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

Answer:

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

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 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:

$z=\frac{X-\mu}{\sigma}$

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:

$0.253=\frac{m-\mu}{\sigma}$

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:

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

And we want this probability:

$P(\bar X<m)$

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

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

$P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{m-\mu}{\frac{\sigma}{\sqrt{16}}})$

$P(Z<\frac{4(m-\mu)}{\sigma}$

If we use the expression obtained for m we got:

$P(Z<\frac{4(0.253\sigma +\mu-\mu)}{\sigma}$

$P(Z<\frac{1.012 \sigma}{\sigma})=P(Z<1.012)=0.844$