Question

An article suggests that substrate concentration (mg/cm^3) of influent to a reactor is normally distributed with μ = 0.60 and σ = 0.08. (Round your answers to four decimal places.)

a. What is the probability that the concentration exceeds 0.50?
b. What is the probability that the concentration is at most 0.20?
c. How would you characterize the largest 5% of all concentration values?

Answer 1

Answer:

a)$P(X>0.5)=P(\frac{X-\mu}{\sigma}>\frac{0.5-\mu}{\sigma})=P(Z>\frac{0.5-0.6}{0.08})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z<-1.25)$

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

$P(z>-1.25)=1-0.106=0.894$

b) $P(X<0.2)$

And we can use the z score formula given by:

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

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

$P(z<-5)=2.87x10^{-7}$

c) $z=1.64<\frac{a-0.6}{0.08}$

And if we solve for a we got

$a=0.6 +1.64*0.08=0.7312$

So the value of height that separates the bottom 95% of data from the top 5% is 0.7312.  

Step-by-step explanation:

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".  

Part a

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

$X \sim N(0.6,0.08)$  

Where $\mu=0.6$ and $\sigma=0.08$

We are interested on this probability

$P(X>0.5)$

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:

$P(X>0.5)=P(\frac{X-\mu}{\sigma}>\frac{0.5-\mu}{\sigma})=P(Z>\frac{0.5-0.6}{0.08})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z<-1.25)$

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

$P(z>-1.25)=1-0.1056=0.8944$

Part b

$P(X<0.2)$

And we can use the z score formula given by:

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

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

$P(z<-5)=2.8665x10^{-7}$

Part c

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

$P(X>a)=0.05$   (a)

$P(X<a)=0.95$   (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 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

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

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95$  

$P(z<\frac{a-\mu}{\sigma})=0.95$

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

$z=1.64<\frac{a-0.6}{0.08}$

And if we solve for a we got

$a=0.6 +1.64*0.08=0.7312$

So the value of height that separates the bottom 95% of data from the top 5% is 0.7312.