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The heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard deviation of 6 meters. What is the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters? Use the portion of the standard normal table given to help answer the question.

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[tex]\mathbb P(X>37)=\mathbb P\left(\dfrac{X-25}6>\dfrac{37-25}6\right)=\mathbb P(Z>2)[/tex]

Since roughly 95% of a normal distribution lies within two standard deviations of the mean, you know that about 5% lies without, and since the distribution is symmetric, you can expect about 2.5% of the distribution to lie above two standard deviations from the mean. So the probability is about [tex]0.025[/tex]

If you want more precision, the actual value is closer to [tex]0.0228[/tex].

You can convert the variate which tracks heights of trees to standard normal variate. Then use z tables to find the probability of the desired area.

The probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is 0.0228

How to convert a normal distribution variate to standard normal distribution variate?

Suppose you  have got a normal distribution variate with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] (let the variate is denoted by X), then we have [tex]X \sim N(\mu, \sigma)[/tex]

To convert it to standard normal distribution with mean 0 and the standard deviation as 1 ( [tex]Z \sim N(0, 1)[/tex] ), we have:

[tex]Z = \dfrac{X - \mu}{\sigma}[/tex]

How to find the probability from z tables for the desired region?

Suppose the above said case when you converted X to Z. Then you can use the conversion for:

[tex]P(X < a) = P(X - \mu< a - \mu) = P(\dfrac{X - \mu}{\sigma} < \dfrac{a-\mu}{\sigma} )= P(Z < \dfrac{a - \mu}{\sigma})[/tex]

The Z value [tex]\dfrac{a-\mu}{\sigma}[/tex]  is denoting the area of the desired event. This can be mapped to probability from the z tables.

For the given case we have:

Let the height of trees in a forest be tracked by random variable X, then we have:

[tex]X \sim N(25, 6)[/tex]

The needed probability is [tex]P(X \geq 37) = 1 - P(X < 37) = 1 - P(Z < \dfrac{37 - 25}{6}) = 1 - P(Z< 2)[/tex]

From the z tables, looking for z = 2, we get the p value as 0.9772

Thus, the needed probability is

[tex]P(X \geq 37) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228[/tex]

Thus,

The probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is 0.0228

Learn more here about normal distribution here:

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