Question

SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. assuming this is true (it is only approximately true, both because of variation from year to year and because scores have decreased since the SAT tests were first developed), answer the following questions.a. what percentage of students score above 625 on the math SAT in any given year?b. what percentage of students score between 400 and 600 on the verbal SAT?c. a college decides to liberalize its admission policy. As a first step, the admissions committee decides to exclude only those applicants scoring below the 20th percentile on the verbal SAT . Translate this percentile into a Z score. Then, calculate the equivalent SAT verbal test score.

Answer 1

Answer:

a) $P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

$P(Z>1.25)=1-P(Z<1.25)=1-0.894=0.106$

b) $P(400<X<600)=P(\frac{400-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{600-\mu}{\sigma})=P(\frac{400-500}{100}<Z<\frac{600-500}{100})=P(-1<Z<1)$

$P(-1<Z<1)=P(Z<1)-P(Z<-1)$

$P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.682$

c) $z=-0.842<\frac{a-500}{100}$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

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

$X \sim N(500,100)$  

Where $\mu=500$ and $\sigma=100$

We are interested on this probability

$P(X>625)$

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:

$P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

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

$P(Z>1.25)=1-P(Z<1.25)=1-0.894=0.106$

Part b

We are interested on this probability

$P(400<X<600)$

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:

$P(400<X<600)=P(\frac{400-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{600-\mu}{\sigma})=P(\frac{400-500}{100}<Z<\frac{600-500}{100})=P(-1<Z<1)$

And we can find this probability with this difference:

$P(-1<Z<1)=P(Z<1)-P(Z<-1)$

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

$P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.682$

Part c

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

$P(X>a)=0.8$   (a)

$P(X<a)=0.2$   (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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

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

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.2$  

$P(z<\frac{a-\mu}{\sigma})=0.2$

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

$z=-0.842<\frac{a-500}{100}$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.  

Answer 2

Percent is quantity per 100 other quantity, whereas probability is quantity per 1 other quantity. The needed percent figures are:

• The percentage of students score above 625 on the math SAT in any given year is 10.56%
• The percentage of students scoring between 400 and 600 is 68.26%
• The needed z-score is -0.84, and the equivalent SAT verbal test score is 416

How to convert percent to probability?

Percent counts the number compared to 100 whereas probability counts it compare to 1.

So, if we have a%, that means for each 100, there are 'a' parts. If we divide each of them with 100, we get:

For each 1, there are a/100 parts.

Thus, 50% = 50/100 = 0.50 (in probability)

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$)

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z-tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

Now, for the given case, if the SAT scores are tracked by a random variable X, then, according o the given data, we get:

$X \sim N(\mu = 500, \sigma = 100)$

The needed percentages are then found as:

• a) Percentage of students score above 625 on the math SAT in any given year

The probability of students scoring above 625 is given by:

P(X > 625)

Using the standard normal distributions, we get:

$P(X > 625) = 1 - P(X \leq 625)\\P(X > 625) = 1 - P(Z \leq \dfrac{625 - \mu=500}{\sigma=100}) = 1 - P(Z \leq \dfrac{625 - 500}{100})\\P(X > 625) = 1 - P(Z \leq 1.25)$

From z-tables, the p-value for Z = 1.25 is obtained as: 0.8944

Thus, we get the needed probabilities as:
$P(X > 625) = 1 - P(Z \leq 1.25)\\P(X > 625) = 1 - 0.8944 = 0.1056$

Thus, the percentage of students score above 625 on the math SAT in any given year is 10.56%

• b) The percentage of students scoring between 400 and 600.

The probability of students scoring between 400 and 600 is denoted as:

$P( 400 \leq X \leq 600)$

or, it can be rewritten as: $P( 400 \leq X \leq 600) = P(X \leq 600) - P(X < 400)$

Using the standard normal distribution, we get the probabilities rewritten as:

$P( 400 \leq X \leq 600) = P(X \leq 600) - P(X < 400)\\P( 400 \leq X \leq 600) = P(Z \leq \dfrac{600 - \mu=500}{\sigma=100}) - P(Z \leq \dfrac{400 - 500}{100})\\\\P( 400 \leq X \leq 600) = P(Z \leq 1) - P(Z \leq -1)$

Using z-tables, the p-value for Z = 1 and -1 are obtained as 0.8413 and 0.1587

Thus, we get:

$P( 400 \leq X \leq 600) = P(Z \leq 1) - P(Z \leq -1)\\\\P( 400 \leq X \leq 600) = 0.8413 - 0.1587 = 0.6826$

Thus, the percentage of students scoring between 400 and 600 is 68.26%

c) Translating lower 20 percentage of scores into z-scores

P(X < x) = 20% = 0.2

$P(X < x) = 0.2\\P(X < x ) = P(Z < z = \dfrac{x - \mu}{\sigma}) \\P(X < x) = P(Z \leq z = \dfrac{x - 500}{100})\\or\\P(Z \leq z = \dfrac{x - 500}{100}) = 0.2$

From the z-tables, the value of Z for which the p-value is obtained around 0.2 is -0.84

Therefore, we get:
$z = \dfrac{x-500}{100} = -0.84\\\\x = 416$

The needed z-score is -0.84, and the equivalent SAT verbal test score is 416

Thus, the needed percent figures are:

• The percentage of students score above 625 on the math SAT in any given year is 10.56%
• The percentage of students scoring between 400 and 600 is 68.26%
• The needed z-score is -0.84, and the equivalent SAT verbal test score is 416

Learn more about z-scores here:
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