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u = 0
E = 100
sd = sqrt((365/4))

# a)

a = 100

ans_a = pnorm( a - 100, mean = u, sd = sd, lower.tail = FALSE)
print(ans_a)
## [1] 0.5
# b) 
b = 110
ans_b = pnorm( b - 100, mean = u, sd = sd, lower.tail = FALSE)
print(ans_b)
## [1] 0.1475849
# c)
c = 120
ans_c = pnorm( c - 100, mean = u, sd = sd, lower.tail = FALSE)
print(ans_c)
## [1] 0.01814355
Calculate the expected value and variance of the binomial distribution using the moment generating function.
The binomial distribution is

\[f(x) = C(n, x)p^x (1-p)^{n-k} \] \[ M(t) = \sum_{x=0}^n e^{tx} C(n, x)p^x(1-p)^{n-x}\] \[ M(t) = \sum_{x=0}^n (pe^t)x} C(n, x)p^x(1-p)^{n-x}\] \[ M(t) = [(1-p) + pe^t]^n \] \[ M'(t) = n (pe^t) [(1-p) + pe^t]^{n-1}\] \[ M(0) = n (pe^0) [(1 - p) + pe^0]^{n-1}= np\]

\[ M''(t) = n(n - 1)(pe^t)^2[(1 - p) + pe^t]^{n - 2} + n(pe^t)[(1 - p) + pe^t]^{n - 1} \] \[ ??^2 = M''(0) - [M'(0)]^2 = n(n - 1)p^2 +np - (np)^2 = np(1 - p)\]

Calculate the expected value and variance of the exponential distribution using the moment generating function.

\[ M(t) = E [e^{tx}] = \int_{0}^{\infty} e^{tx} \lambda e^{-\lambda x}dx \] \[ = \lambda \int_{0}^{\infty} e^{(t- \lambda)x}dx = \frac{\lambda}{t- \lambda}[e^{(t-\lambda)x}]^{\infty}_0 \] \[ = \frac {\lambda}{\lambda - t}\]

\[ M(0) = \frac {1}{\lambda} \] \[??^2 = M''(0) - [M'(0)]^2 = \frac {1}{\lambda^2}\]