Chapter 2: Q. 10 (page 139)

Questions T2.9 and T2.10 refer to the following setting. Until the scale was changed in $1995$ , SAT scores were based on a scale set many years ago. For Math scores, the mean under the old scale in the $1990 s$ was $470$ and the standard deviation was $110$ . In $2009$ , the mean was $515$ and the standard deviation was $116$ .
T2.10. Jane took the SAT in $1994$ and scored $500$ . Her sister Colleen took the SAT in $2009$ and scored $530$ . Who did better on the exam, and how can you tell?
(a) Colleen-she scored $30$ points higher than Jane.
(b) Colleen-her standardized score is higher than Jane's.
(c) Jane-her standardized score is higher than Colleen's.
(d) Jane-the standard deviation was bigger in $2009 .$
(e) The two sisters did equally well-their $z$ -scores are the same.

Short Answer

Expert verified

Jane, because she has a better standardised score than Colleen. As a result, option (c) is the correct answer.

Step by step solution

Given information

Jane took the SAT in $1994$ and scored $500$ . Her sister Colleen took the SAT in $2009$ and scored $530$ .

Explanation

According to the information, the mean score was $μ = 470$

The standard deviation was $σ = 110$ for Math scores.

Consider the new scale in $2009$ , the mean score was $μ_{2} = 515$ .

The standard deviation was $σ_{2} = 116$ for Math scores.
Jane took the SAT in $1994$ and scored $500$ . Her sister Colleen took the SAT in 2009 and scored $530 .$
Jane's standardized score ( $z$ -score) is:
$\begin{aligned} z = \frac{x_{1} - μ_{1}}{σ_{1}} \end{aligned} = \frac{500 - 470}{110} = 0.27$