Question

consider the following exponential probability density function. $f(x)=\frac{1}{3}e^{-x/3}$ for $xgeq0$. if needed, round your answers to four decimal digits. (a) choose the correct formula for $p(xleq x_0)$. (i) $p(xleq x_0)=1 - e^{-x_0/3}$ (ii) $p(xleq x_0)=1 + e^{x_0/3}$ (iii) $p(xleq x_0)=1 - e^{x_0/3}$ (iv) $p(xleq x_0)=1 + e^{-x_0/3}$ - select your answer - (b) find $p(xleq2)$. (c) find $p(xgeq3)$. (d) find $p(xleq5)$. (e) find $p(2leq xleq5)$.

Explanation:

Step1: Recall cumulative - distribution formula for exponential distribution

For an exponential probability density function $f(x)=\lambda e^{-\lambda x}, x\geq0$, the cumulative - distribution function $F(x_0)=P(X\leq x_0)=1 - e^{-\lambda x_0}$. Here $\lambda=\frac{1}{3}$.

Step2: Answer part (a)

The correct formula for $P(x\leq x_0)$ is $P(x\leq x_0)=1 - e^{-x_0/3}$, so the answer is (i).

Step3: Calculate $P(x\leq2)$ for part (b)

Substitute $x_0 = 2$ into $P(x\leq x_0)=1 - e^{-x_0/3}$. We get $P(x\leq2)=1 - e^{-2/3}\approx1 - 0.5134 = 0.4866$.

Step4: Calculate $P(x\geq3)$ for part (c)

Since $P(x\geq3)=1 - P(x\lt3)$ and $P(x\lt3)=P(x\leq3)$ for a continuous distribution. Using $P(x\leq x_0)=1 - e^{-x_0/3}$ with $x_0 = 3$, we have $P(x\leq3)=1 - e^{-3/3}=1 - e^{-1}\approx1 - 0.3679 = 0.6321$. Then $P(x\geq3)=e^{-1}\approx0.3679$.

Step5: Calculate $P(x\leq5)$ for part (d)

Substitute $x_0 = 5$ into $P(x\leq x_0)=1 - e^{-x_0/3}$. We get $P(x\leq5)=1 - e^{-5/3}\approx1 - 0.1889 = 0.8111$.

Step6: Calculate $P(2\leq x\leq5)$ for part (e)

Using the property $P(2\leq x\leq5)=P(x\leq5)-P(x\leq2)$. We know $P(x\leq5)\approx0.8111$ and $P(x\leq2)\approx0.4866$. So $P(2\leq x\leq5)=0.8111 - 0.4866 = 0.3245$.

Answer:

(a) (i)
(b) $0.4866$
(c) $0.3679$
(d) $0.8111$
(e) $0.3245$