Question

mr. emmer gave a test in his chemistry class. the scores were normally distributed with a mean of 82 and a standard deviation of 4. what percent of students would you expect to score between 78 and 36?

Answer 1

`P(36\le X\le78)=P((X1-\mu)/(\sigma)\le Z\le(X2-\mu)/(\sigma))`

Where:

`\begin{gathered} X1=36 \\ X2=78 \\ \mu=82 \\ \sigma=4 \end{gathered}`

So:

`\begin{gathered} P(36\le X\le78)=P((78-82)/(4)\le Z\le(86-82)/(4)) \\ P(36\le X\le78)=P(-1\le Z\le1) \\ P(36\le X\le78)=0.6827\approx0.68 \end{gathered}`

As a percentage: 68%