Question-18

Suppose a random variable \(X\) follows an exponential distribution with mean 5. If \(P(X \leq k) = 0.95\), then find the value of \(k\).

• \(\dfrac{1}{5}\hspace{0.5mm}(\ln 20)\)

• \(5\hspace{0.5mm}(\ln 20)\)

• \(\dfrac{1} {5}\hspace{0.5mm}( \ln2)\)

• \(5\hspace{0.5mm}(\ln 2)\)

Hint

If \(X \sim Exp(\lambda),\) then \(P(X \leq a) = 1- e^{-\lambda a}\)

Answer

• \(\dfrac{1}{5}\hspace{0.5mm}(\ln 20)\)

• \(5\hspace{0.5mm}(\ln 20)\)

• \(\dfrac{1} {5}\hspace{0.5mm}(\ln 2)\)

• \(5\hspace{0.5mm}(\ln 2)\)

Solution

\(P(X \leq k) = 0.95\)
\(1- e^{-k/5} = 0.95\)
\(e^{-k/5} = 0.05\)

\(e^{k/5} = \dfrac{1}{0.05} = 20\)

\(k/5 = \ln(20)\)
\(k = 5\ln(20)\)