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\{\frac{X - 10.5}2 > \frac{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 \frac{x_(76) - 10.5}2\right\} = 0.76`

`\frac{x_(76) - 10.5}2 = \Phi^(-1)(0.76)`

`\frac{x_(76) - 10.5}2 \approx 0.7063`

`x_(76) - 10.5 \approx 1.4126`

`x_(76) \approx \boxed{11.9126}`