Probability and Statistics Lab 4 Part 1

Info

Objectives

• Understand and apply discrete probability distributions using R.

Prerequisites

• Probability distribution in R studio
• Cumulative Distribution Function
• Probability and Statistics Lecture 8
• Document P4-part1.pdf

Handwritten Solutions

Question 1

X	1	2	3	4	5	6	7
P(X)	k	2k	3k	k²	k² + k	2k²	4k²

We need to:

1. Find kk by ensuring the sum of all probabilities is 1.
2. **Compute $P(X<5)P(X<5)andP(1\le X\le 5)P(1\le X\le 5).$
3. Write an R program to plot the probability distribution and cumulative distribution.

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$$k+2k+3k+k^2+k^2+k+2k^2+4k^2=1$$

Let’s solve this equation step by step:

Combine like terms:

$$k+2k+3k+k+k^2+k^2+2k^2+4k^2=1$$ $$7k+8k^2=1$$

Rearrange the equation to standard quadratic form:

$$8k^2+7k-1=0$$

Now, solve the quadratic equation using the quadratic formula ( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):

$$a=8,b=7,c=-1$$ $$k=\frac{-7\pm \sqrt{7^2-4\cdot 8\cdot (-1)}}{2\cdot 8}$$ $$k=\frac{-7\pm \sqrt{49+32}}{16}$$ $$k=\frac{-7\pm \sqrt{81}}{16}$$ $$k=\frac{-7\pm 9}{16}$$

So the solutions are:

$$k=\frac{2}{16}=\frac{1}{8}\text{or}k=\frac{-16}{16}=-1$$

Since probability cannot be negative, we discard ( k = -1 ).

Thus, the valid solution is:

$$k=\frac{1}{8}$$

k %3C- 1/8
 
# Define the PDF of X
x <- c(1, 2, 3, 4, 5, 6, 7)
P_X <- c(k, 2*k, 3*k, k^2, k^2 + k, 2*k^2, 4*k^2)
 
# Calculate P(X < 5) and P(X <= 5)
P_X_less_than_5 <- sum(P_X[x < 5])
P_X_less_than_equal_5 <- sum(P_X[x <= 5])
 
# Print the results
cat("P(X < 5) =", P_X_less_than_5, "\n")
cat("P(X <= 5) =", P_X_less_than_equal_5, "\n")
 
# Plot the probability distribution
barplot(P_X, names.arg=x, main="Probability Distribution of X", xlab="X", ylab="P(X)")
 
# Plot the cumulative distribution
cumulative_P_X <- cumsum(P_X)
plot(x, cumulative_P_X, type="s", main="Cumulative Distribution of X", xlab="X", ylab="F(X)", ylim=c(0,1))
points(x, cumulative_P_X, pch=19)

 # Print the results
> cat("P(X < 5) =", P_X_less_than_5, "\n")
P(X < 5) = 0.765625 
> cat("P(X <= 5) =", P_X_less_than_equal_5, "\n")
P(X <= 5) = 0.906252
 

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Question 2

 
 
k <- 0.05  # chosen so that all probabilities are positive and sum < 1
 
# Define the probability values with both fixed values and terms using k
p <- c(0.1, k, 0.2, 2*k, 0.3, k, 0.2, 2*k, 0.3, 3*k)
 
# Normalize the probabilities to ensure they sum to 1 (very important)
p <- p / sum(p)
 
# Define the corresponding values of the discrete random variable X
x <- 1:10
 
# Calculate the cumulative distribution using pdiscrete()
# This returns cumulative probabilities F(x)
pl <- pdiscrete(x, probs = p, values = x)
 
# Plot the cumulative distribution
plot(pl,
     main = "Cumulative Distribution",
     xlab = "X",
     ylab = "Cumulative Probability",
     col = "red",
     pch = 19,
     type = "o")  # type = "o" means lines + points
 

Output

Question 3

 
# Define values of the discrete random variable
x <- 1:5
 
# Define corresponding probabilities
p <- c(1/5, 2/5, 2/15, 1/15, 1/5)
 
# Confirm that probabilities sum to 1
total_prob <- sum(p)
cat("Sum of probabilities:", total_prob, "\n")  # Should be 1
 
# Compute PMF and CDF using e1071 functions
pmf <- ddiscrete(x, probs = p, values = x)
cdf <- pdiscrete(x, probs = p, values = x)
 
# Plot the Probability Mass Function (PMF)
plot(pmf, type = "o", col = "coral",
     main = "Probability Distribution",
     xlab = "X", ylab = "P(X)", pch = 19)
 
# Plot the Cumulative Distribution Function (CDF)
plot(cdf, type = "o", col = "blue",
     main = "Cumulative Distribution",
     xlab = "X", ylab = "Cumulative Probability", pch = 19)
 

Output

Sum of probabilities: 1

### Question 4 ```R # Define values of the discrete random variable x <- c(-3, -2, -1, 0, 1, 2)

Define corresponding probabilities

p_x ← c(0.05, 0.1, 0.2, 0.3, 0.2, 0.15)

Compute the expected value (mean) of X

mean_x ← sum(x * p_x) cat(“Mean of X:”, mean_x, “\n”)

Compute the second moment E[X^2]

Ex2 ← sum((x^2) * p_x) cat(“E[X^2]:”, Ex2, “\n”)

Compute variance of X

variance_x ← Ex2 - (mean_x^2) cat(“Variance of X:”, variance_x, “\n”)

Plot the Probability Mass Function (PMF)

plot(x, p_x, col = “blue”, type = “o”, main = “Probability Distribution”, xlab = “X”, ylab = “P(X = x)”, pch = 19)

Compute Cumulative Distribution Function (CDF)

cdf ← cumsum(p_x)

Plot the CDF

plot(x, cdf, type = “o”, col = “red”, main = “Cumulative Distribution Function”, xlab = “X”, ylab = “F(X)”, pch = 19)

```bash
Mean of X: -0.05
E[X^2]: 1.85
Variance of X: 1.8475

Question 5

References