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 z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 730$ (the score of interest), $\mu = 528$ (mean) and $\sigma=117$ (standard deviation). So $z=\frac{730 - 528}{117}=\frac{202}{117}\approx1.7265$.

Step2: Find the proportion of students above the z - score

We want to find $P(X\geq730)$, which is equivalent to $P(Z\geq1.7265)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z\geq z)=1 - P(Z < z)$. Looking up $P(Z < 1.7265)$ in the standard normal table (or using a calculator with a normal - distribution function), we know that $P(Z < 1.7265)\approx0.9573$. 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. So the percentage is $0.0427\times100 = 4.2700\%$.

Answer:

4.2700%