Question

In a recent year, the ACT scores for the reading portion of the test were normally distributed, with a mean of 21.3 and a standard deviation of 6.2. Find the probability that a randomly selected high school student who took the reading portion of the ACT has a score that is (a) less than 15, (b) between 18 and 25, and (c) more than 34, and (d) identify any unusual events. Explain your reasoning

Answer 1

a)
`P(X<15)=P((X-\mu)/(\sigma)<(15-\mu)/(\sigma))=P(Z<(15-21.3)/(6.2))=P(z<-1.016)`

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

`P(z<-1.016)=0.155`

b)
`P(18<X<25)=P((18-\mu)/(\sigma)<(X-\mu)/(\sigma)<(25-\mu)/(\sigma))=P((18-21.3)/(6.2)<Z<(25-21.3)/(6.2))=P(-0.532<z<0.597)`

And we can find this probability with this difference:

`P(-0.532<z<0.597)=P(z<0.597)-P(z<-0.532)`

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

`P(-0.532<z<0.597)=P(z<0.597)-P(z<-0.532)=0.725-0.297=0.428`

c)
`P(X>34)=P((X-\mu)/(\sigma)>(34-\mu)/(\sigma))=P(Z>(34-21.3)/(6.2))=P(z>2.048)`

And we can find this probability using the complement rule and the normal standard table or excel and we got:

`P(z> 2.048)=1-P(Z<2.048) = 1- 0.980=0.02`

d)
`Lower = 21.3 -2*6.2 = 8.9`

`Upper = 21.3 +2*6.2 = 33.7`

If a value is lower than 8.9 or higher than 33.7 would be considered as unusual.

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 scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(21.3,6.2)`

Where
`\mu=21.3` and
`\sigma=6.2`

We are interested on this probability

`P(X<15)`

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<15)=P((X-\mu)/(\sigma)<(15-\mu)/(\sigma))=P(Z<(15-21.3)/(6.2))=P(z<-1.016)`

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

`P(z<-1.016)=0.155`

Part b

`P(18<X<25)=P((18-\mu)/(\sigma)<(X-\mu)/(\sigma)<(25-\mu)/(\sigma))=P((18-21.3)/(6.2)<Z<(25-21.3)/(6.2))=P(-0.532<z<0.597)`

And we can find this probability with this difference:

`P(-0.532<z<0.597)=P(z<0.597)-P(z<-0.532)`

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

`P(-0.532<z<0.597)=P(z<0.597)-P(z<-0.532)=0.725-0.297=0.428`

Part c

`P(X>34)=P((X-\mu)/(\sigma)>(34-\mu)/(\sigma))=P(Z>(34-21.3)/(6.2))=P(z>2.048)`

And we can find this probability using the complement rule and the normal standard table or excel and we got:

`P(z> 2.048)=1-P(Z<2.048) = 1- 0.980=0.02`

Part d

For this case we can use the rule of thumb that we expect about 95% of the data values between two deviations and we can find the normal limits like this:

`Lower = 21.3 -2*6.2 = 8.9`

`Upper = 21.3 +2*6.2 = 33.7`

If a value is lower than 8.9 or higher than 33.7 would be considered as unusual.