Questions: If Y has an exponential distribution and P(Y>2)=0.0821, find the value of each of the following: a. E(Y)= b. P(Y ≤ 1.7)=

Solution Steps

Given that \( P(Y > 2) = 0.0821 \), we can derive the rate parameter \( \lambda \) using the relationship:

\[ P(Y > 2) = 1 - F(2) = 0.0821 \]

This implies:

\[ F(2) = 1 - 0.0821 = 0.9179 \]

The cumulative distribution function (CDF) for an exponential distribution is given by:

\[ F(x; \lambda) = 1 - e^{-\lambda x} \]

Setting \( x = 2 \):

\[ 1 - e^{-2\lambda} = 0.9179 \]

Thus, we have:

\[ e^{-2\lambda} = 0.0821 \]

Taking the natural logarithm:

\[ -2\lambda = \ln(0.0821) \implies \lambda = -\frac{\ln(0.0821)}{2} \approx 1.2499 \]

The expected value \( E(Y) \) for an exponential distribution is given by:

\[ E(Y) = \frac{1}{\lambda} \]

Substituting the calculated value of \( \lambda \):

\[ E(Y) = \frac{1}{1.2499} \approx 0.8001 \]

To find \( P(Y \leq 1.7) \), we use the CDF:

\[ P(Y \leq 1.7) = F(1.7) = 1 - e^{-1.2499 \cdot 1.7} \]

Calculating \( F(1.7) \):

\[ F(1.7) \approx 1 - e^{-2.12483} \approx 0.8805 \]

Final Answer