Mathematics, 18.07.2020 18:01 Sebs1

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.

Answers: 3

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Answer from: dejhuanstafford8512

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 $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

Mean of 12 cm, thus $\mu = 12$ Standard deviation of 0.04 cm, thus $\sigma = 0.04$ .

Item a:

Sample of 16, thus n = 16 and $s = \frac{0.04}{\sqrt{16}} = 0.01$

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 n = 64 and $s = \frac{0.04}{\sqrt{64}} = 0.005$

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:

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

By the Central Limit Theorem:

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

$Z = \frac{11.95 - 12}{0.01}$

Z = -5

Z = -5 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

Answer from: jrios4983

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:

$\mu_x=\mu=12\ cm\\\sigma_x=\frac{\sigma}{\sqrt{n} }=\frac{0.04}{\sqrt{16} }=0.01$

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:

$\mu_x=\mu=12\ cm\\\sigma_x=\frac{\sigma}{\sqrt{n} }=\frac{0.04}{\sqrt{64} }=0.005\ cm$

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:

$z=\frac{x-\mu_x}{\sigma_x} =\frac{x-\mu}{\sigma/\sqrt{n} } \\z=\frac{11.95-12}{0.04/\sqrt{16} }\\ z=-5$

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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