Question

Scores on the ACT college entrance exam follow a bell-shaped distribution with a mean $18$ and a standard deviation $6$. Wayne’s standardized score on the ACT was $-0.7$ What was Wayne’s actual ACT score? $\left(a\right)4.2\left(b\right)-4.2\left(c\right)13.8\left(d\right)17.3\left(e\right)22.2$

Short answer

The correct option is (c).

Step-by-step solution

Step 1. Given

The average population $\left(\mu \right)=18$

The standard deviation of the population $\left(\sigma \right)=6$

A score that is standardized $\left(z\right)=-0.7$

Step 2. Concept

The z score is calculated using the following formula: $z=X-\mu /\sigma$

Step 3. Calculation

Consider that $X$ reflects Wayne's real ACT score, which follows a normal distribution with a mean of $18$and a standard deviation of$6$.

Wayne's real ACT score is determined as follows:

$z=\frac{X-\mu }{\sigma }$

$-0.7=\frac{X-18}{6}$

$-4.2+18=X$

$X=13.8$

Hence, the correct option is $\left(c\right)13.8$