Question

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=77.7 and σ=9.6.

NOTE: Assume for the sake of this problem that the score is a continuous variable. A score can thus take on any value on the continuum. (In real life, scores are often treated as if they were continuous values but are actually discrete in most cases.)

(a) Write the event ''a score over 65.7'' in terms of X:

(b) Find the probability of this event:

(c) Find the probability that a randomly chosen score is greater than 86.7:

(d) Find the probability that a randomly chosen score is between 65.7 and 86.7:

Answer 1

a)
`P(X>65.7)`

b)
`P(X>65.7)=P((X-\mu)/(\sigma)>(65.7-\mu)/(\sigma))=P(Z>(65.7-77.7)/(9.6))=P(z>-1.25)`

And we can find this probability using the complement rule:

`P(z>-1.25)=1-P(z<-1.25)=1-0.106=0.894`

c)
`P(X>86.7)=P((X-\mu)/(\sigma)>(86.7-\mu)/(\sigma))=P(Z>(86.7-77.7)/(9.6))=P(z>0.9375)`

And we can find this probability using the complement rule:

`P(z>0.9375)=1-P(z<0.9375)=1-0.826=0.174`

d)
`P(65.7<X<86.7)=P((65.7-\mu)/(\sigma)<(X-\mu)/(\sigma)<(86.7-\mu)/(\sigma))=P((65.7-77.7)/(9.6)<Z<(86.7-77.7)/(9.6))=P(-1.25<z<0.9375)`

And we can find this probability with this difference:

`P(-1.25<z<0.9375)=P(z<0.9375)-P(z<-1.25)`

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

`P(-1.25<z<0.9375)=P(z<0.9375)-P(z<-1.25)=0.826-0.106=0.720`

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".

Solution to the problem

Let X the random variable that represent the scores for the MAT112 of a population, and for this case we know the distribution for X is given by:

`X \sim N(77.7,9.6)`

Where
`\mu=77.7` and
`\sigma=9.6`

Part a

We can write the event ''a score over 65.7'' like this:

`P(X>65.7)`

Part b

We are interested on this probability

`P(X>65.7)`

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>65.7)=P((X-\mu)/(\sigma)>(65.7-\mu)/(\sigma))=P(Z>(65.7-77.7)/(9.6))=P(z>-1.25)`

And we can find this probability using the complement rule:

`P(z>-1.25)=1-P(z<-1.25)=1-0.106=0.894`

Part

`P(X>86.7)`

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>86.7)=P((X-\mu)/(\sigma)>(86.7-\mu)/(\sigma))=P(Z>(86.7-77.7)/(9.6))=P(z>0.9375)`

And we can find this probability using the complement rule:

`P(z>0.9375)=1-P(z<0.9375)=1-0.826=0.174`

Part d

`P(65.7<X<86.7)=P((65.7-\mu)/(\sigma)<(X-\mu)/(\sigma)<(86.7-\mu)/(\sigma))=P((65.7-77.7)/(9.6)<Z<(86.7-77.7)/(9.6))=P(-1.25<z<0.9375)`

And we can find this probability with this difference:

`P(-1.25<z<0.9375)=P(z<0.9375)-P(z<-1.25)`

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

`P(-1.25<z<0.9375)=P(z<0.9375)-P(z<-1.25)=0.826-0.106=0.720`