n = 364 #rest of year
m = 0 #mean
EX = 100 #expected value
var = 1/4 #σ^2
barvar = n * var
barsd = sqrt(barvar)

\(>= 100\):

z = (100 - EX) / barsd
pnorm(z, lower.tail=FALSE)
## [1] 0.5

\(>= 110\):

z = (110 - EX) / barsd
pnorm(z, lower.tail=FALSE)
## [1] 0.1472537

\(>= 120\)

z = (120 - EX) / barsd
pnorm(z, lower.tail=FALSE)
## [1] 0.01801584

Given the binomial PMF:

\(P_x = (\frac{n}{x})p^x(1 - p)^{n - x}\)

Where \(P_x\) is the probability of \(x\) successes within \(n\) trials and () is the number of combinations of \(x\) in \(n\), and the Moment Generating Function:

\(M_x(t) = E(e^{tx})\)

Which is the expected value of \(x\) at moment \(t\), combining with the PMF of the binomial distribution yields the functions:

\(M(t) = \sum^{n}_{x=0}e^{tx} * (\frac{n}{x})p^x(1 - p)^{n - x}\)

Which simplifies to \(M(t) = (pe^t + 1 - p)^n\) when \(t\) is a real number.

To get the first moment, take the first derivative and evaluate at 0.

First derivative: \(M'(t) = npe^t(pe^t - p + 1)^{n-1}\)

Evaluate at 0: \(M'(0) = npe^0(pe^0 - p + 1)^{n-1} \Longrightarrow np(p - p + 1)^{n-1} \Longrightarrow np(1)^{n-1} \Longrightarrow np\)

So the expected value is \(np\).

For the variance, take the second derivative and evaluate at 0:

\(M"(t) = npe^t(pe^t - p + 1)^{n-2}(npe^t - p + 1)\)

Evaluate at 0:

\(M"(0) = npe^0(pe^0 - p + 1)^{n-2}(npe^0 - p + 1) \Longrightarrow np(np - p + 1)\)

So the variance is \(np(np - p + 1)\).

For the exponential distribution, given the PDF:

\(f(x) = λe^{-λx}\)

And the Moment Generating Function:

\(M(t)=\int^\inf_0 e^{tx}*λ e^{-λ x} = \frac{λ}{λ-t}, t < λ\)

We take the first derivative for the expected value at 0:

\(M'(t) = \frac{λ}{(λ-t)^2} \Longrightarrow\)

\(E[X] = M'(0) = \frac{λ}{(λ-0)^2} \Longrightarrow \frac{λ}{λ^2}\Longrightarrow \frac{1}{λ}\)

So the expected value is \(\frac{1}{λ}\).

We take the second derivative for the variance to get the second moment of the distribution:

\(E[X^2] = M''(t) = \int^∞_0 t^2λε^{-λχ}= \frac{2}{λ^2}\)

And the variance of the continuous random variable is caluclated as:

\(VAR[X] = E[X^2] - E[X]^2 \Longrightarrow\)

\(VAR[X] = \frac{2}{λ^2}- \frac{1}{λ^2} \Longrightarrow \frac{1}{λ^2}\)

So the variance is \(\frac{1}{λ^2}\).