Question

In 2007, when she was a high-school senior, Eleanor scored 680 on the mathematics part of the SAT. The distribution of SAT math scores in 2007 was Normal with mean 515 and standard deviation 114. Gerald took the ACT mathematics test in the same year, and scored 27. ACT math scores for 2007 were Normally distributed with mean 21 and standard deviation 5.1.

1. Find the standardized scores for both students using the R like a calculator.

2. Assuming that both tests measure the same kind of ability, who scored higher?

3. It’s possible to score higher than 2400 on the SAT, but scores 2400 and above are reported as 2400. If the distribution of SAT scores (combining math and reading) are close to Normal with mean 1780 and standard deviation 351. What proportion of SAT scores are reported as 2400 (this includes the scores that are also higher than 2400)?

4. Which score divides the top 27% from the bottom 73% (using SAT information for Eleanor)?

Answer 1

Let X be a randomly selected SAT math score of 2007. X has a normal distribution with mean and standard deviation

Let Y be a randomly selected ACT math score of 2007. Y has a normal distribution with mean and standard deviation

1) Eleanor scored 680 in SAT math. the standardized score is

Gerald scored 27 in SAT math. the standardized score is

Using the following R code we get

#standardized score of Eleanor
zEle<-(680-515)/114
sprintf("The standardized score of Eleanor is %.4f",zEle)
#standardized score of Gerald
zGer<-(27-21)/5.1
sprintf("The standardized score of Gerald is %.4f",zGer)

#get this

2. The standardized score of Eleanor in Math is higher than that of Gerald. Hence Eleanor scored higher.

3. Let W be the combined math and reading SAT scores. W has a normal distribution with mean and standard deviation

The proportion of SAT scores are reported as 2400 is same as the probability that a randomly selected SAT score is  2400 or higher.

R code to do the above

#standardized value of 2400
z<-(2400-1780)/351
#get the value of P(Z>z)
p<-1-pnorm(z,0,1)
sprintf('The probability P(W>2400) is %.4f',p)

#get this

ans:  the proportion of SAT scores are reported as 2400 is 0.0387

4. SAT information of Eleanor: Let X be a randomly selected SAT math score of 2007. X has a normal distribution with mean and standard deviation

Let X=x be the score that divides the top 27% from the bottom 73% . that means

P(X>x) = 0.27 -- The top 27%

P(X<x) = 0.73 -- The bottom 73%

First we need to find the z value for which

P(Z<z) = 0.73

Using the standard normal tables we can get that for z=0.62, P(Z<0.62)= 0.73

After getting the value of z, we equate it to the z score of x.

that is

Using the following R code we get

#z score of P(Z<z)=0.73
z<-qnorm(0.73,0,1)
x<-515+114*z
sprintf('The score which divides the top 27%% from the bottom 73%% is %.2f',x)

#get this