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 \(\mu\) and standard deviation \(\sigma\) (let the variate is denoted by X), then we have \(X \sim N(\mu, \sigma)\)
To convert it to standard normal distribution with mean 0 and the standard deviation as 1 ( \(Z \sim N(0, 1)\) ), we have:
\(Z = \dfrac{X - \mu}{\sigma}\)
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:
\(P(X < a) = P(X - \mu< a - \mu) = P(\dfrac{X - \mu}{\sigma} < \dfrac{a-\mu}{\sigma} )= P(Z < \dfrac{a - \mu}{\sigma})\)
The Z value \(\dfrac{a-\mu}{\sigma}\) 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:
\(X \sim N(25, 6)\)
The needed probability is \(P(X \geq 37) = 1 - P(X < 37) = 1 - P(Z < \dfrac{37 - 25}{6}) = 1 - P(Z< 2)\)
From the z tables, looking for z = 2, we get the p value as 0.9772
Thus, the needed probability is
\(P(X \geq 37) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228\)
Thus,
The probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is 0.0228
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