Question

The diameter of a brand of​ ping-pong balls is approximately normally​ distributed, with a mean of 1.32 inches and a standard deviation of 0.08 inch. A random sample of 4 ​ping-pong balls is selected. Complete parts​ (a) through​ (d). a. What is the sampling distribution of the​ mean? A. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will also be approximately normal. B. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 can not be found. C. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will not be approximately normal. D. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will be the uniform distribution.

Answer 1

Answer:

a) A. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will also be approximately normal.

b) $P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)$

And using a calculator, excel or the normal standard table we have that:

$P(Z<-1)=0.159$

c) $P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)$

And using a calculator, excel or the normal standard table we have that:

$P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

Let X the random variable that represent the diameter of a brand of ping pong of a population, and for this case we know the distribution for X is given by:

$X \sim N(1.32,0.08)$  

Where $\mu=1.32$ and $\sigma=0.08$

And we select a sample of size n =4 and we want to find the distribution for $\bar X$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

So the best option for this case would be:

A. Because the population diameter of​ Ping-Pong balls is approximately normally​ distributed, the sampling distribution of samples of 4 will also be approximately normal.

Part b

$P(\bar X<1.28)$

We can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

$P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)$

And using a calculator, excel or the normal standard table we have that:

$P(Z<-1)=0.159$

Part c

$P(1.28< \bar X<1.34)$

$P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)$

And using a calculator, excel or the normal standard table we have that:

$P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532$