ENGR 200 – Chapter 3 Solutions (Set 1

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)

λ = 2.3 × 5 = 11.5

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

μ = 20 × 0.05 = 1.0

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.

Var(X) = λ = 4

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

ENGR 200 – Chapter 3 Solutions (Set 1 – Required, Contextual)