Question

A random sample of n = 4 scores is obtained from a normal population with µ = 30 and LaTeX: \sigmaσ= 8. What is the probability that the sample mean will be smaller than M = 22? (four decimals)

Answer 1

Answer:

$P(\bar X < 22)$

And we can use the z score given by:

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

And replacing we got:

$z = \frac{22-30}{\frac{8}{\sqrt{2}}}= -2$

So we can rewrite the probability like this:

$P(\bar X < 22)= P(Z<-2)$

And using the normal standard distribution or excel we got:

$P(\bar X < 22)= P(Z<-2)=0.0228$

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

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

$X \sim N(30,8)$  

Where $\mu=30$ and $\sigma=8$

We select a sample size of n =4 and since the distribution for X is normal, then the distribution for the sample mean $\bar X$ is given by:

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

And we want this probability:

$P(\bar X < 22)$

And we can use the z score given by:

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

And replacing we got:

$z = \frac{22-30}{\frac{8}{\sqrt{2}}}= -2$

So we can rewrite the probability like this:

$P(\bar X < 22)= P(Z<-2)$

And using the normal standard distribution or excel we got:

$P(\bar X < 22)= P(Z<-2)=0.0228$