A representative sample of oak trees in a forest shows the trees to have a mean circumference of 19.7 inches and a standard deviation of 1.3 inches. Assuming the
circumferences are normally distributed, estimate the proportion of oak trees in this forest that have a circumference greater than 21 inches.

Explain or show your reasoning.

15.87% portion of oak trees in this forest have a circumference greater than 21 inches

Step-by-step explanation:

Given that

Mean = μ = 19.7 inches

SD = σ = 1.3 inches

We have to find the probability of circumference greater than 21 inches

Let x = 21 inches

First of all we have to find the z-score of the value

z-score is given by the formula

Putting the values

We will use the z-score table to find the probability of x<21

P(z<1)  = 0.8413

The value gives us the area under the curve up till that value

To find the greater than probability, we have to subtract less than probability from 1.

P(z>1) = 1 - P(z<1) = 1 - 0.8413 = 0.1587 => 15.87%

Hence,

15.87% portion of oak trees in this forest have a circumference greater than 21 inches

the answer is 99.