Mixed Distribution Problem You design an insurance policy that pays a random amount Payment $= 1000 \cdot A$, where $A$ denotes the age at death. $A$ is assumed to have a continuous uniform distribution on $[50,110]$ (that is, a constant density function). Modify this policy to pay an amount Modified payment = $\begin{cases} 1000 \cdot A\; \text{if} \;A<90\\ 100,000 \;\text{if}\; A \geq 90 \end{cases}$
a) Find $E[X]$ and $\sigma$ of the original payment.
b) Find $E[X]$ and $\sigma$ of the modified payment.
I know the for the original payment the function jumps from 0-50 (like the step function) but I don't know how to compute $E[X]$ because my thinking for the 1st one was to just do $\dfrac{50}{110}(1000)\cdot 25.5 + \dfrac{60}{110}\cdot 1000 \displaystyle\int_{50}^{110} x$ which will give me my $E[X]$ for part a but that didn't work. So need some help understanding these sorts of problems.
 A: An unreasonable distribution for the time of death!
We do the modified version, since it is a little more complicated. We let $T$be the age at time of death. By assumption the density function $f(t)$ of $T$ is $\frac{1}{60}$ on the interval $[50,110]$ and $0$ elsewhere. 
The payment $X$ is $1000T$ if $50\le T\lt 90$, and $100000$ if $90\le T\le 110$. Let $g(t)=1000t$ if  $50\le T\lt 90$, and let $g(t)=100000$ if $90\le t\le 110$. then $X=g(T)$, and therefore
$$E(X)=\int_{50}^{110} g(t)f(t)\,dt.$$
Because of the shape of $g(t)$, we break up the integral into two parts, and get
$$E(X)=\int_{50}^{90} \frac{1000 t}{60}\,dt +\int_{90}^{110} \frac{100000}{60}\,dt.$$
The integrations are very easy. 
To find the standard deviation $\sigma$, we first find the variance $\sigma^2$ of $X$, and then take the square root. Recall that $\text{Var}(X)=E(X^2)-(E(X))^2$. We already know $E(X)$, so only need $E(X^2)$. This is
$$\int_{50}^{110} (g(t))^2f(t)\,dt.$$
Just like before, we need to break up the integral, and we obtain
$$E(X^2)=\int_{50}^{90} \frac{(1000 t)^2}{60}\,dt +\int_{90}^{110} \frac{(100000)^2}{60}\,dt.$$
The first problem is done in much the same way, except that there is no need to break up the integral. For $E(X)$, we integrate $\frac{1000 t}{60}$ from $50$ to $110$, and go through a similar procedure for $E(X^2)$. 
