Normal Distribution

Question

Suppose the heights of a certain type of tree is normally distributed with a mean of 10m and a standard deviation of 2m

a) What is the probability a randomly chosen tree is between 9.6m and 10.2m?

b) What is the probability a randomly chosen tree is less than 9m given it is less than 11m

Answer 1

Hello!

This is a great normal distribution question - I'd be happy to further elaborate upon the following steps in approaching this problem (or any other similar problems), just let me know!

part (a.)

1. We are given information regarding the parameters defining the distribution of the heights (in meters) of a specific type of tree. A normal distribution can be expressed as follows:

X ∼ N(μ,σ)

1. Here, the parameter "mu", μ, describes the mean of the distribution and the parameter "sigma", σ, describes the standard deviation or how spread out the observations are from one another in the dataset. For our problem, the distribution can be written as:

X ∼ N(10,2)

1. Sketching the distribution, marking the center of the curve with our mean value and incrementing towards the tails according to the value of our standard deviation is a great habit to develop as it helps orient us and facilitates a more spatial understanding when we use our z-score formula in the following steps.
2. Using the z-score formula we can compute standardized values for the two tree height values associated with the range the part (a.) inquires about, which were 9.6m and 10.2m.

z₁ = (9.6 - 10) / 2

= -0.2

z₂ = (10.2 - 10) / 2

= 0.1

1. Taking these z-scores to our z table (technologies such as the ti-83/84 calculators, Excel, Desmos graphing calculator, or R programming can also be utilized to streamline the process if you feel comfortable with understanding the material - if you wanted to learn these tools I am happy to introduce you to them, just let me know!) to retrieve cumulative/leftward probabilities associated with each z-score and then carry out a subtraction calculation on those probabilities in order to determine the value or part (a.):

P(-0.2 ≤ z ≤ 0.1) = P(z < 0.1) - P(z < -0.2)

= 0.539828 - 0.420740

part (b.)

1. This part of the problem is interested in a conditional probability which is a probability conditioned on something. The part where the problem says, "...probability a randomly chosen tree is less than 9m given it is less than 11m.", can actually be restated so that it might be a little clearer as to what we need to do here:

"...of the less than 11m probability, what is the less than 9m probability?"

1. With it rephrased as such, and relating this to a more familiar/relatable scenario such as taking a 10 point test, getting 9 points right, and posing the question "..of the 10 points, what is the proportion that you got right?", which would be 9/10 = 0.90. So, given this line of thinking, we gather a more intuitive picture of what is happening with a conditional probability. Really, it is just the shift of focus to a different subset of the data space where "...given that..." is essentially us updating the denominator according to this condition and then we put up in the numerator the value of interest contained within that updated/refocused denominator value. Here, specifically, we would have a solid calculation if we took the area under the curve associated with being less than 9m then dividing by the less than 11m area:

P(less than 9m | less than 11m) = P(less than 9m) / P(less than 11m)

1. So, for part (b), to employ our reasoning in finding the conditional probability that a randomly selected tree has a height less than 9m given that we are only considering the 11m and less subgroup we have to apply what we did from part (a.) in calculated these z-scores and then acquire their respective cumulative areas of which you will then be able to divide the one probability by the other in order to retrieve a value for the conditional probability.

I hope this helps in guiding you as you work through this problem. Those are the thoughts you might want to have in problem solving this, but if you would need more clarification on them or want additional practice do let me know, I am happy to help you!

-- Ryan