Question

In a recent year, the ACT scores for the math portion of the test were normally distributed, with a mean of 21.1 and a standard deviation of 5.3. Find the probability that a randomly selected high school student who took the math portion of the ACT has a score that is (a) less than 16, (b) between 19 and 24, and (c) more than 26, and (d) identify any unusual events. Explain your reasoning

Answer 1

Answer:

a) $P(X$

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

$P(z$

b) $P(19$

And we can find this probability with this difference:

$P(-0.396$

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

$P(-0.396$

c) $P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)$

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

$P(z>0.925)=1- P(Z$

d) We can consider unusual events values above or below 2 deviations from the mean

$Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5$

A value below 10.5 can be consider as unusual

$Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7$

A value abovr 31.7 can be consider 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:

$X \sim N(21.1,5.3)$  

Where $\mu=21.1$ and $\sigma=5.3$

We are interested on this probability

$P(X$

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:

$P(X$

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

$P(z$

Part b

$P(19$

And we can find this probability with this difference:

$P(-0.396$

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

$P(-0.396$

Part c

$P(X>26)=P(\frac{X-\mu}{\sigma}>\frac{26-\mu}{\sigma})=P(Z>\frac{26-21.1}{5.3})=P(z>0.925)$

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

$P(z>0.925)=1- P(Z$

Part d

We can consider unusual events values above or below 2 deviations from the mean

$Lower = \mu -2*\sigma = 21.1 -2*5.3 =10.5$

A value below 10.5 can be consider as unusual

$Upper = \mu +2*\sigma = 21.1 +2*5.3 =31.7$

A value abovr 31.7 can be consider as unusual