Module 4 Practice Problems
20 Comprehensive Problems • Covers All Module 4 Topics
How to Use These Practice Problems
- Work through problems on paper before checking solutions
- Use the hints if you're stuck
- Problems are organized by topic and difficulty
- Show all your work - understanding the process is key!
Problems 1-5: Random Variables & Probability Distributions
Problem 1 Discrete vs ContinuousEasy
Classify each random variable as discrete (D) or continuous (C):
- Number of customers entering a store in one hour
- Weight of a randomly selected apple
- Number of defective items in a batch of 50
- Time until the next phone call
- Number of correct answers on a 20-question quiz
Solution:
a) Discrete - counting customers (0, 1, 2, ...)
b) Continuous - measuring weight (any positive value)
c) Discrete - counting defects (0, 1, 2, ..., 50)
d) Continuous - measuring time (any positive value)
e) Discrete - counting correct answers (0, 1, 2, ..., 20)
Problem 2 Valid Distribution?Easy
Is this a valid probability distribution? X: 0, 1, 2, 3 with P(X=x): 0.1, 0.4, 0.3, 0.2
Solution:
Yes, this is valid.
Check 1: All probabilities between 0 and 1? (0.1, 0.4, 0.3, 0.2 all satisfy this)
Check 2: Do they sum to 1? 0.1 + 0.4 + 0.3 + 0.2 = 1.0
Both conditions satisfied → valid distribution
Problem 3 Calculate Expected ValueMedium
Given the distribution: X: 1, 2, 3, 4 with P(X=x): 0.2, 0.3, 0.4, 0.1. Find E(X).
Solution:
E(X) = Σ [x · P(X=x)]
= 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1)
= 0.2 + 0.6 + 1.2 + 0.4
= 2.4
Interpretation: The expected (average) value is 2.4
Problem 4 Variance CalculationMedium
Using the same distribution from Problem 3, calculate Var(X) and σ.
Solution:
μ = 2.4 (from Problem 3)
Var(X) = (1-2.4)²(0.2) + (2-2.4)²(0.3) + (3-2.4)²(0.4) + (4-2.4)²(0.1)
= 1.96(0.2) + 0.16(0.3) + 0.36(0.4) + 2.56(0.1)
= 0.392 + 0.048 + 0.144 + 0.256
= 0.84
σ = √0.84 = 0.917
Problem 5 Create a DistributionMedium
A game involves rolling a die. You win $x where x is the number shown. Find the probability distribution and E(X).
Solution:
Probability Distribution:
| x | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| P(X=x) | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
Expected Winnings:
E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6)
= (1+2+3+4+5+6)/6 = 21/6 = $3.50
Problems 6-10: Binomial Conditions
Problem 6 Identify BinomialEasy
Which scenario is binomial? Explain why or why not.
A: Draw 10 cards from a deck without replacement, count aces.
B: Roll a die 15 times, count number of 6's.
Solution:
Scenario A: NOT binomial
- Fixed n = 10
- Two outcomes (ace/not)
- NOT independent (cards not replaced)
- Probability changes (P changes after each draw)
Scenario B: BINOMIAL
- Fixed n = 15
- Two outcomes (6 or not 6)
- Independent rolls
- Constant p = 1/6
Problem 7 Identify ParametersEasy
A basketball player makes 75% of free throws and shoots 12 times. If X = number made, identify n and p.
Solution:
n = 12 (number of trials/shots)
p = 0.75 (probability of success/make)
This is binomial: Fixed trials, two outcomes (make/miss), independent shots, constant 75% success rate.
Problem 8 Binomial ConditionsMedium
A student guesses on a 20-question true/false test. Verify this is binomial and state n and p.
Solution:
Verification:
- Fixed: n = 20 questions
- Two outcomes: correct or incorrect
- Independent: answering one doesn't affect others
- Constant: p = 0.5 (guessing gives 50% chance)
Yes, this is binomial with n = 20, p = 0.5
Problem 9 NOT BinomialMedium
Flip a coin until you get heads. Count number of flips. Is this binomial? Why or why not?
Solution:
NOT binomial.
Reason: The number of trials is NOT fixed. You keep flipping until you get heads, which could be 1 flip, 10 flips, or 100 flips. This violates the "fixed n" condition.
Note: This follows a geometric distribution instead.
Problem 10 Quality Control ScenarioMedium
A factory produces items with a 4% defect rate. Inspector randomly selects 30 items for testing. Is X = number of defects binomial? Identify n and p.
Solution:
Yes, binomial (assuming random selection with replacement or very large population).
n = 30 (items inspected)
p = 0.04 (defect rate)
Conditions satisfied: Fixed 30 trials, two outcomes (defect/OK), independent selections, constant 4% rate.
Problems 11-15: Binomial Probability Calculations
Problem 11 Exact ProbabilityMedium
Flip a fair coin 4 times. Find P(X = 2 heads).
Solution:
n = 4, p = 0.5, k = 2
C(4,2) = 4!/(2!2!) = 6
P(X = 2) = 6 × (0.5)² × (0.5)²
= 6 × 0.25 × 0.25
= 6 × 0.0625
= 0.375 or 37.5%
Problem 12 Multiple Choice TestHard
A 10-question multiple choice test has 4 choices per question. If a student guesses randomly, find P(exactly 3 correct).
