Question

a two - year college will accept any student with a z - score of 0.605 and above. if the test is normally distributed with a mean of 500 and a standard deviation of 100, what is the cutoff score for acceptance? (lowest score)
a 500
b 400
c 560.5
d 66

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $z$ is the z - score, $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation. We want to find $x$, and we know that $z = 0.605$, $\mu=500$, and $\sigma = 100$.

Step2: Rearrange the formula to solve for $x$

Starting from $z=\frac{x - \mu}{\sigma}$, we can multiply both sides by $\sigma$: $z\sigma=x-\mu$. Then add $\mu$ to both sides to get $x=\mu + z\sigma$.

Step3: Substitute the given values

Substitute $\mu = 500$, $z = 0.605$, and $\sigma=100$ into the formula $x=\mu + z\sigma$. So $x=500+0.605\times100$.

Step4: Calculate the value of $x$

$x=500 + 60.5=560.5$.

Answer:

C. 560.5