Question

A geologist manages a large museum collection of minerals, whose mass (in grams) is known to be normally distributed, with some mean µ and standard deviation σ. She knows that 60% of the minerals have mass less than a certain amount m, and needs to select a random sample of n = 16 specimens for an experiment. With what probability will their average mass be less than the same amount m?

Answer 1

`P(\bar X <m)=0.844`

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 mass of a population, and for this case we know the distribution for X is given by:

`X \sim N(\mu,\sigma)`

And the z score is defined as:

`z=(X-\mu)/(\sigma)`

We know that for some amount m we have this:

`P(X>m)=0.4` (a)

`P(X<m)=0.6` (b)

We can use the z table or excel in order to find a quantile that satisfy the two conditions. The excel code would be:

"=NORM.INV(0.6,0,1)" and using condition (b) we have that Z=0.253

So we have this:

`0.253=(m-\mu)/(\sigma)`

If we solve for m from the last equation we got:

`m=0.253\sigma +\mu`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

And we want this probability:

`P(\bar X<m)`

We can apply the z score formula for this case given by:

`z=(X-\mu)/((\sigma)/(√(n)))`

`P((X-\mu)/((\sigma)/(√(n)))<(m-\mu)/((\sigma)/(√(16))))`

`P(Z<(4(m-\mu))/(\sigma)`

If we use the expression obtained for m we got:

`P(Z<(4(0.253\sigma +\mu-\mu))/(\sigma)`

`P(Z<(1.012 \sigma)/(\sigma))=P(Z<1.012)=0.844`