Solution:
n = 10, p = 0.25, k = 3
C(10,3) = 10!/(3!7!) = 120
P(X = 3) = 120 × (0.25)³ × (0.75)⁷
= 120 × 0.015625 × 0.1335
= 120 × 0.002086
≈ 0.250 or 25.0%
Problem 13 Mean and SDMedium
For a binomial distribution with n = 50 and p = 0.3, find μ and σ.
Solution:
Mean: μ = np = 50 × 0.3 = 15
Standard Deviation:
σ = √(np(1-p)) = √(50 × 0.3 × 0.7) = √10.5 ≈ 3.24
Interpretation: Expect about 15 successes on average, with typical variation of 3.24.
Problem 14 At MostHard
Roll a die 3 times. Find P(at most 1 six). [P(X ≤ 1)]
Solution:
n = 3, p = 1/6 ≈ 0.1667
P(X = 0): C(3,0) × (1/6)⁰ × (5/6)³ = 1 × 1 × 125/216 = 125/216
P(X = 1): C(3,1) × (1/6)¹ × (5/6)² = 3 × (1/6) × (25/36) = 75/216
P(X ≤ 1): 125/216 + 75/216 = 200/216 ≈ 0.926 or 92.6%
Problem 15 At LeastHard
A treatment is 80% effective. If 5 patients receive it, find P(at least 4 improve). [P(X ≥ 4)]
Solution:
n = 5, p = 0.8
P(X = 4): C(5,4) × (0.8)⁴ × (0.2)¹ = 5 × 0.4096 × 0.2 = 0.4096
P(X = 5): C(5,5) × (0.8)⁵ × (0.2)⁰ = 1 × 0.32768 × 1 = 0.32768
P(X ≥ 4): 0.4096 + 0.32768 ≈ 0.737 or 73.7%
Problems 16-20: Applications
Problem 16 Technology UseMedium
For binomial with n=100, p=0.15, use technology notation to find P(X ≥ 20).
Solution:
TI-84: 1 - binomcdf(100, 0.15, 19)
Excel: =1 - BINOM.DIST(19, 100, 0.15, TRUE)
R: 1 - pbinom(19, 100, 0.15)
Result: ≈ 0.113 or 11.3%
Problem 17 Normal Approximation CheckMedium
Can you use normal approximation for n=25, p=0.6? Check conditions.
Solution:
Check: np = 25 × 0.6 = 15 ≥ 10
Check: n(1-p) = 25 × 0.4 = 10 ≥ 10
Yes, normal approximation is appropriate.
Use X ~ N(μ=15, σ=√6 ≈ 2.45)
Problem 18 Expected DefectsEasy
A factory produces 1000 items per day with 2% defect rate. How many defects expected per day?
Solution:
n = 1000, p = 0.02
μ = np = 1000 × 0.02 = 20 defects
σ = √(1000 × 0.02 × 0.98) = √19.6 ≈ 4.43 defects
Expect about 20 defects per day, with typical variation of 4-5 defects.
Problem 19 Probability RangeHard
For binomial n=8, p=0.4, find P(2 ≤ X ≤ 5).
Solution:
P(2 ≤ X ≤ 5) = P(X=2) + P(X=3) + P(X=4) + P(X=5)
Using binomial formula or technology:
P(X=2) = C(8,2)(0.4)²(0.6)⁶ ≈ 0.209
P(X=3) = C(8,3)(0.4)³(0.6)⁵ ≈ 0.279
P(X=4) = C(8,4)(0.4)⁴(0.6)⁴ ≈ 0.232
P(X=5) = C(8,5)(0.4)⁵(0.6)³ ≈ 0.124
Sum: 0.209 + 0.279 + 0.232 + 0.124 ≈ 0.844 or 84.4%
Problem 20 Comprehensive ProblemHard
A pharmaceutical company tests a drug on 60 patients. Historical data shows 70% success rate.
a) Verify binomial conditions and state n and p
b) Find expected number of successes and standard deviation
c) Find P(at least 45 patients improve)
Solution:
Part a: Binomial conditions verified:
- Fixed n = 60 patients
- Two outcomes (improve/don't improve)
- Independent (assuming each patient responds independently)
- Constant p = 0.7
Part b: Expected value and SD
μ = np = 60 × 0.7 = 42 patients
σ = √(np(1-p)) = √(60 × 0.7 × 0.3) = √12.6 ≈ 3.55 patients
Part c: P(X ≥ 45)
Check normal approximation: np=42≥10, n(1-p)=18≥10
Using technology: P(X ≥ 45) = 1 - binomcdf(60, 0.7, 44) ≈ 0.200 or 20.0%
Interpretation: About 20% chance that 45 or more patients improve.
Completed All 20 Problems!
Great work! You've practiced all major concepts from Module 4. Key skills mastered:
- Identifying discrete vs continuous random variables
- Calculating expected value and variance
- Verifying binomial conditions
- Using the binomial probability formula
- Finding binomial mean and standard deviation
- Applying technology and normal approximation
Ready for the quiz? Make sure you understand these solutions before moving on!