Question

Suppose tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches. What is the probability that a randomly selected tree will be at least 10 inches in diameter?

Answer 1

`P(X>10)=P((X-\mu)/(\sigma)>(10-\mu)/(\sigma))=P(Z>(10-8.8)/(2.8))=P(z>0.429)`

And we can find this probability with this difference and using the normal standard table or excel:

`P(z>0.49)=1-P(z<0.429)=1-0.666=0.334`

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

`X \sim N(8.8,2.8)`

Where
`\mu=8.8` and
`\sigma=2.8`

We are interested on this probability :

`P(X>10)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

`P(X>10)=P((X-\mu)/(\sigma)>(10-\mu)/(\sigma))=P(Z>(10-8.8)/(2.8))=P(z>0.429)`

And we can find this probability with this difference and using the normal standard table or excel:

`P(z>0.49)=1-P(z<0.429)=1-0.666=0.334`