Question

2. In 2012, the mean SAT math score was 514, and the standard deviation was 117. For the purposes of this question, assume that the scores were normally distributed.

Using a graphing calculator, and without using zz-scores, find the probability (rounded to the nearest thousandth), and explain how the answer was determined that a randomly
selected SAT math student in 2012 scored
a. Between 400 and 480.
b. Less than 350.

Answer 1

a)
`P(400 < X < 480) = 0.221`

We can use the following excel code to find the answer:

"=NORM.DIST(480,514,117,TRUE)-NORM.DIST(400,514,117,TRUE)"

The other possibility is with the command normalcdf on the ti84 plus calculator, by this procedure:

Press 2nd > press VARS > DISTR > select normalcdf > Enter the following:

normalcdf(400 ,480,514,117)

And press equal.

b)
`P(X < 350) = 0.0805`

We can use the following excel code to find the answer:

"=NORM.DIST(350,514,117,TRUE)"

The other possibility is with the command normalcdf on the ti84 plus calculator, by this procedure:

Press 2nd > press VARS > DISTR > select normalcdf > Enter the following:

normalcdf(-1000000,350,514,117)

And press equal.

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".

Solution to the problem

For this case we have that our random variable is given by:

`X \sim N (\mu = 514, \sigma =117)`

And we want to find these probabilities:

Part a

`P(400 < X < 480) = 0.221`

We can use the following excel code to find the answer:

"=NORM.DIST(480,514,117,TRUE)-NORM.DIST(400,514,117,TRUE)"

The other possibility is with the command normalcdf on the ti84 plus calculator, by this procedure:

Press 2nd > press VARS > DISTR > select normalcdf > Enter the following:

normalcdf(400 ,480,514,117)

And press equal.

Part b

`P(X < 350) = 0.0805`

We can use the following excel code to find the answer:

"=NORM.DIST(350,514,117,TRUE)"

The other possibility is with the command normalcdf on the ti84 plus calculator, by this procedure:

Press 2nd > press VARS > DISTR > select normalcdf > Enter the following:

normalcdf(-1000000,350,514,117)

And press equal.