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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.
%

Sagot :

LammettHashidn

a. This information is given to you.

b. We want to find

[tex]\mathrm{Pr}\{X > 8.9\}[/tex]

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

[tex]X = \mu + \sigma Z[/tex]

where [tex]\mu,\sigma[/tex] are the mean/s.d. of [tex]X[/tex]. Now,

[tex]\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}[/tex]

where [tex]\Phi(z)[/tex] is the CDF for [tex]Z[/tex].

c. The 76th percentile is the value of [tex]X=x_{76}[/tex] such that

[tex]\mathrm{Pr}\{X \le x_{76}\} = 0.76[/tex]

Transform [tex]X[/tex] to [tex]Z[/tex] and apply the inverse CDF of [tex]Z[/tex].

[tex]\mathrm{Pr}\left\{Z \le \dfrac{x_{76} - 10.5}2\right\} = 0.76[/tex]

[tex]\dfrac{x_{76} - 10.5}2 = \Phi^{-1}(0.76)[/tex]

[tex]\dfrac{x_{76} - 10.5}2 \approx 0.7063[/tex]

[tex]x_{76} - 10.5 \approx 1.4126[/tex]

[tex]x_{76} \approx \boxed{11.9126}[/tex]

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