Question

Scores on the $ACT$ college entrance exam follow a bell-shaped distribution with a mean $18$ and standard deviation $6$.

Wayne's standardized score on the $ACT$ was $-0.5$.

What was Wayne's actual $ACT$score?

• A. $5.5$
• B. $12$
• C. $15$
• D. $17.5$
• E. $21$

Answer 1

The correct option is C

$15$

Explanation for the correct answer:

Step-1:Given information:

Given data, Mean $=18$
Standard deviation $=6$
Standardized score $=z=-0.5$

Step-2: ACT score evaluation process:
The score $\mathrm{x}$ decreased by the mean and divided by the standard deviation gives the standardized score $\mathrm{z}$.
$\mathrm{z}=\frac{\mathrm{x}-\mathrm{Mean}}{\mathrm{Standard}\mathrm{deviation}}$
By multiplying both sides by the standard deviation
$\mathrm{z}\times \mathrm{Standard}\mathrm{deviation}=\mathrm{x}-\mathrm{Mean}$

$\mathrm{z}\times \mathrm{Standard}\mathrm{deviation}=\mathrm{x}-\mathrm{Mean}+\mathrm{Mean}$

$\mathrm{Mean}+\mathrm{z}\times \mathrm{Standard}\mathrm{deviation}=\mathrm{x}$
Interchange the equation
$\mathrm{x}=\mathrm{Mean}+\mathrm{z}\times \mathrm{Standard}\mathrm{deviation}$
Step-3: Calculate ACT scores

Substitute the known values and evaluate the expression
$$\begin{gathered} \mathrm{x}=\mathrm{Mean}+\mathrm{z}\times \mathrm{Standard}\mathrm{deviation} \\ \Rightarrow \mathrm{x}=18+(-0.5)\times 6 \\ \Rightarrow \mathrm{x}=18-3 \\ \Rightarrow \mathrm{x}=15 \end{gathered}$$

Thus Wayne's actual ACT score is $15.$

Hence, Option $\left(C\right)$ is the correct answer.