Q.Two coins are to be flipped. The first coin will land on heads with probability $0.6$, the second with probability $0.5$. Assume that the results of the flips are independent, and let $X$ equal the total number of heads that result. The value of $E(X)$ is

Computes mean and variance of the given uniform-type density on $(0,l)$ using $E[X]$ and $E[X^2]-\left(E[X]\right)^2$.

1. $\text{Mean}=E[X]=\displaystyle\int_0^lx\,f(x)\,dx=\int_0^lx\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^2}2\right]_0^l=\dfrac1l\cdot\dfrac{l^2}2=\dfrac l2$.
2. $E[X^2]=\displaystyle\int_0^lx^2\,f(x)\,dx=\int_0^lx^2\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^3}3\right]_0^l=\dfrac1l\cdot\dfrac{l^3}3=\dfrac{l^2}3$. …

A constant has no spread around its mean, so its variance is identically zero, regardless of the constant's value.

1. Variance measures the expected squared deviation from the mean: $\text{Var}(X)=E[(X-E[X])^2]$. …

Since $\text{Var}(X)=E(X^2)-[E(X)]^2$, with $\text{Var}(X)=4$ and $E(X)=2$, we get $E(X^2)=8$.

1. The variance formula is $\text{Var}(X)=E(X^2)-[E(X)]^2$.
2. Given $\text{Var}(X)=4$ and $E(X)=2$. …

Solving the two expectation equations gives $C=-2$.

1. By linearity of expectation, $E(X+C)=E(X)+C=8$ and $E(X-C)=E(X)-C=12$.
2. Subtracting the second equation from the first: $(E(X)+C)-(E(X)-C)=8-12\Rightarrow 2C=-4$.
3. So $C=-2$. …