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

Answer:

a) $P(X<15)=P(\frac{X-\mu}{\sigma}<\frac{15-\mu}{\sigma})=P(Z<\frac{15-21.3}{6.2})=P(z<-1.016)$

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

$P(z<-1.016)=0.155$

b) $P(18<X<25)=P(\frac{18-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{25-\mu}{\sigma})=P(\frac{18-21.3}{6.2}<Z<\frac{25-21.3}{6.2})=P(-0.532<z<0.597)$

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) $P(X>34)=P(\frac{X-\mu}{\sigma}>\frac{34-\mu}{\sigma})=P(Z>\frac{34-21.3}{6.2})=P(z>2.048)$

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.

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".  

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:

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

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

$P(X<15)=P(\frac{X-\mu}{\sigma}<\frac{15-\mu}{\sigma})=P(Z<\frac{15-21.3}{6.2})=P(z<-1.016)$

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

$P(z<-1.016)=0.155$

Part b

$P(18<X<25)=P(\frac{18-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{25-\mu}{\sigma})=P(\frac{18-21.3}{6.2}<Z<\frac{25-21.3}{6.2})=P(-0.532<z<0.597)$

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

$P(X>34)=P(\frac{X-\mu}{\sigma}>\frac{34-\mu}{\sigma})=P(Z>\frac{34-21.3}{6.2})=P(z>2.048)$

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.