STATS 17 Practice Quiz

60 Questions Covering All Learning Objectives

Author

Dr. Marcela Alfaro Córdoba

Published

28 April 2026

0.1 📚 How to Use This Practice Quiz

  • Answer each question before revealing the solution
  • Check the explanation to understand the concept
  • Track your progress - try to get at least 80% correct!
  • Focus on weak areas - multiple choice questions are organized by learning objective, and there’s a section at the end to calculate probabilities.
  • Write your asnwers on paper - the exam will include a free response section where you will have to show your work.
  • Use the reference formulas provided in the Midterm Guide - they are the same as the ones you will have access in the exam.
  • Take it multiple times - repetition helps retention!

1 Section 1: Basic Concepts & Definitions

1.1 Question 1

Learning Objective: Basic Concepts

You’re analyzing data from a TikTok survey about favorite pizza toppings. The average number of toppings chosen by all 50,000 respondents is:

  1. A statistic
  2. A parameter
  3. A sample
  4. A population

Correct Answer: A

Explanation: This is a statistic because it’s calculated from the sample of 50,000 respondents, not the entire population of all possible people.

1.2 Question 2

Learning Objective: Sampling Methods

Netflix wants to study binge-watching habits. They divide users by age group (Gen Z, Millennial, Gen X, Boomer) and randomly select 100 from each. This is:

  1. Simple random sampling
  2. Stratified sampling
  3. Cluster sampling
  4. Systematic sampling

Correct Answer: B

Explanation: Stratified sampling divides the population into groups (strata) and samples from each group.

1.3 Question 3

Learning Objective: Types of Variables

Which variable is discrete quantitative?

  1. Speed of a car in mph
  2. Number of Instagram likes on a post
  3. Temperature of coffee in °F
  4. Time spent scrolling TikTok

Correct Answer: B

Explanation: Likes are countable whole numbers, making them discrete.

1.4 Question 4

Learning Objective: Research Ethics

A dating app secretly tests a new matching algorithm on half its users without telling them. This violates:

  1. Random assignment
  2. Statistical significance
  3. Informed consent
  4. Placebo effect

Correct Answer: C

Explanation: Informed consent requires participants be informed about experiments and give permission.

1.5 Question 5

Learning Objective: Sources of Bias

To study college students’ stress levels, you survey people at the gym at 6am. The main concern is:

  1. Measurement bias
  2. Selection bias
  3. Response bias
  4. Sampling error

Correct Answer: B

Explanation: Selection bias - 6am gym-goers aren’t representative of all students.

1.6 Question 6

Learning Objective: Sources of Bias

A researcher studying sleep patterns asks: “You don’t get enough sleep, do you?” This is an example of:

  1. Selection bias
  2. Nonresponse bias
  3. Response bias
  4. Undercoverage

Correct Answer: C

Explanation: Response bias - the leading question influences how people respond.

1.7 Question 7

Learning Objective: Parameter vs. Statistic

The true proportion of all Gen Z adults who prefer text over calls is a:

  1. Statistic
  2. Parameter
  3. Variable
  4. Sample

Correct Answer: B

Explanation: A parameter describes the entire population (all Gen Z adults).

1.8 Question 8

Learning Objective: Sampling Methods

A fitness tracker study divides a city into neighborhoods, randomly selects 5 neighborhoods, then surveys everyone in those 5. This is:

  1. Stratified sampling
  2. Cluster sampling
  3. Simple random sampling
  4. Systematic sampling

Correct Answer: B

Explanation: Cluster sampling randomly selects entire groups (clusters) to survey completely.

2 Section 2: Data Displays

2.1 Question 9

Learning Objective: Interpreting Displays

A histogram of video game scores shows most scores clustered on the left with a long tail to the right. This is:

  1. Right-skewed
  2. Left-skewed
  3. Symmetric
  4. Bimodal

Correct Answer: A

Explanation: When the tail extends to the right, the distribution is right-skewed.

2.2 Question 10

Learning Objective: Choosing Displays

Which plot would best show the relationship between hours studied and exam scores?

  1. Histogram
  2. Scatterplot
  3. Pie chart
  4. Bar chart

Correct Answer: B

Explanation: Scatterplots show relationships between two quantitative variables.

2.3 Question 11

Learning Objective: Boxplot Interpretation

A boxplot shows a longer whisker above the box than below. This suggests:

  1. Left-skewed distribution
  2. Right-skewed distribution
  3. Symmetric distribution
  4. Uniform distribution

Correct Answer: B

Explanation: A longer upper whisker indicates right skewness.

