Question

The random variable X is normally distributed with mean 82 and standard deviation 7.4. Find the value of q such that P(82-q 11 years ago

Answer 1

\[\begin{align*}P\left(82-q<X<82+q\right)&=P\left(\frac{82-q-82}{7.4}<\frac{X-82}{7.4}<\frac{82+q-82}{7.4}\right)\\
&=P\left(-\frac{q}{7.4}<Z<\frac{q}{7.4}\right)\\
&=2P\left(0<Z<\frac{q}{7.4}\right)\\
&=2\left[P\left(Z<\frac{q}{7.4}\right)-\frac{1}{2}\right]\\
0.44&=2P\left(Z<\frac{q}{7.4}\right)-1\\
1.44&=2P\left(Z<\frac{q}{7.4}\right)\\
0.72&=P\left(Z<\frac{q}{7.4}\right)
\end{align*}\]
The probability occurs for a \(z\) value of approximately \(0.583\), so you must have
\[\frac{q}{7.4}=0.583~~\iff~~q=4.314\]