Question

Suppose that X has a normal distribution such that \({\bf{Pr}}\left( {{\bf{X < 116}}} \right){\bf{ = 0}}{\bf{.20}}\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {{\bf{X < 328}}} \right){\bf{ = 0}}{\bf{.90}}\,\)Determine X's mean and variance.

Short answer

\(\begin{array}{l}\sigma = 99.53\\\mu = 199.60\end{array}\)

Step-by-step solution

Given information

X is a random variable that follows a normal distribution\(X \sim N\left( {\mu ,\sigma } \right)\). The probability value\(\Pr \left( {X < 116} \right) = 0.20\,\,and\,\,\Pr \left( {X < 328} \right) = 0.90\,\).

Converting to Z-score

Z score calculation:

\(\begin{array}{l} = \Pr \left( {X < 116} \right) = 0.20\,\\ = \Pr \left( {\frac{{X - \mu }}{\sigma } < \frac{{116 - \mu }}{\sigma }} \right) = 0.20\\ = \Pr \left( {Z < \frac{{116 - \mu }}{\sigma }} \right) = 0.20 \ldots \left( 1 \right)\end{array}\)

The z-score that satisfies the equation (1) from the Z-table is -0.84.

\(\frac{{116 - \mu }}{\sigma } = - 0.84 \ldots \left( 2 \right)\)

Similarly, we calculate the z-score for the other probability

\(\begin{array}{l} = \Pr \left( {X < 328} \right) = 0.90\,\\ = \Pr \left( {\frac{{X - \mu }}{\sigma } < \frac{{328 - \mu }}{\sigma }} \right) = 0.90\\ = \Pr \left( {Z < \frac{{328 - \mu }}{\sigma }} \right) = 0.90 \ldots \left( 3 \right)\end{array}\)

The z-score that satisfies the equation (3) from the Z-table is 1.29.

\(\frac{{328 - \mu }}{\sigma } = 1.29 \ldots \left( 4 \right)\)

Solving (2) and (4),

\(\begin{array}{l}\sigma = 99.53\\\mu = 199.60\end{array}\)