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# the Central Limit Theorem: Exercises 
# (from file Exercises_01-The central limit theorem.pdf)
###############################################################

# Problems "number of samples for a probability alpha of sample mean exceeding threshold"
# X - N(mu, sigma^2), X = mu + sigma * Z (Z standard normal)
# compute the 1-alpha quantile of Z, z.alpha
# the 1-alpha quantile of X will be x.alpha = mu + sigma * z.alpha
# taking the sample mean X bar with n samples, X bar - N(mu, sigma^2/n)
# therefore x.bar.alpha = mu + (sigma / sqrt(n)) * z.alpha
# setting threshold = mu + delta,
# we want to find n s.t.  mu + delta = x.bar.alpha
# solving the equation, we have n = ((sigma * z.alpha) / delta) ^ 2

mu <- 7
sigma <- 2

alpha <- 0.05
threshold <- 7.1
delta <- threshold - mu

z.alpha <- qnorm(1-alpha)
z.alpha
x.alpha <- mu + sigma * z.alpha
x.alpha
n <- 10
x.bar.alpha <- mu + (sigma / sqrt(n)) * z.alpha
x.bar.alpha

n = ((sigma * z.alpha) / delta) ^ 2
n

qnorm(1-alpha, mean=mu, sd=sigma/sqrt(n))

# EXERCISE 6
###############

mu <- 1.3
sigma <- 0.1
n <- 200

# CLT
var.x.bar <- sigma^2/n
sd.x.bar  <- sqrt(var.x.bar)

# a) Probability that the sample mean warpage exceeds 1.305 mm
1 - pnorm(1.305, mean = mu, sd = sd.x.bar)

# b) 25th percentile of the sample mean
qnorm(0.25, mean = mu, sd = sd.x.bar)

# c) Number of wafers to sample for a 0.05 probability of exceeding 1.305
w <- (qnorm(0.95, mean = 0, sd = sigma) / (1.305 - mu))^2
w
pnorm(1.305, mean=mu, sd=sigma/sqrt(w))

# EXERCISE 8
###############

mu <- 30.01
sd <- 0.1
n <- 50

# a) Probability that the total amount of solution in 50 drums is more than 1500 L
1 - pnorm(1500, mean = n * mu, sd = sd * sqrt(n))

# b) Probability that 80 drums can be filled without running out given the total amount is 2401 L
n <- 80
pnorm(2401, mean = n * mu, sd = sd * sqrt(80))

# c) Solution amount needed for a 0.9 probability that 80 drums can be filled without running out
qnorm(0.9, mean = n * mu, sd = sd * sqrt(80))


# EXERCISE 13
###############

# Given parameters
lambda.A <- 20  # mean rate for mass A
lambda.B <- 25  # mean rate for mass B
time.period <- 5

# a) Probability that fewer than 200 particles are emitted by both masses together in a five-minute time period

lambda.total <- time.period * (lambda.A + lambda.B)
lambda.total

# Exaxt solution
ppois(199, lambda = lambda.total)

# Normal approximation via CLT
sd.total <- sqrt(time.period * (lambda.A + lambda.B))  # Using normal approximation

pnorm(200, mean = lambda.total, sd = sd.total)


# b) Probability that mass B emits more particles than mass A in a two-minute time period
lambda.diff <- 2 * (lambda.B - lambda.A)
sd.diff <- sqrt(2 * (lambda.A + lambda.B))  # Using normal approximation

1 - pnorm(0, mean = lambda.diff, sd = sd.diff)


# EXERCISE 16
###############

mu <- 40
sd <- 5
n <- 100
sd.x.bar <- sd / sqrt(n)

# a) Probability that X is less than or equal to 36.7 hours
pnorm(36.7, mean=mu, sd=sd.x.bar)

# d) Probability that X is less than or equal to 39.8 hours
pnorm(39.8, mean=mu, sd=sd.x.bar)


# EXERCISE 18
###############

mu1 <- 0.5
sd1 <- 0.4
mu2 <- 0.6
sd2 <- 0.5
n <- 100

# a) Probability that the total time used by machine 1 is greater than 55 hours
mu.t1 <- n * mu1
sd.t1 <- sqrt(n) * sd1

1 - pnorm(55, mean = mu.t1, sd = sd.t1)

# b) Probability that the total time used by machine 2 is less than 55 hours
mu.t2 <- n * mu2
sd.t2 <- sqrt(n) * sd2

pnorm(55, mean = mu.t2, sd = sd.t2)

# c) Probability that the total time used by both machines together is greater than 115 hours
mu.sum <- n * (mu1 + mu2)
sd.sum <- sqrt(n) * sqrt(sd1^2 + sd2^2)

1 - pnorm(115, mean = mu.sum, sd = sd.sum)

# d) Probability that the total time used by machine 1 is greater than the total time used by machine 2
mean.diff <- mu.t1 - mu.t2
sd.diff <- sqrt(sd.t1^2 + sd.t2^2)

1 - pnorm(0, mean = mean.diff, sd = sd.diff)