Respuesta :
Answer:
(a) [tex]X\sim N(0.10,\ 0.02^{2})[/tex].
(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%.
Step-by-step explanation:
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. Â
A. X ~ N( , ) Â
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.
Solution:
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.
Thus, [tex]X\sim N(0.10,\ 0.02^{2})[/tex].
(b)
Compute the value of P (X > 0.11) as follows:
[tex]P(X>0.11)=P(\frac{X-\mu}{\sigma}>\frac{0.11-0.10}{0.02})\\=P(Z>0.50)\\=1-P(Z<0.50)\\=1-0.69146\\=0.30854\\\approx 0.31[/tex]
*Use a z-table.
Thus, 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 pth percentile is a data value such that at least p% of the data-set is less than or equal to this data value and at least (100 - p)% of the data-set are more than or equal to this data value.
Let x represent the 60th percentile value.
The, P (X < x) = 0.60.
⇒ P (Z < z) = 0.60
The value of z for this probability is,
z = 0.26.
Compute the value of x as follows:
[tex]z=\farc{x-\mu}{\sigma}\\0.26=\farc{x-0.10}{0.02}\\x=0.10+(0.26\times 0.02)\\x=0.1052\\x\approx 0.11[/tex]
Thus, the 60th percentile is 11%.
