Question

A park ranger at a large national park wants to estimate the mean diameter of all the aspen trees in the park. The park ranger believes that due to environmental changes, the aspen trees are not growing as large as they were in 1975. Data collected in 1975 indicate that the distribution for aspen trees in this park was approximately normal with a mean of 8 inches and a standard deviation of 2.5 inches. Find the approximate probability that a randomly selected aspen tree in this park in 1975 would have a diameter less than 5.5 inches.

Answer 1

`P(X<5.5)=P((X-\mu)/(\sigma)<(5.5-\mu)/(\sigma))=P(Z<(5.5-8)/(2.5))=P(z<1)`

We can find the probability using the normal distribution table or excel and we got:

`P(z<1)=0.841`

Explanation:

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

`X \sim N(8,2.5)`

Where
`\mu=8` and
`\sigma=2.5`

We are interested on this probability

`P(X<5.5)`

We can solve this problem using the z score formula given by:

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

Replacing the info we got:

`P(X<5.5)=P((X-\mu)/(\sigma)<(5.5-\mu)/(\sigma))=P(Z<(5.5-8)/(2.5))=P(z<1)`

We can find the probability using the normal distribution table or excel and we got:

`P(z<1)=0.841`