2. Suppose that the durations of high school baseball games are approximately normally distributed with mean 105 minutes and standard deviation 11 minutes.
Use a table of standard normal curve areas to find the probability that a randomly selected high school baseball game lasts
a. Less than 115 minutes.
b. More than 100 minutes.
c. Between 90 and 110 minutes.

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 durationsof high scholl baseball games of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(105,11)[/tex]

Where [tex]\mu=105[/tex] and [tex]\sigma=11[/tex]

We are interested on this probability

[tex]P(X<115)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X<115)=P(\frac{X-\mu}{\sigma}<\frac{115-\mu}{\sigma})=P(Z<\frac{115-105}{11})=P(Z<0.909)[/tex]

And we can find this probability using excel or the normal standard table:

[tex]P(Z<0.909)=0.818[/tex]

With the following code "=NORM.DIST(0.909,0,1,TRUE)"

Part b

We are interested on this probability

[tex]P(X>100)[/tex]

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>100)=P(\frac{X-\mu}{\sigma}>\frac{100-\mu}{\sigma})=P(Z>\frac{100-105}{11})=P(Z>-0.455)[/tex]

And we can find this probability using excel or the normal standard table:

[tex]P(Z>-0.455)=1-P(Z<-0.455)= 1-0.325=0.675[/tex]

With the following code "=1-NORM.DIST(-0.455,0,1,TRUE)"

Part c

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

[tex]P(90<X<110)=P(\frac{90-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{110-\mu}{\sigma})=P(\frac{90-105}{11}<Z<\frac{110-105}{11})=P(-1.36<Z<0.45)[/tex]

And we can find this probability on this way:

[tex]P(-1.36<Z<0.45)=P(z<0.45)-P(z<-1.36)=0.675-0.0863=0.589[/tex]