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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 $1990s$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.

Step by step solution

01

Given information

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

02

Explanation

According to the information, the mean score was $\mathrm{μ}=470$

The standard deviation was $\mathrm{σ}=110$for Math scores.

Consider the new scale in $2009$, the mean score was ${\mathrm{μ}}_2=515$.

The standard deviation was${\mathrm{σ}}_2=116$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:
$$\begin{gathered} \begin{array}{r}z=\frac{x_1−{\mathrm{μ}}_1}{{\mathrm{σ}}_1}\end{array} \\ =\frac{500-470}{110} \\ =0.27 \end{gathered}$$

Colleen's standardized score ($z$-score) is:

$$\begin{gathered} \begin{array}{r}z=\frac{x_1−{\mathrm{μ}}_1}{{\mathrm{σ}}_1}\end{array} \\ =\frac{530-515}{116} \\ =0.13 \end{gathered}$$.

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