The heights of pecan trees are normally distributed with a mean of 10 feet and a standard deviation of 2 feet. Show all work.

(a) What is the probability that a randomly selected pecan tree is between 9 and 12 feet tall? (round the answer to 4 decimal places)
(b) Find the 80th percentile of the pecan tree height distribution. (round the answer to 2 decimal places)

Solution

Let $X$ denote the heights of pecan trees. Given that $\mu = 10$ and $\sigma = 2$.

$X\sim N(10, 2^2)$.

a. The probability that a randomly selected pecan tree is between 9 and 12 feet tall is

$$ \begin{aligned} P(9 \leq X\leq 12) &=P(9 \leq X\leq 12)\\ &=P\bigg(\frac{9-10}{2}\leq \frac{X-\mu}{\sigma} \leq \frac{12-10}{2}\bigg)\\ &=P\bigg(-0.5 \leq Z \leq 1\bigg)\\ &= P(Z < 1) -P(Z < -0.5)\\ &=0.8413-0.3085\\ &= 0.5328 \end{aligned} $$

normal area
normal area

b. Let the the 80th percentile of the pecan tree height distribution be $a$.

$$ \begin{aligned} & P(X < a) =0.8\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma} < \frac{a-10}{2}\big)=0.8\\ &\Rightarrow P(Z < \frac{a-10}{2}\big)=0.8\\ &\Rightarrow \frac{a-10}{2}= 0.842\\ & \qquad \text{(From normal statistical table)}\\ &\Rightarrow a = 10 + 0.842* 2\\ &\Rightarrow a = 11.684\\ &\Rightarrow a \approx 11.68 \end{aligned} $$

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