Answer: 0.4
Step-by-step explanation:
Given: Cumulative exponential distribution:
[tex]P(x<k)=1-e^{\frac{-x}{k}}[/tex]
As per given,
[tex]P(x<2)=1-e^{\frac{-2}{k}}=0.381\\\\\Rightarrow\ e^{\frac{-2}{k}}=1-0.381\\\\\Rightarrow\ e^{\frac{-2}{k}}=0.619\\\\\Rightarrow\ {\frac{-2}{k}}=\ln(0.619) \ \ \ [\text{Taking natural log on both sides }]\\\\\Rightarrow\ {\frac{-2}{k}}=-0.47965\\\\\Rightarrow\ k=\dfrac{2}{0.47965}\\\\\Rightarrow\ k=4.17[/tex]
[tex]P(0.5 < X < 3) \\\\=P(X<3)-P(X<0.5) \\\\=(1-e^{-\frac{3}{4.17}})-(1-e^{-\frac{0.5}{4.17}}) \\\\=e^{-\frac{0.5}{4.17}}-e^{-\frac{3}{4.17}}\\\\= 0.4[/tex]
Hence, the required probability = 0.4
