Discrete and Continuous Distributions

In this project you will explore the calculation of probabilities for different distributions

1 Binomial Distributions

1.1 \(X\sim \text{Binom}(n=10, p=0.8)\)

Suppose \(X\sim \text{Binom}(n=10, p=0.8)\), then calculate the following: 1) \(P(X \le 3)\)

x <- pbinom(q = 3, size = 10, prob = 0.8 )
vals <- 0:3
y <- sum(dbinom(vals, size = 10, p=0.8))
x

## [1] 0.0008643584

2. \(P( 3\le X \le 7)\)

vals <- 3:7
y <- sum(dbinom(vals, size = 10, p=0.8))
y

## [1] 0.3221225

3. \(P (X\ge 4 )\)

vals <- 4:10
y <- sum(dbinom(vals, size = 10, p=0.8))
y

## [1] 0.9991356

4. Verify that \(P(X\ge 4) + P(X\le 3) =1\).

vals1 <- 4:10
vals2 <- 0:3
y1 <- sum(dbinom(vals1, size = 10, p=0.8))
y2 <- sum(dbinom(vals2, size = 10, p=0.8))

y1+y2

## [1] 1

1.2 \(X\sim \text{Binom}(n=15, p=0.7)\)

Suppose \(X\sim \text{Binom}(n=15, p=0.7)\), then calculate the following: 1) \(P(X \le 4)\)

vals = 0:4
x = sum(dbinom(vals, size = 15, p = 0.7))
x

## [1] 0.0006722341

2. \(P( 5\le X \le 9)\)

vals = 5:9
x= sum(dbinom(vals, size = 15, p = 0.7))
x

## [1] 0.2777063

3. \(P (X\ge 4 )\)

vals = 4:15
x = sum(dbinom(vals, size = 15, p = 0.7))
x

## [1] 0.9999083

4. Verify that \(P(X\ge 2) = 1 - P(X\le 1)\).

vals1 = 2:15
vals2 = 0:1
LHS = sum(dbinom(vals1, size = 15, p = 0.7))
RHS = 1 - sum(dbinom(vals2, size = 15, p = 0.7))
LHS - RHS

## [1] 0

2 Poisson Distribution

2.1 \(X\sim \text{Pois}(\lambda = 8)\)

Suppose \(X\sim \text{Pois}(\lambda=8)\), then calculate the following: 1) \(P(X \le 4)\)

vals <- 0:4
y<- sum(dpois(vals, lambda = 8))
y

## [1] 0.0996324

2. \(P( 5\le X \le 9)\)

vals <- 5:9
y<- sum(dpois(vals, lambda = 8))
y

## [1] 0.6169919

3. \(P (X\ge 4 )\)

vals <- 0:3
y<- sum(dpois(vals, lambda = 8))
1-y #need to calculate 1-P(Xle 3)

## [1] 0.9576199

2.2 \(X\sim \text{Pois}(\lambda = 5)\)

Suppose \(X\sim \text{Pois}(\lambda=5)\), then calculate the following: 1) \(P(X \le 3)\)

vals <- 0:3
y<- sum(dpois(vals, lambda = 5))
y

## [1] 0.2650259

2. \(P( 3\le X \le 7)\)

vals <- 3:7
y<- sum(dpois(vals, lambda = 5))
y

## [1] 0.7419763

3. \(P (X\ge 4 )\)

vals <- 0:3
y<- sum(dpois(vals, lambda = 5))
1-y

## [1] 0.7349741

3 Normal Distribution

3.1 \(X\sim N(\mu = 10, \sigma = 2)\)

Calculate the following probabilities: 1) \(P( 5\le X \le 9)\)

pnorm(9, mean = 10, sd= 2) - pnorm(5, mean = 10, sd = 2)

## [1] 0.3023279

2. \(P(X \ge 12)\)

1 - pnorm(12, mean =10, sd =2)

## [1] 0.1586553

3. \(P(X \le 7)\)

pnorm(7, mean = 10, sd = 2)

## [1] 0.0668072

4. Verify the empirical rule for this normal distribution. That is, show that \(P(8\le X \le 12) \equiv 0.68\), \(P(6\le X \le 14) \equiv 0.95\), and \(P(4\le X \le 16) \equiv 0.997\).

pnorm(12, mean = 10, sd = 2) - pnorm(8, mean = 10, sd=2)

## [1] 0.6826895

pnorm(14, mean = 10, sd = 2) - pnorm(6, mean = 10, sd=2)

## [1] 0.9544997

pnorm(16, mean = 10, sd = 2) - pnorm(4, mean = 10, sd=2)

## [1] 0.9973002

5. Calculate the critical value of for \(\alpha =0.05, 0.95, 0.001\)

qnorm(0.05, mean = 10, sd = 2, lower.tail = FALSE)

## [1] 13.28971

qnorm(0.95, mean = 10, sd = 2, lower.tail = FALSE)

## [1] 6.710293

qnorm(0.001, mean = 10, sd = 2, lower.tail = FALSE)

## [1] 16.18046

3.2 \(X\sim N(\mu = 15, \sigma = 2)\)

Calculate the following probabilities: 1) \(P( 11\le X \le 16)\)

pnorm(16, mean = 15, sd= 2) - pnorm(11, mean = 15, sd = 2)

## [1] 0.6687123

2. \(P(X \ge 12)\)

1 - pnorm(12, mean =15, sd =2)

## [1] 0.9331928

3. \(P(X \le 13)\)

pnorm(13, mean = 15, sd = 2)

## [1] 0.1586553

4. Verify the empirical rule for this normal distribution. That is, show that \(P(13\le X \le 17) \equiv 0.68\), \(P(11\le X \le 19) \equiv 0.95\), and \(P(9\le X \le 21) \equiv 0.997\).

pnorm(17, mean = 15, sd = 2) - pnorm(13, mean = 15, sd=2)

## [1] 0.6826895

pnorm(19, mean = 15, sd = 2) - pnorm(11, mean = 15, sd=2)

## [1] 0.9544997

pnorm(21, mean = 15, sd = 2) - pnorm(9, mean = 15, sd=2)

## [1] 0.9973002

5. Calculate the critical value of for \(\alpha =0.05, 0.95, 0.001\)

qnorm(0.05, mean = 15, sd = 2, lower.tail = FALSE)

## [1] 18.28971

qnorm(0.95, mean = 15, sd = 2, lower.tail = FALSE)

## [1] 11.71029

qnorm(0.001, mean = 15, sd = 2, lower.tail = FALSE)

## [1] 21.18046