ENGR 200 – Chapter 3 Solutions (Set 2

Goal: Each problem is phrased in an engineering or student-life context. You must infer the correct model and inputs before computing. The solutions show: identify model → extract parameters → compute.

Excel (Microsoft 365): use POISSON.DIST, BINOM.DIST, NEGBINOM.DIST. Older Excel may use legacy names like POISSON or BINOMDIST.

Poisson • parameter from textQ1. The campus IT help desk logs ticket arrivals. Their data show an average of about 3 tickets per hour during midterms. During a randomly chosen hour, what is the probability exactly 5 tickets arrive?

Solution

Identify model: Random independent arrivals at a constant rate ⇒ Poisson.

Extract inputs: mean per hr λ = 3; we want P(X=5).

Poisson pmf: P(X=k) = e^(−λ) · λ^k / k!

P(X=5) = e^(−3) · 3^5 / 5!
       = e^(−3) · 243 / 120
       ≈ 0.049787068 · 2.025
       = 0.10081881

Excel: =POISSON.DIST(5, 3, FALSE)

Poisson tailQ2. For the same help desk hour, what is the probability of at least 5 tickets?

Solution

Model: Poisson(λ=3).

Tail via CDF: P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − Σ (k=0..4) [ e^(−3) · 3^k / k! ]

P(X≤4) = 0.049787068 + 0.149361205 + 0.224041807 + 0.224041807 + 0.168031355
       = 0.815263245
P(X≥5) = 1 − 0.815263245
       = 0.18473676

Excel: =1 - POISSON.DIST(4, 3, TRUE)

Binomial (Exact) + Poisson (Approx)Q3. A student team inspects 100 3D-printed clips. Past data show a defect rate of about 2% per clip, independently. What is the probability the batch has 3 or more defects?

Solution

Identify model: Fixed n with independent “defect / no defect” ⇒ Binomial(n=100, p=0.02). Poisson(λ=np=2) is a common approximation when p is small and n is large.

Exact (Binomial): P(X ≥ 3) = 1 − [P(0)+P(1)+P(2)].

Binomial pmf: P(X=k) = C(n,k)·p^k·(1−p)^(n−k)

P(0) = C(100,0)(0.02)^0(0.98)^100 = 0.13261956
P(1) = C(100,1)(0.02)^1(0.98)^99 = 0.27085420
P(2) = C(100,2)(0.02)^2(0.98)^98 = 0.27321222
Sum  = 0.67668598
P(X≥3) = 1 − 0.67668598 = 0.32331402

Excel — Binomial (Modern):
=1 - BINOM.DIST(2, 100, 0.02, TRUE)
Or directly: =1 - (BINOM.DIST(0,100,0.02,FALSE) + BINOM.DIST(1,100,0.02,FALSE) + BINOM.DIST(2,100,0.02,FALSE))

Excel — Binomial (Legacy):
=1 - BINOMDIST(2, 100, 0.02, TRUE)

Approximation (Poisson, λ = np = 2): P(X ≥ 3) ≈ 1 − [P(0)+P(1)+P(2)].

P(0) = e^(−2) · 2^0/0! = 0.135335283
P(1) = e^(−2) · 2^1/1! = 0.270670566
P(2) = e^(−2) · 2^2/2! = 0.270670566
Sum  = 0.676676415
P(X≥3) ≈ 1 − 0.676676415 = 0.32332358
(very close to the exact binomial)

Excel — Poisson (Modern):
=1 - POISSON.DIST(2, 2, TRUE)

Excel — Poisson (Legacy):
=1 - POISSON(2, 2, TRUE)

GeometricQ4. A robotics team tests a flaky sensor by flipping a fair coin until the first head to simulate a “success” event. What is the probability it takes exactly 4 flips?

Solution

Model: Geometric with success p=1/2 each flip.

Geometric pmf: P(T=k) = (1−p)^(k−1) · p

P(T=4) = (1/2)^3 · (1/2) = (1/2)^4 = 0.0625

Excel: =NEGBINOM.DIST(4-1, 1, 0.5, FALSE)

Geometric expectationQ5. You try to log in to a busy lab server; each try succeeds with probability 0.2 (independent). What is the expected number of tries until the first success?

Solution

Model: Geometric(p=0.2).

Expectation: E[T] = 1/p

E[T] = 1 / 0.2 = 5 tries

Excel: =1/0.2

Negative BinomialQ6. An electronics lab repeats a pass/fail soldering test with success probability 0.4 each attempt, independently. What is the probability the 3rd success occurs on the 7th attempt?

Solution

Model: NegBin for r=3 successes; P(T=7) = C(6,2)p^3(1-p)^4.

C(6,2) = 15
p^3    = 0.4^3 = 0.064
(1-p)^4= 0.6^4 = 0.1296
P(T=7) = 15 × 0.064 × 0.1296 = 0.12441600

Excel: =NEGBINOM.DIST(4, 3, 0.4, FALSE) // failures before 3rd success
Excel: =COMBIN(6,2)*(0.4^3)*(0.6^4)

BinomialQ7. A quality-control die used in a game lab lands on six with probability 0.1 (independent across rolls). In 15 rolls, what is the probability of exactly 2 sixes?

Solution

Model: Binomial(n=15, p=0.1).

Binomial pmf: P(X=k) = C(n,k) · p^k · (1−p)^(n−k)

P(X=2) = C(15,2) · 0.1^2 · 0.9^13
       = 105 · 0.01 · 0.254186582
       = 0.26689591

Excel: =BINOM.DIST(2, 15, 0.1, FALSE)

Poisson CDFQ8. The campus call center logs a mean of 12 calls per hour around registration deadlines. During a random hour that week, what is the probability there are fewer than 10 calls?

Solution

Model: Poisson(λ=12). “Fewer than 10” means X ≤ 9.

CDF to 9: P(X ≤ 9) = Σ (k=0..9) [ e^(−12) · 12^k / k! ]

P(X ≤ 9) = 0.24239216   (from summation / CDF)

Excel: =POISSON.DIST(9, 12, TRUE)

Binomial momentsQ9. In a programming quiz of 20 auto-graded items, a student answers each correctly with probability 0.4 (independent). What are the mean and variance of the total correct?

Solution

Model: Binomial(n=20, p=0.4).

Moments: E[X] = np ; Var[X] = np(1−p)

μ = 20 × 0.4        = 8.0
σ² = 20 × 0.4 × 0.6 = 4.8

Excel mean: =20*0.4 | Excel variance: =20*0.4*(1-0.4)

Poisson • mean=varianceQ10. A laser cutter queue averages 9 jobs per hour. Assuming a Poisson model, what is the probability the next hour sees exactly 9 job submissions?

Solution

Model: Poisson(λ=9).

P(X=k) = e^(−λ) · λ^k / k!

P(X=9) = e^(−9) · 9^9 / 9!
       = 0.0001234098 · (387,420,489 / 362,880)
       = 0.13175564

Excel: =POISSON.DIST(9, 9, FALSE)

ENGR 200 – Chapter 3 Solutions (Set 2 – Advanced, Contextual)