(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%.
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 > [tex]\frac{0.11-0.10}{0.02} \\[/tex])
= 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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