Question

A set of final examinations grades in an introductory statistics course is normally distributed, with a mean of 75 and a standard deviation of 7. Complete parts (a) through (d).

a.) What is the probability that a student scored below 87 on this exam?

b.) What is the probability that a student scored between 68 and 90?

c.) The probability is 25% that a student taking the test scores higher than what grade?

d.) If the professor grades on a curve (For example, the professor could give A

Answer 1

a)
`P(X<87)=P((X-\mu)/(\sigma)<(87-\mu)/(\sigma))=P(Z<(87-75)/(7))=P(Z<1.714)`

And we can find this probability using the z table or excel and we got:

`P(Z<1.714)=0.957`

b)
`P(68<X<90)=P((68-\mu)/(\sigma)<(X-\mu)/(\sigma)<(90-\mu)/(\sigma))=P((68-75)/(7)<Z<(90-75)/(7))=P(-1<z<2.143)`

And we can find this probability on this way:

`P(-1<z<2.143)=P(z<2.143)-P(z<-1)`

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

`P(-1<z<2.143)=P(z<2.143)-P(z<-1)=0.984-0.159=0.825`

c)
`a=75 + 0.674*7=79.718`

So the value of height that separates the bottom 75% of data from the top 25% is 79.718.

d) We can find the percentiles per each case for the original case using the z score first:

`z =(81-75)/(7)=0.857`

P(Z<0.857) =0.804[/tex]

That represent approximated the 80 percentile

And for the new case:

`z =(68-62)/(3)=2`

P(Z<2) =0.977[/tex]

That represent approximated the 97 percentile

So is better the second case since we are on a percentile higher respect the other people.

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".

2) Part a

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

`X \sim N(75,7)`

Where
`\mu=75` and
`\sigma=7`

We are interested on this probability

`P(64.2<X<67.2)`

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<87)=P((X-\mu)/(\sigma)<(87-\mu)/(\sigma))=P(Z<(87-75)/(7))=P(Z<1.714)`

And we can find this probability using the z table or excel and we got:

`P(Z<1.714)=0.957`

Part b

`P(68<X<90)=P((68-\mu)/(\sigma)<(X-\mu)/(\sigma)<(90-\mu)/(\sigma))=P((68-75)/(7)<Z<(90-75)/(7))=P(-1<z<2.143)`

And we can find this probability on this way:

`P(-1<z<2.143)=P(z<2.143)-P(z<-1)`

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

`P(-1<z<2.143)=P(z<2.143)-P(z<-1)=0.984-0.159=0.825`

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<a)=0.75` (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<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.75`

`P(z<(a-\mu)/(\sigma))=0.75`

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

`z=0.674 =<(a-75)/(7)`

And if we solve for a we got

`a=75 + 0.674*7=79.718`

So the value of height that separates the bottom 75% of data from the top 25% is 79.718.

Part d: Assuming this question" If the professor grades on a curve (i.e. gives A's to the top 10% of the class, regardless of the test score), are you better off with a grade of 81 on this exam or a grade of 68 on a different exam, where the mean is 62 and the standard deviation is 3?"

We can find the percentiles per each case for the original case using the z score first:

`z =(81-75)/(7)=0.857`

P(Z<0.857) =0.804[/tex]

That represent approximated the 80 percentile

And for the new case:

`z =(68-62)/(3)=2`

`P(Z<2) =0.977`

That represent approximated the 97 percentile

So is better the second case since we are on a percentile higher respect the other people.