2.4 Question 12

Learning Objective: Choosing Displays

You want to show the proportion of students choosing each major. The best display is:

  1. Histogram
  2. Scatterplot
  3. Pie chart or bar chart
  4. Boxplot

Correct Answer: C

Explanation: Pie charts and bar charts show proportions of categorical data effectively.

3 Section 3: Measures of Central Tendency

3.1 Question 13

Learning Objective: Central Tendency

Daily coffee sales: 45, 50, 52, 55, 200. Which measure best represents typical sales?

  1. Mean
  2. Median
  3. Mode
  4. Range

Correct Answer: B

Explanation: The median is resistant to the outlier (200).

3.2 Question 14

Learning Objective: Mode

Dataset: 8, 10, 10, 12, 15. What is the mode?

  1. 11
  2. 10
  3. 12
  4. No mode

Correct Answer: B

Explanation: The mode is 10 (appears twice).

3.3 Question 15

Learning Objective: Median

For the data: 3, 7, 9, 11, 15, the median is:

  1. 7
  2. 9
  3. 11
  4. Cannot determine

Correct Answer: B

Explanation: The median is the middle (3rd) value: 9.

3.4 Question 16

Learning Objective: Properties of Mean

If every value in a dataset is increased by 5, the mean:

  1. Stays the same
  2. Increases by 5
  3. Increases by 5n
  4. Doubles

Correct Answer: B

Explanation: Adding a constant shifts the mean by that constant.

3.5 Question 17

Learning Objective: Sigma Notation

What does Σ(xᵢ - x̄) always equal for any dataset?

  1. The variance
  2. The standard deviation
  3. Zero
  4. The mean

Correct Answer: C

Explanation: The sum of deviations from the mean always equals zero.

4 Section 4: Measures of Spread

4.1 Question 18

Learning Objective: Standard Deviation

Two classes have mean score 80. Class A has SD = 15, Class B has SD = 5. Which is true?

  1. Class A has more consistent scores
  2. Class B has more consistent scores
  3. Both have equal consistency
  4. Cannot determine

Correct Answer: B

Explanation: Smaller SD means more consistency.

4.2 Question 19

Learning Objective: Range

For data: 20, 25, 30, 35, 40, what is the range?

  1. 10
  2. 15
  3. 20
  4. 40

Correct Answer: C

Explanation: Range = 40 - 20 = 20.

4.3 Question 20

Learning Objective: IQR

The IQR (Interquartile Range) represents:

  1. The middle 25% of data
  2. The middle 50% of data
  3. The middle 75% of data
  4. All the data

Correct Answer: B

Explanation: IQR = Q3 - Q1 captures the middle 50%.

4.4 Question 21

Learning Objective: Properties of SD

If all values in a dataset are multiplied by 3, the standard deviation:

  1. Stays the same
  2. Is multiplied by 3
  3. Is multiplied by 9
  4. Is divided by 3

Correct Answer: B

Explanation: Multiplying all values by a constant multiplies the SD by that constant.

4.5 Question 22

Learning Objective: IQR Calculation

A dataset has Q1 = 10 and Q3 = 30. The IQR is:

  1. 10
  2. 20
  3. 30
  4. 40

Correct Answer: B

Explanation: IQR = 30 - 10 = 20.

4.6 Question 23

Learning Objective: Skewness

For a left-skewed distribution, which relationship typically holds?

  1. Mean < Median
  2. Mean > Median
  3. Mean = Median
  4. Cannot determine

Correct Answer: A

Explanation: In left-skewed distributions, the mean is pulled toward the tail (left).

5 Section 5: Probability Fundamentals

5.1 Question 24

Learning Objective: Probability Basics

The probability of any event must be:

  1. Between -1 and 1
  2. Between 0 and 1
  3. Between 0 and 100
  4. Greater than 0

Correct Answer: B

Explanation: Probabilities must be between 0 and 1, inclusive.

5.2 Question 25

Learning Objective: Complement Rule

If P(A) = 0.7, then P(not A) equals:

  1. 0.3
  2. 0.7
  3. -0.7
  4. 1.4

Correct Answer: A

Explanation: P(not A) = 1 - 0.7 = 0.3.

5.3 Question 26

Learning Objective: Mutually Exclusive Events

Events A and B are mutually exclusive. This means:

  1. They can both happen
  2. They cannot both happen
  3. One causes the other
  4. P(A|B) = P(A)

Correct Answer: B

Explanation: Mutually exclusive events cannot occur simultaneously.

