Question

in the class of 2019, more than 1.6 million students took the sat. the distribution of scores on the math section (out of 800) is approximately normal with a mean of 528 and standard deviation of 117. the university of michigan has a recommended sat math score of at least 730. what percent of students who took the sat math test meet this requirement? round your answer to 4 decimal places and then convert to a percentage.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 730$, $\mu=528$, and $\sigma = 117$.
$z=\frac{730 - 528}{117}=\frac{202}{117}\approx1.7265$

Step2: Find the proportion of the area to the right of the z - score

We know that the total area under the standard normal curve is 1. The cumulative - distribution function of the standard normal distribution $\varPhi(z)$ gives the area to the left of $z$. We want the area to the right of $z = 1.7265$, which is $P(Z\geq1.7265)=1 - P(Z < 1.7265)$.
Looking up $P(Z < 1.7265)$ in the standard normal table (or using a calculator with a normal - distribution function), $P(Z < 1.7265)\approx0.9573$.
So $P(Z\geq1.7265)=1 - 0.9573 = 0.0427$

Step3: Convert to percentage

To convert the proportion to a percentage, we multiply by 100.
$0.0427\times100 = 4.2700\%$

Answer:

$4.2700\%$