A geologist manages a large museum collection of minerals, whose mass (in grams) is known to be normally distributed. She knows that 60% of the minerals have mass less than 5000 g, and needs to select a random sample of n = 16 specimens for an experiment. With what probability will their average mass be less than 5000 g?

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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

[tex]X \sim N(\mu,\sigma)[/tex]

Where [tex]\mu=?[/tex] and [tex]\sigma=?[/tex]

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

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

2) Solution to the problem

For this case we know this condition given :

[tex]P(X<5000)=0.6[/tex]

We can use the Z score given by this formula:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

And using this formula we got:

[tex]P(Z<\frac{5000-\mu}{\sigma})=0.6[/tex]

And we can find a value on the normal standard distribution that accumulates 0.6 of the are aon the left and 0.4 of the area on the right, on this case the value is Z=0.2533. And we can use the following excel code to find it :"=NORM.INV(0.6,0,1)"

So then we can do this:

[tex]0.2533=\frac{5000-\mu}{\sigma}[/tex] (1)

By the other hand when we find the z score for the sample mean we have this:

[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And we want to find this probability:

[tex]P(\bar X < 5000)[/tex]

And if we use the z score formula we got:

[tex]P(Z< \frac{5000 -\mu}{\frac{\sigma}{\sqrt{16}}})=P(Z<\sqrt{16} \frac{5000-\mu}{\sigma})[/tex] (2)

And replacing condition (1) into equation (2) we got:

[tex]P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

And we can use the following excel code to find it: "=NORM.DIST(1.0132,0,1,TRUE)"