Question

The average THC content of marijuana sold on the street is 9.3%. Suppose the THC content is normally distributed with standard deviation of 1%. 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( 9.3 Correct, 1 Correct) b. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 9.2. 0.4601 Incorrect c. Find the 75th percentile for this distribution. 8

Answer 1

Answer:

a) $X \sim N(9.3,1)$  

b) $P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

c) The value of height that separates the bottom 75% of data from the top 25% is 9.9745.  

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(9.3,1)$  

Where $\mu=9.3$ and $\sigma=1$

3) Part b

We are interested on this probability

$P(X>9.2)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>9.2)=P(\frac{X-\mu}{\sigma}>\frac{9.2-\mu}{\sigma})=P(Z>\frac{9.2-9.3}{1})=1-P(Z$

4) Part c

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$   (a)

$P(X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.6745. On this case P(Z<0.6745)=0.75 and P(z>0.6745)=0.25

If we use condition (b) from previous we have this:

$P(X$  

$P(Z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.6745$

And if we solve for a we got

$a=9.3 +1*0.6745=9.9745$

So the value of height that separates the bottom 75% of data from the top 25% is 9.9745.