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$Let zeta be a point selected at random uniformly from the unit interval [0,1 Consider the random variable X=e^(zeta). (a) Sketch X as a function of zeta. (b) Find and plot the CDF of X. (c) Find the probability of the events {X > sqrte} and {e^(0.25) < X <= e^(0.75)}.$

Let $\zeta$ be a point selected at random uniformly from the unit interval $[0,1$ Consider the random variable $X=e^{\zeta}$ . $\newline$ (a) Sketch $X$ as a function of $\zeta$ . $\newline$ (b) Find and plot the CDF of $X$ . $\newline$ (c) Find the probability of the events $\{X>\sqrt{e}\}$ and $\left\{e^{0.25}<X \leq e^{0.75}\right\}$ .

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Write and solve direct variation equations

Full solution

Q. Let

\zeta

be a point selected at random uniformly from the unit interval

[0,1

Consider the random variable

X=e^{\zeta}

.

\newline

(a) Sketch

X

as a function of

\zeta

.

\newline

(b) Find and plot the CDF of

X

.

\newline

(c) Find the probability of the events

\{X>\sqrt{e}\}

and

\left\{e^{0.25}<X \leq e^{0.75}\right\}

.

Sketch X function of zeta: Step $1$ : Sketch $X$ as a function of $\zeta$ . Since $X = e^{\zeta}$ and $\zeta$ ranges from $0$ to $1$ , $X$ will increase exponentially from $e^0$ (which is $1$ ) to $e^1$ (which is $\zeta$ $0$ ).
Find and plot CDF: Step $2$ : Find and plot the CDF of $X$ . The CDF of $X$ , $F_X(x)$ , is $P(X \leq x)$ . Since $X = e^{\zeta}$ , we need to find $P(e^{\zeta} \leq x)$ . This is equivalent to finding $P(\zeta \leq \log(x))$ , because the exponential function is increasing. Thus, $F_X(x) = \log(x)$ for $x$ in $[1, e]$ .
Calculate $P(X > \sqrt{e})$ : Step $3$ : Calculate the probability $P(X > \sqrt{e})$ . We know that $\sqrt{e} = e^{0.5}$ . So, $P(X > \sqrt{e}) = 1 - P(X \leq \sqrt{e}) = 1 - F_X(e^{0.5}) = 1 - 0.5 = 0.5$ .
Calculate $P(e^{0.25} < X \leq e^{0.75})$ : Step $4$ : Calculate the probability $P(e^{0.25} < X \leq e^{0.75})$ . Using the CDF, $P(e^{0.25} < X \leq e^{0.75}) = F_X(e^{0.75}) - F_X(e^{0.25}) = 0.75 - 0.25 = 0.5$ .

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