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

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

We will use the formula

`\begin{gathered} Z=(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=(x-\mu)/(\sigma) \\ Z_(81)=(81-85)/(4) \\ =(-4)/(4) \\ =-1 \end{gathered}`

Z for 89, we have

`\begin{gathered} Z_(89)=(89-85)/(4) \\ =(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