# Solved: The mean number of students taking the Scholastic Aptitude test (SAT) has risen to an all-time high of more than 1.5 mill

#### ByDr. Raju Chaudhari

Oct 13, 2020

The mean number of students taking the Scholastic Aptitude test (SAT) has risen to an all-time high of more than 1.5 million (College Board, August 26, 2008). Students are allowed to repeat the test in hopes of improving the score that is sent to college and university admission office. The number of times the SAT was taken and the number of students are as follows

No. of times No. of Students
1 721,769
2 601,325
3 166,736
4 22,299
5 6,730

(a) Let $x$ be a random variable indicating the number of times a student take the SAT. Show the probability distribution for this random variable.
(b) What is the probability that a student takes the SAT more than one time?
(c) What is the probability that a student takes the SAT three or more time?
(d) What is the expected value of the number of times the SAT is taken? What is your interpretation of the expected value?
(e) What is the variance and standard deviation of the number of times the SAT is taken?

#### Solution

(a) Total Student = 1,518,859

 $$\begin{eqnarray*} x = 1, & & f(1) = 721,769/1,518,859 = 0.4752\\ x = 2, & & f(2) = 601,325/1,518,859 = 0.3959\\ x = 3, & & f(3) = 166,736/1,518,859 = 0.1098\\ x = 4, & & f(4) = 22,299/1,518,859 = 0.0147\\ x = 5, & & f(5) = 6730/1,518,859 = 0.0044 \end{eqnarray*}$$

(b) $P(x > 1) = 1 - f(1) = 1 - 0.4752 = 0.5248$.

Over 50% of the students take the SAT more than 1 time.

(c) The probability that a student takes the SAT three or more time

 \begin{aligned} P(x \geq 3) &= f(3) + f(4) + f(5)\\ &= 0.1098 + 0.0147 + 0.0044 \\ &= 0.1289 \end{aligned}

(d) The mean number of times a student take SAT is

$x$ $f (x)$ $x f (x)$ $x - \mu$ $(x - \mu)^2$ $(x - \mu)^2 f (x)$
1 0.4752 0.4752 -0.6772 0.4586 0.2179
2 0.3959 0.7918 0.3228 0.1042 0.0412
3 0.1098 0.3293 1.3228 1.7497 0.1921
4 0.0147 0.0587 2.3228 5.3953 0.0792
5 0.0044 0.0222 3.3228 11.0408 0.0489
Total 1.6772 0.5794

$E(x) = \sum_x xf(x) = 1.67772$ approximately 1.7 times.

(e) Variance of the number of times the SAT is taken

$V(x) = \sigma^2 = \sum_x (x-\mu)^2 f(x) = 0.5794$.

The standard deviation is $\sigma =\sqrt{V(x)} = \sqrt{0.5794} = 0.7612$.