ENGR 200 – Chapter 3 Solutions (Set 1 – Required, Contextual)
  How to use: Each problem is written in an engineering or student-life context. First decide the model (Poisson, Binomial, etc.), then extract parameters from the story, then compute. All solutions include step-by-step math and Excel equivalents.
 
  Poisson • infer λQ1. A materials engineer measures surface flaws on wire. Historical data indicate an average of 2.3 flaws per mm. For a 5 mm segment, what is the expected count parameter λ?
  Solution
    Model: Poisson; parameter = rate × length.
    λ = (2.3 flaws/mm) × (5 mm)
    
    Excel: =2.3*5    →  11.5
   
  Poisson • P(exactly k)Q2. Using the same wire segment as Q1, what is the probability of exactly 10 flaws?
  Solution
    Inputs: λ = 11.5, k = 10.
    Poisson pmf:  P(X=k) = e^(−λ) · λ^k / k!
    
P(X=10) = e^(−11.5) · 11.5^10 / 10!
        ≈ 0.1129350709
Excel: =POISSON.DIST(10, 11.5, FALSE)  →  0.11293507
   
  Binomial meanQ3. A 3D printer produces parts with independent defect probability 5% per part. If a batch has 20 parts, what is the expected number of defects?
  Solution
    Model: Binomial(n=20, p=0.05).
    Mean of Binomial:  μ = n·p
    
    Excel: =20*0.05  →  1
   
  Binomial • P(exactly k)Q4. For the same 20-part batch, what is the probability that exactly 2 parts are defective?
  Solution
    Inputs: n = 20, p = 0.05, k = 2.
    Binomial pmf:  P(X=k) = C(n,k) · p^k · (1−p)^(n−k)
    
P(X=2) = C(20,2) · 0.05^2 · 0.95^18
       = 190 · 0.0025 · 0.3773536
       = 0.1886768013
Excel: =BINOM.DIST(2, 20, 0.05, FALSE)  →  0.18867680
   
  Binomial meanQ5. A fair die is rolled 10 times during a lab demo. What is the expected number of sixes?
  Solution
    Model: Binomial(n=10, p=1/6).
    μ = n·p
    
μ = 10 × (1/6) = 1.6666666667
Excel: =10*(1/6)  →  1.666666667
   
  Binomial • P(exactly k)Q6. In the same demo, what is the probability of exactly 3 sixes?
  Solution
    Inputs: n = 10, p = 1/6, k = 3.
    P(X=k) = C(n,k) · p^k · (1−p)^(n−k)
    
P(X=3) = C(10,3) · (1/6)^3 · (5/6)^7
       = 120 · (1/216) · (78125/279936)
       ≈ 0.1550453596
Excel: =BINOM.DIST(3, 10, 1/6, FALSE)  →  0.15504536
   
  Bernoulli guessingQ7. A student guesses on a 5-question multiple-choice quiz with 4 options each (one correct). What is the probability of getting all 5 correct by guessing?
  Solution
    Model: Independent Bernoulli trials, p = 1/4 per item.
    All 5 correct = (1/4)^5
    
P = (1/4)^5 = 1 / 1024 = 0.0009765625
Excel: =(1/4)^5  →  0.0009765625
   
  Binomial • P(exactly k)Q8. For the same quiz, what is the probability of getting exactly 2 correct by guessing?
  Solution
    Inputs: n = 5, p = 1/4, k = 2.
    P(X=k) = C(n,k) · p^k · (1−p)^(n−k)
    
P(X=2) = C(5,2) · (1/4)^2 · (3/4)^3
       = 10 · (1/16) · (27/64)
       = 270 / 1024
       = 0.263671875
Excel: =BINOM.DIST(2, 5, 1/4, FALSE)  →  0.263671875
   
  Poisson momentsQ9. A campus arrivals process is modeled as Poisson with mean 4. What is the variance?
  Solution
    Fact: For Poisson(λ), mean = variance = λ.
    
    Excel: =4  →  4
   
  Binomial momentsQ10. Suppose X is the number correct on a 10-item clicker quiz, with independent success probability 0.3 each question. What are the mean and variance of X?
  Solution
    Model: Binomial(n=10, p=0.3).
    E[X] = n·p,   Var[X] = n·p·(1−p)
    
μ = 10 × 0.3        = 3.0
σ² = 10 × 0.3 × 0.7 = 2.1
      Excel (mean): =10*0.3  →  3
      Excel (variance): =10*0.3*(1-0.3)  →  2.1