SOLUTION: 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 rand

Question 1200311: 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 10.9.
c. Find the 78th percentile for this distribution.

Answer by GingerAle(43) (Show Source): You can put this solution on YOUR website!
**a. Distribution of X**
* X follows a normal distribution with mean (μ) = 10.5% and standard deviation (σ) = 2%.
**Therefore, X ~ N(10.5, 2²)**
**b. Probability of THC content greater than 10.9%**
1. **Calculate the z-score:**
* z = (X - μ) / σ
* z = (10.9 - 10.5) / 2 = 0.2
2. **Find the probability using a standard normal distribution table or calculator:**
* P(X > 10.9) = P(Z > 0.2) = 1 - P(Z ≤ 0.2)
* Using a standard normal table, we find P(Z ≤ 0.2) ≈ 0.5793
* Therefore, P(X > 10.9) ≈ 1 - 0.5793 = 0.4207
**The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 10.9% is approximately 0.4207.**
**c. Find the 78th percentile**
* The 78th percentile is the value of X such that 78% of the data falls below it.
* **Use the inverse standard normal function (also known as the quantile function):**
* Find the z-score corresponding to the 78th percentile using a standard normal table or calculator.
* z ≈ 0.77
* **Calculate X:**
* X = μ + zσ
* X = 10.5 + (0.77 * 2)
* X = 10.5 + 1.54 = 12.04
**The 78th percentile for this distribution is 12.04%.**

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