Suppose that the mean and standard deviation of the scores on a statistics exam 80.22 and 6.899 and are approximately normally distributed. Calculate the proportion of scores between 81.4 and 82.1.

Question

Suppose that the mean and standard deviation of the scores on a statistics exam 80.22 and 6.899 and are approximately normally distributed.  Calculate the proportion of scores between 81.4 and 82.1.

anonymous · Answer

$$\begin{align*}
P(81.4<X<82.1)&=P\left(\frac{81.4-80.22}{6.899}<\frac{X-80.22}{6.899}<\frac{82.1-80.22}{6.899}\right)\\
&=P\left(0.1711<Z<0.2725\right)\\
&=P(Z<0.2725)-P(Z<0.1711)
\end{align*}$$