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IT 223 -- Feb 23, 2026

Review Exercises

1. If you know the expected value of a random variable x, what is the expected value of the sum of n outcomes of this random variable x?
2. If you know the variance of a random variable x, what is the expected value of the sum of n independent outcomes of this random variable x?
3. What is the Law of Averages? What is the officlal name of the Law of Averages?

Averages of Random Variables

• The expected value of the average of n outcomes of a random variable:
  E(x) = E[(x₁ + ... + x_n) / n] = E(x₁ / n + ... + x_n / n)
      = E(x₁ / n) + ... + E(x_n / n) = E(x₁) / n + ... + E(x_n) / n
      = nE(x) / n = E(x)
  

  E(x) = E(x)

• The variance of the sum of n independent outcomes of a random variable:
  Var(x) = Var[(x₁ + ... + x_n) / n] = Var[x₁ / n + ... + x_n / n]
      = Var(x₁ / n) + ... + Var(x_n / n) = Var(x₁) / n² + ... + Var(x_n) / n²
      = = nVar(x) / n² = nVar(x)
  

  Var(x) = nVar(x)

• The standard deviation of the sum of n independent outcomes of a random variable:
  σ_x = √[Var(x)] = √[Var(x) / n] = σ_x / √n
  

  σx = σx / √n

• Recall that E(x) = E(x). How does σ_x compare to σ_x?

The Central Limit Theorem

• Even if a random variable x is not normally distributed, both the sum S and average x of n independent outcomes of a random variable are approximately normally distributed if n is large. Usually we say that n is large if n > 30.
• Example 24:   Flip a coin 1,000 times and obtain 514 heads. Find a 95% confidence interval for the true value p of obtaining a head. This will give us an idea if the coin is fair.
  Ans: Flipping a coin once is a Bernoulli random variable with P(X = 1) = p, so E(x) = p and σ_x = √p(1 - p). The sum of n independent Bernoulli random variables S is the number of heads; E(S) = np and σ_S = √np(1 - p). For computing σ_S, we don't know the value of p, so we estimate it with p^{^} = S / n = 514/1000 = 0.514. Therefore the estimated standard deviation is
  σ^{^}_S = √1000 * 0.514 * (1 - 0.514) = √249.804 = 15.81.
  Now a 95% confidence interval for z is -2 ≤ z ≤ 2. (More precisely, -1.96 ≤ z ≤ 1.96). We can solve for p:
       -2 ≤ z ≤ 2
       -2 ≤ (S - np) / σ^{^}_S ≤ 2
       -2 ≤ (514 - 1,000p) / 22.36 ≤ 2
       -44.72 ≤ 514 - 1,000p ≤ 44.72
       -44.72 - 514 ≤ - 1,000p ≤ 44.72 - 514
       -558.72 ≤ - 1,000p ≤ -469.28
  0.55872 ≥ p ≥ 0.46928
  This means that the 95% confidence interval is [0.46928, 0.55872] or [47%, 56%].

Factorials and Counting Combinations

• The number of ways of drawing k objects from an urn that contains n distinct objects is
  
• The binomial coefficient defined in the previous bullet point is also written as nCk and read "n choose k."
• The binomial coefficients can also be obtained from Pascal's Triangle:
  

                      1
                   1     1
                1     2     1
              1    3     3     1
           1    4     6     4     1
        1     5    10    10    5     1
      1    6    15    20    15    6     1
    1   7    21    35    35    21    7     1
  1   8   28    56    70    56    28    8     1
  

• Each entry in Pascal's triangle can be obtained recursively by adding the two entries above it.
• nCk can be obtained from in Pascal's Triangle as entry k in row n.
• Practice Problems: How many ways are there of choosing
  
  1. 2 items from 3?
     Answer: 3C2 = 3! / 2!(3 - 2)! = 3*2*1 / ((2*1)*1) = 3.
     The ways of choosing 2 items from 3 is (12), (13), (23): 3 ways.
     Using R:
     

     > choose(3, 2)
     [1] 3
     

  2. 3 items from 5?
     Answer: 5C3 = 5! / (3!(5 - 3)!) = (5*4*3*2*1) / ((3*2*1)(2*1) = 5*2 = 10.
     The ways of choosing 3 items from 5 is
     (123) (124) (125) (134) (135) (145) (234) (235) (245) (345):  10 ways.
     Using R:
     

     > choose(5, 3)
     [1] 10
     

  3. 5 items from 8?
     Answer: 8C5 = 8! / (5!(8 - 5)!) = (8*7*6*5*4*3*2*1) / ((5*4*3*2*1)(3*2*1)) = 8 * 7 = 56,
     Using R:
     

     > choose(8, 5)
     [1] 56
     

  4. 13 items from 20?
     Answer: 20C13 = 20! / (13!(20 - 13)!) = (20*19*18*17*16*15*14) / (7*6*5*4*3*2*1) = 19*17*16*15 = 77520,
     Using R:
     

     > choose(20, 13)
     [1] 77520
     

The Binomial Theorem

• Suppose you have a Bernoulli distribution with p = probability of success. What is the probability of k successes out of n trials? Ans:
  
  
  
• This formula can also be written as
  P(successes in k trials) = nCk p^k (1 - p)^n-k
• Practice Problems:
  
  1. Flip 5 coins. What is the probability of obtaining exactly 3 heads?
  2. Roll a die 10 times. What is the probability of rolling exactly 5 ones?
  3. Shoot 50 free throws, where the probability of success is 80%. What is the probability of hitting 45 or more free throws?
  4. The probability that the Cubs can beat Team A in a single game is 60%. What is the probability that they win 7 or more games in a 10 game series?

