Continuous uniform distribution question The location of a mollusc is uniformly distributed with minimum -1 and maximum +2 metres from the high-tide line. Their daily energy intake (in kcals) is 1 + 25% of their squared location. **(a) What is the cdf of the daily energy intake?** I'm assuming that to calculate this, we need to find out (b) first, then use $F(x) = \int f(x)$ ? **(b) What is the pdf of the daily energy intake?** Knowing that the pdf of the distribution of mollusc is $f(x) = 1/(2-(-1)) = 1/3$ , how do I find the pdf of the daily energy intake? **(c) What is the expected daily energy intake?** Because the probability to find a mollusc with energy intakes that are in the 0 to 1 (and consequently, 0 to -1) zone is higher, this means that the expectation isn't simply $(2-(-1))/2$ , right?

Guide:

Let the energy be $$E= 1 + \frac14 X^2$$

Let $y \in (1,2)$,

Compute $Pr(E \le y)=Pr(1+\frac14X^2 \le y)$

Convert that expression to function of CDF of $X$ to evaluate it.

Edit:

\begin{align} Pr(E \le y) &= Pr(1+\frac14 X^2 \le y)\\\ &=Pr(X^2 \le 4(y-1)) \\\ &= Pr (-2\sqrt{y-1} \le X \le 2 \sqrt{y-1}) \\\ &= Pr(X \le 2 \sqrt{y-1})-Pr(X <-2\sqrt{y-1}) \\\ &= \begin{cases} Pr(X \le 2 \sqrt{y-1})-Pr(X <-2\sqrt{y-1}) & , y \in (1, \frac54) \\\ Pr(X \le 2 \sqrt{y-1}) - 0 &, y \in [ \frac54, 2)\end{cases} \\\ &= \begin{cases} \frac{2 \sqrt{y-1}+1}3-\frac{-2\sqrt{y-1}}3 & , y \in (1, \frac54) \\\ \frac{2 \sqrt{y-1}+1}3 &, y \in [ \frac54, 2)\end{cases} \\\ &= \begin{cases} \frac{2 \sqrt{y-1}+1}3-\frac{-2\sqrt{y-1}+1}3 & , y \in (1, \frac54) \\\ \frac{2 \sqrt{y-1}+1}3 &, y \in [ \frac54, 2)\end{cases} \\\ &= \begin{cases} \frac{4 \sqrt{y-1}}3 & , y \in (1, \frac54) \\\ \frac{2 \sqrt{y-1}+1}3 &, y \in [ \frac54, 2)\end{cases} \\\ \end{align}