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.

Answer 1

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