Question

a set of high school student heights are normally distributed with a mean of 161 centimeters and a standard deviation of 15 centimeters. mikio is a high school student with a height of 191 centimeters. what proportion of student heights are higher than mikios height? you may round your answer to four decimal places.

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 161$, $\sigma=15$, and $x = 191$. So, $z=\frac{191 - 161}{15}=\frac{30}{15}=2$.

Step2: Find the proportion of values higher than 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 proportion of values less than or equal to $z$. We want $P(Z>2)$. Since $P(Z\leq z)+P(Z > z)=1$, and from the standard normal table, $P(Z\leq2)\approx0.9772$. Then $P(Z > 2)=1 - P(Z\leq2)=1 - 0.9772 = 0.0228$.

Answer:

$0.0228$