within .01cm of 12cm? Explain.

Using the normal distribution and the central limit theorem, it is found that:

a) The sampling distribution is approximately normal, centered at 12 cm and with a standard deviation of 0.01 cm.

b) The sampling distribution is approximately normal, centered at 12 cm and with a standard deviation of 0.005 cm.

c) 100% probability that the average diameter of piston rings from a sample size 16 is more than 11.95 cm .

d) Due to the lower standard error, the sample of 64 is more likely to be within 0.01 cm of 12 cm.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

In this problem:

Item a:

Sample of 16, thus and

The sampling distribution is approximately normal, centered at 12 cm and with a standard deviation of 0.01 cm.

Item b:

Sample of 64, thus and

The sampling distribution is approximately normal, centered at 12 cm and with a standard deviation of 0.005 cm.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 11.95, thus:

By the Central Limit Theorem:

has a p-value of 0.

1 - 0 = 1.

100% probability that the average diameter of piston rings from a sample size 16 is more than 11.95 cm .

Item d:

Due to the lower standard error, the sample of 64 is more likely to be within 0.01 cm of 12 cm.

A similar problem is given at link

The answer is below

Step-by-step explanation:

Given that:

mean value (μ) = 12 cm and standard deviation (σ) = 0.04 cm

a) Since a random sample (n) of 16 rings is taken, therefore the mean (μx) ans standard deviation (σx) of the sample mean Xbar is given by:

The sampling distribution of Xbar is centered about 12 cm and the standard deviation of the Xbar distribution is 0.01 cm

b) Since a random sample (n) of 64 rings is taken, therefore the mean (μx) ans standard deviation (σx) of the sample mean Xbar is given by:

The sampling distribution of Xbar is centered about 12 cm and the standard deviation of the Xbar distribution is 0.005 cm

c) n = 16 and the raw score (x) = 11.95 cm

The z score equation is given by:

P(x > 11.95 cm) = P(z > -5) = 1 - P(z < -5) = 1 - 0.000001 ≅ 1 ≅ 100%

d) for n = 64, the standard deviation is 0.01 cm, therefore it is more likely to be within .01cm of 12cm

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step-by-step explanation:

hopefully im right

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09