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Section 8.5 : Probability
- Let,
\[f\left( x \right) = \left\{ {\begin{array}{ll}{\displaystyle \frac{3}{{37760}}{x^2}\left( {20 - x} \right)}&{{\mbox{if }}2 \le x \le 18}\\0&{{\mbox{otherwise}}}\end{array}} \right.\]
- Show that \(f\left( x \right)\) is a probability density function.
- Find \(P\left( {X \le 7} \right)\).
- Find \(P\left( {X \ge 7} \right)\).
- Find \(P\left( {3 \le X \le 14} \right)\).
- Determine the mean value of \(X\).
- For a brand of light bulb the probability density function of the life span of the light bulb is given by the function below, where t is in months.
\[f\left( t \right) = \left\{ {\begin{array}{ll}{0.04{{\bf{e}}^{ - \,\,\frac{t}{{25}}}}}&{{\mbox{if }}t \ge 0}\\0&{{\mbox{if }}t < 0}\end{array}} \right.\]
- Verify that \(f\left( t \right)\) is a probability density function.
- What is the probability that a light bulb will have a life span less than 8 months?
- What is the probability that a light bulb will have a life span more than 20 months?
- What is the probability that a light bulb will have a life span between 14 and 30 months?
- Determine the mean value of the life span of the light bulbs.
- Determine the value of \(c\) for which the function below will be a probability density function. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{c\left( {8{x^3} - {x^4}} \right)}&{{\mbox{if }}0 \le x \le 8}\\0&{{\mbox{otherwise}}}\end{array}} \right.\] Solution