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

Answer:

$P(X>10)=P(\frac{X-\mu}{\sigma}>\frac{10-\mu}{\sigma})=P(Z>\frac{10-8.8}{2.8})=P(z>0.429)$

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

$P(z>0.49)=1-P(z$

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 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:

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

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

$P(X>10)=P(\frac{X-\mu}{\sigma}>\frac{10-\mu}{\sigma})=P(Z>\frac{10-8.8}{2.8})=P(z>0.429)$

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

$P(z>0.49)=1-P(z$