Finding a Confidence Interval for a Proportion

• Example 1:   Flip a coin 1,000 times and obtain 514 heads. Find a 95% confidence interval for the true value p of obtaining a head. This will give us an idea if the coin is fair.
  Ans: Flipping a coin once is a Bernoulli random variable with unknown probability of obtaining a head p. We also know that the sum S = 514, and the estimated probability of success
        p^{^} = S / n = 514 / 1,0000 = 0.514.
  We know that
        E(S) = nE(x) = np.
  Furthermore, we know that
        σ_S = √[np(1-p)].
  Since p is unknown we estimate it with p^{^} and obtain
      σ_S  = √[np^{^}(1-p^{^})] = √[1000(0.514)(1 - 0.514)] = 15.81.
  Finally, we compute the z-score
        z = (S - E(S)) / σ_S = (514 - np) / 15.81.
  To obtain z, we have scaled S by subtracting its expectation and dividing by its standard error.  This means that E(z) = 0 and σ_z = 1. By the Central Limit Theorem, z is a standard normal random variable if n > 30, so -2 ≤ z ≤ 2 95% of the time. (More precisely, -1.96 ≤ z ≤ 1.96 95% of the time. We can then solve this inequality for for p:
       -2 ≤ z ≤ 2
       -2 ≤ (S - np) / σ_S ≤ 2
       -2 ≤ (514 - 1,000p) / 15.81 ≤ 2
       -31.62 ≤ 514 - 1,000p ≤ 31.62
       -31.62 - 514 ≤ - 1,000p ≤ 31.62 - 514
       -545.62 ≤ - 1,000p ≤ -475.62
       0.54562 ≥ p ≥ 0.47562
  This means that the 95% confidence interval is
       [0.47562, 0.54562] or [48%, 55%].

Project 4

• Look at the Project 4 Description.
• Simulating repeated trials from a Bernoulli random variable using R:
       Method 1: Generate n values using rbinom(n, 1, p)
       Method 2: Generate one value using rbinom(1, n, p). The one value is the number of ones obtained, which is the sum of the outcomes.

Tests of Hypotheses for Proportions

• A test of hypothesis can be used to determine if a coin or die is fair.
• For testing if a coin is fair, we test the null hypothesis
        H₀: p = 0.5
  against the alternative hypothesis
        H₁: p ≠ 0.5.
• Here are the four steps for testing if the probability of success p for a Bernoulli random variable is equal to p₀:
  
  1. Write down the null and alternative hypothesis:
           H₀: p = p₀
           H₁: p ≠ p₀.
  2. Write down the test statistic, assuming the null hypothesis:
           z = (S - E(S)) / σ_S
  3. Write down a 95% confidence interval for z:
           I = [-1.96, 1.96]
  4. If z ∉ I, reject H₀; if z ∈ I, accept H₁.
• Example 2:   You flip a coin 100 times and obtain 43 heads. Is the coin fair?
  
  1. H₀: p = 0.5; H₁: p ≠ 0.5
  2. S = 43, E(S) = np, and σ_S = √[np(1-p)] = √[100(0.5)(1-0.5)] = 5.0, so
         z = (S - E(S)) / σ_S = (43 - 100(0.5)) / 5.0 = -1.4
  3. A 95% confidence interval for z is [-2, 2].
  4. -1.4 ∈ [-2, 2], so accept the null hypothesis. Not enough evidence to conclude that the coin is not fair.
• Example 4:   Use R to simulate 500 flips of a fair coin. Perform a test of hypotheses to see if the random number generator is fair. In addition to using R to generate S with rbinom, also use it to compute SE for S and z as well.
• Example 3: To test if a four-faced die (shape of a tetrahedron) is fair, such a die is rolled 1600 times. 421 aces are obtained. Perform a test of hypothesis to test whether the die is fair.
  Here are the steps to perform the test of hypothesis:
  
  1. Write down the null hypothesis:    H₀: p = 1/4
  2. Compute the text statistic:    z = (S - E(S)) / SE_S
  3. Write down a 95% confidence interval:    I = (-1.96, 1.96)
  4. If z ∈ I, accept the null hypothesis that the die is fair. If z ∉ I reject the null hypothesis.
• Answer:
  
  1. H₀: p = 1/4.
     
  2. S = 421   n = 1,600   SE(S) = sqrt(1,600 * (1/4) * (1-1/4)) = 17.32
     z = (S - np) / SE(S) = (421 - 1,600 (1/4)) / 17.32 = 1.212
     
  3. The 95% confidence interval using the standard normal table is (-1.96, 1.96)
     
  4. 1.212 ∈ (-1.96, 1.96), so accept the null hypothesis that the die is fair; we don't have enough evidence to reject the null hypothesis.
• Use R to simulate the following situations. Use the random number generator rbinom to simulate 500 outcomes of a Bernouilli random variable with p = 0.5.
• For each of the cases in Problem 9, test whether the random number generator is fair for the given probability.