Question

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.

Answer 1

Answer:

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