A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator. Is it appropriate to use the normal distribution to find probabilities for x?

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". The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]

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:

[tex]X \sim N(61,14)[/tex]

Where [tex]\mu=61[/tex] and [tex]\sigma=14[/tex]

Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is also normal and given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

[tex] \mu_{\bar X} = 61[/tex]

[tex]\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715[/tex]

So then is appropiate use the normal distribution to find the probabilities for [tex]\bar X[/tex]