Question

The daily high temperature in Chicago for the month of August is approximately normal with mean 78 degrees F, and standard deviation 9 degrees F.
a. What is the probability that a randomly selected day in August will have a high temperature greater than the mean daily high temperature of 78 degrees F?
b. What is the percentile for a day in August with a high temperature of 75 degrees F?
c. What is the 75th percentile for the daily high temperature for the month of August?
d. What is the interquartile range for the daily high temperature for the month of August?

Answer 1

Answer:

a) $P(X>78) = P(Z> \frac{78-78}{9}) = P(Z>0)= 0.5$

b) $P(X$

So then 75 F correspond to approximately the 37 percentile

c) $z=0.674$

And if we solve for a we got

$a=78 +0.674*9=84.07$

So the value of height that separates the bottom 75% of data from the top 25% is 84.07 F.  

d) $IQR = 84.07-71.93= 12.14$

See explanation below.

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 daily high temperature in Chicago for the month of August of a population, and for this case we know the distribution for X is given by:

$X \sim N(78,9)$  

Where $\mu=78$ and $\sigma=9$

We are interested on this probability

$P(X>78)$

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

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

Using the z score we got:

$P(X>78) = P(Z> \frac{78-78}{9}) = P(Z>0)= 0.5$

Part b

For this case we can find the percentile with the following probability:

$P(X$

If we use the z score formula we got:

$P(X$

So then 75 F correspond to approximately the 37 percentile

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=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=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:

$z=0.674$

And if we solve for a we got

$a=78 +0.674*9=84.07$

So the value of height that separates the bottom 75% of data from the top 25% is 84.07 F.  

Part d

For this case we know that $IQR = Q_3 - Q_1 = P_{75}-P_{25}$

So then we just need to find the percentile 25.

$P(X>a)=0.25$   (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 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

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:

$z=-0.674$

And if we solve for a we got

$a=78 -0.674*9=71.93$

So the value of height that separates the bottom 25% of data from the top 75% is 71.93 F.  

So then the interquartile range would be:

$IQR = 84.07-71.93= 12.14$