Question

The average THC content of marijuana sold on the street is 10.5%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to 4 decimal places where possible,
a. What is the distribution of X? X ~ N(
,
)

b. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 8.9.


c. Find the 76th percentile for this distribution.
%

Answer 1

a. This information is given to you.

b. We want to find

$\mathrm{Pr}\{X > 8.9\}$

so we first transform $X$ to the standard normal random variable $Z$ with mean 0 and s.d. 1 using

$X = \mu + \sigma Z$

where $\mu,\sigma$ are the mean/s.d. of $X$. Now,

$\mathrm{Pr}\left\{\dfrac{X - 10.5}2 > \dfrac{8.9 - 10.5}2\right\} = \mathrm{Pr}\{Z > -0.8\} \\\\~~~~~~~~= 1 - \mathrm{Pr}\{Z\le-0.8\} \\\\ ~~~~~~~~ = 1 - \Phi(-0.8) \approx \boxed{0.7881}$

where $\Phi(z)$ is the CDF for $Z$.

c. The 76th percentile is the value of $X=x_{76}$ such that

$\mathrm{Pr}\{X \le x_{76}\} = 0.76$

Transform $X$ to $Z$ and apply the inverse CDF of $Z$.

$\mathrm{Pr}\left\{Z \le \dfrac{x_{76} - 10.5}2\right\} = 0.76$

$\dfrac{x_{76} - 10.5}2 = \Phi^{-1}(0.76)$

$\dfrac{x_{76} - 10.5}2 \approx 0.7063$

$x_{76} - 10.5 \approx 1.4126$

$x_{76} \approx \boxed{11.9126}$