Question

1The grades on a math quizare normally distributed witha mean of 85 and a standarddeviation of 4. What percent ofthe students scored between81 and 89?

Answer 1

SOLUTION

To solve this, we get the Z for 81 and 89

We will use the formula

$\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ Where\text{ x = sample mean, that is 81 and 89} \\ \mu=population\text{ mean = 85} \\ \sigma=standard\text{ deviation = 4} \end{gathered}$

Z for 81, we have

$\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ Z_{81}=\frac{81-85}{4} \\ =\frac{-4}{4} \\ =-1 \end{gathered}$

Z for 89, we have

$\begin{gathered} Z_{89}=\frac{89-85}{4} \\ =\frac{4}{4} \\ =1 \end{gathered}$

Using the Zscore calculator for probability between two Zscores, we have

[tex]P(-1

Hence the answer is 0.68 to the nearest hundredth

Or 68.27% to the nearest hundredth