Question

The average THC content of marijuana sold on the street is 10.4%. 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 9.5.

c. Find the 73rd percentile for this distribution. %

Answer 1

The average THC content of marijuana sold on the street is 10.4%. Suppose the THC content is normally distributed with a standard deviation of 2%.

Let X be the THC content for a randomly selected bag of marijuana that is sold on the street.

a) The distribution of X is N(10.4, 2)

Mean = $\mu$ = 10.4

Standard deviation = $\sigma$ = 2

b) We have to find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 9.5.

That is we have to find P( X > 9.5)

For finding this probability we have to find z score.

$z =\frac{x-\mu}{\sigma} =\frac{9.5-10.4}{2}=-0.45$

That is we have to find P(Z > - 0.45)

P(Z > - 0.45) = 1 - P(Z < - 0.45) = 1 - 0.3264 = 0.6736 ( Using z table)

So the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 9.5 is 0.6736

c) We have to find the 73rd percentile for this distribution.

First, we have to find z value corresponding to 0.73

By using z table we get z = 0.6138

x = $\mu+z*\sigma$ = 10.4 + 0.6138*2 = 11.63

So the 73rd percentile for this distribution is 11.63%