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

Answer:

a) $P(X<87)=P(\frac{X-\mu}{\sigma}<\frac{87-\mu}{\sigma})=P(Z<\frac{87-75}{7})=P(Z<1.714)$

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

$P(Z<1.714)=0.957$

b) $P(68<X<90)=P(\frac{68-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{90-\mu}{\sigma})=P(\frac{68-75}{7}<Z<\frac{90-75}{7})=P(-1<z<2.143)$

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:

$z =\frac{81-75}{7}=0.857$

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

That represent approximated the 80 percentile

And for the new case:

$z =\frac{68-62}{3}=2$

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.

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".  

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:

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

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

$P(X<87)=P(\frac{X-\mu}{\sigma}<\frac{87-\mu}{\sigma})=P(Z<\frac{87-75}{7})=P(Z<1.714)$

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

$P(Z<1.714)=0.957$

Part b

$P(68<X<90)=P(\frac{68-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{90-\mu}{\sigma})=P(\frac{68-75}{7}<Z<\frac{90-75}{7})=P(-1<z<2.143)$

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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75$  

$P(z<\frac{a-\mu}{\sigma})=0.75$

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

$z=0.674 =<\frac{a-75}{7}$

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:

$z =\frac{81-75}{7}=0.857$

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

That represent approximated the 80 percentile

And for the new case:

$z =\frac{68-62}{3}=2$

$P(Z<2) =0.977$

That represent approximated the 97 percentile

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