Question

The average THC content of marijuana sold on the street is 10.9%. 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

Answer 1

(a) X ≈ N (0.10 , 0.02²)

(b) The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.

(c) The 60th percentile is 11%.

The complete question is:

The average THC content of marijuana sold on the street is 10%. 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 two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.  

A. X ~ N( , )  

B. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11%.

C. Find the 60th percentile for this distribution.

According to the question,

The random variable X is defined as the THC content for a bag of marijuana that is sold on the street.

(a)The random variable X is Normally distributed with mean, μ = 0.10 and standard deviation, σ = 0.02.

Therefore, X ≈ N (0.10 , 0.02²)

(b) The value of P (X > 0.11) as follows:

P (X > o.11_ = P(X -μ/a > $\frac{0.11-0.10}{0.02}$)

= P(Z > 0.50)

= 1 -P (Z <0.50)

= 1- 0.69146

= 0.30854 ≈ 0.31

Hence,

the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.

(c)Let x represent the 60th percentile value.

Then, P (X < x) = 0.60.

⇒ P (Z < z) = 0.60

The value of z is,

z = 0.26.

Now put the value of x as follows:

z = x - μσ

0.26 = x - 0.100.02

x = 0.10+ ( 0.26 x 0.02)

x = 0.1052

x ≈ 0.11

Therefore,

The 60th percentile is 11%

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