Question

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.

Answer 1

a)
`P(X<115)= P(Z<0.909)=0.818`

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

b)
`P(X>100)=P(Z>-0.455)=1-P(Z<-0.455)= 1-0.325=0.675`

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

c)
`P(90<X<110)=P(-1.36<Z<0.45)=P(z<0.45)-P(z<-1.36)=0.675-0.0863=0.589`

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

`X \sim N(105,11)`

Where
`\mu=105` and
`\sigma=11`

We are interested on this probability

`P(X<115)`

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

`z=(x-\mu)/(\sigma)`

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

`P(X<115)=P((X-\mu)/(\sigma)<(115-\mu)/(\sigma))=P(Z<(115-105)/(11))=P(Z<0.909)`

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

`P(Z<0.909)=0.818`

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

Part b

We are interested on this probability

`P(X>100)`

`z=(x-\mu)/(\sigma)`

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

`P(X>100)=P((X-\mu)/(\sigma)>(100-\mu)/(\sigma))=P(Z>(100-105)/(11))=P(Z>-0.455)`

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

`P(Z>-0.455)=1-P(Z<-0.455)= 1-0.325=0.675`

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:

`P(90<X<110)=P((90-\mu)/(\sigma)<(X-\mu)/(\sigma)<(110-\mu)/(\sigma))=P((90-105)/(11)<Z<(110-105)/(11))=P(-1.36<Z<0.45)`

And we can find this probability on this way:

`P(-1.36<Z<0.45)=P(z<0.45)-P(z<-1.36)=0.675-0.0863=0.589`