Question

The extract of a plant native to Taiwan has been tested as apossible treatment for Leukemia. One of the chemical compoundsproduced from the plant was analyzed for a particular collagen. Thecollagen amount was found to be normally distributed with a mean of79 and standard deviation of 88 grams per mililiter. 1.What is the probability that the amount of collagen isgreater than 69 grams per mililiter?2.What is the probability that the amount of collagen is lessthan 81 grams per mililiter?3.What percentage of compounds formed from the extract of thisplant fall within 3 standard deviations of themean?

Answer 1

Answer:

1. $P(X>69)=P(Z>-1.136)=1-P(Z$

2. $P(X$

3. $P(52.6 \leq X \leq 105.4)=P(-3$

Step-by-step explanation:

Assuming the value for the mean 79 and the deviation 8.8

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 amount of collagen of a population, and for this case we know the distribution for X is given by:

$X \sim N(79,8.8)$  

Where $\mu=79$ and $\sigma=8.8$

We are interested on this probability

$P(X>69)$

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:

$P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z-1.136)$

And we can find this probability on this way:

$P(Z>-1.136)=1-P(Z$

3) Part b

$P(X$

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:

$P(X$

And we can find this probability on this way:

$P(Z$

4) Part c

We need to find first the limits for 3 deviations within the mean

$\mu -3 \sigma = 79-3(8.8)=52.6$

$\mu +3 \sigma = 79+3(8.8)=105.4$

$P(52.6$

And we can find this probability on this way:

$P(-3$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-3$