Question

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule. (a) What proportion of the students scored at least 25 points on this test, rounded to five decimal places? (b) What is the 63 percentile of the distribution of test scores, rounded to three decimal places?

Answer 1

Answer:

a) $P(X>25)=P(\frac{X-\mu}{\sigma}>\frac{25-\mu}{\sigma})=P(Z>\frac{25-22}{2})=P(z>1.5)$

And we can find this probability using the complement rule and we got:

$P(z>1.5)=1-P(z<1.5)=1-0.933193=0.06681$

b) $z=0.332<\frac{a-22}{2}$

And if we solve for a we got

$a=22 +0.332*2=22.664$

So the value of height that separates the bottom 63% of data from the top 37% is 22.664.  

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:

$X \sim N(22,2)$  

Where $\mu=22$ and $\sigma=2$

We are interested on this probability

$P(X>25)$

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:

$P(X>25)=P(\frac{X-\mu}{\sigma}>\frac{25-\mu}{\sigma})=P(Z>\frac{25-22}{2})=P(z>1.5)$

And we can find this probability using the complement rule and we got:

$P(z>1.5)=1-P(z<1.5)=1-0.933193=0.06681$

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.37$   (a)

$P(X<a)=0.63$   (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 0.63 of the area on the left and 0.37 of the area on the right it's z=0.332. On this case P(Z<0.332)=0.63 and P(z>0.332)=0.37

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.63$  

$P(z<\frac{a-\mu}{\sigma})=0.63$

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

$z=0.332<\frac{a-22}{2}$

And if we solve for a we got

$a=22 +0.332*2=22.664$

So the value of height that separates the bottom 63% of data from the top 37% is 22.664.