5.4 Question 27

Learning Objective: Independent Events

Two events are independent if:

  1. P(A AND B) = 0
  2. P(A AND B) = P(A) × P(B)
  3. P(A OR B) = P(A) + P(B)
  4. They are mutually exclusive

Correct Answer: B

Explanation: Independence: P(A AND B) = P(A) × P(B).

5.5 Question 28

Learning Objective: Addition Rule

P(A) = 0.5, P(B) = 0.3, P(A AND B) = 0.1. What is P(A OR B)?

  1. 0.6
  2. 0.7
  3. 0.8
  4. 0.9

Correct Answer: B

Explanation: P(A OR B) = 0.5 + 0.3 - 0.1 = 0.7.

5.6 Question 29

Learning Objective: Multiplication Rule

You flip a fair coin twice. P(two heads) = ?

  1. 0.25
  2. 0.5
  3. 0.75
  4. 1.0

Correct Answer: A

Explanation: 0.5 × 0.5 = 0.25.

5.7 Question 30

Learning Objective: Conditional Probability

If P(A AND B) = 0.2 and P(B) = 0.4, what is P(A|B)?

  1. 0.2
  2. 0.4
  3. 0.5
  4. 0.8

Correct Answer: C

Explanation: P(A|B) = 0.2 / 0.4 = 0.5.

5.8 Question 31

Learning Objective: Basic Probability

Rolling a die, P(rolling less than 3) = ?

  1. 1/6
  2. 2/6
  3. 3/6
  4. 4/6

Correct Answer: B

Explanation: Values {1, 2} give P = 2/6.

6 Section 6: Contingency Tables

6.1 Question 32

Learning Objective: Contingency Tables

In a 2×2 table with totals 100, if the top-left cell is 30 and row total is 60, the top-right cell is:

  1. 20
  2. 30
  3. 40
  4. 70

Correct Answer: B

Explanation: 60 - 30 = 30.

6.2 Question 33

Learning Objective: Marginal Probability

A marginal probability is found by:

  1. Dividing a cell by row total
  2. Dividing a cell by column total
  3. Dividing a row/column total by grand total
  4. Multiplying probabilities

Correct Answer: C

Explanation: Marginal probabilities use row/column totals.

6.3 Question 34

Learning Objective: Joint Probability

A joint probability in a contingency table represents:

  1. P(Row category)
  2. P(Column category)
  3. P(Row AND Column)
  4. P(Row | Column)

Correct Answer: C

Explanation: Joint = cell / grand total.

6.4 Question 35

Learning Objective: Conditional Probability from Tables

To find P(Purchase | Urban) from a table, you:

  1. Divide Purchase total by grand total
  2. Divide Urban total by grand total
  3. Divide (Urban AND Purchase) by Urban total
  4. Divide (Urban AND Purchase) by Purchase total

Correct Answer: C

Explanation: Conditional on Urban = cell / Urban row total.

7 Section 7: Binomial Distribution

7.1 Question 36

Learning Objective: Binomial Distribution

Which scenario fits a binomial distribution?

  1. Time until next customer arrives
  2. Number of heads in 10 coin flips
  3. Height of random students
  4. Number of customers per hour

Correct Answer: B

Explanation: Fixed trials, two outcomes, constant probability, independence.

7.2 Question 37

Learning Objective: Expected Value - Binomial

For X ~ Binomial(n=15, p=0.4), what is E(X)?

  1. 4
  2. 6
  3. 9
  4. 15

Correct Answer: B

Explanation: E(X) = 15 × 0.4 = 6.

7.3 Question 38

Learning Objective: Standard Deviation - Binomial

For X ~ Binomial(n=20, p=0.5), what is SD(X)?

  1. 2.24
  2. 5
  3. 10
  4. 20

Correct Answer: A

Explanation: SD(X) = √(20×0.5×0.5) = √5 ≈ 2.24.

7.4 Question 39

Learning Objective: Choosing Distribution

80% of customers are satisfied. You survey 12 customers. Which distribution models number satisfied?

  1. Binomial(12, 0.8)
  2. Poisson(12)
  3. Normal(12, 0.8)
  4. Uniform(0, 12)

Correct Answer: A

Explanation: Binomial(12, 0.8) fits perfectly.

7.5 Question 40

Learning Objective: Binomial Properties

In a binomial distribution, as n increases while p stays constant, what happens to E(X)?

  1. Decreases
  2. Stays the same
  3. Increases
  4. Becomes 0

Correct Answer: C

Explanation: E(X) = np increases with n.

8 Section 8: Poisson Distribution

8.1 Question 41

Learning Objective: Poisson Distribution

Which scenario best fits a Poisson distribution?

  1. Number of heads in 10 flips
  2. Number of emails per hour
  3. Whether a customer buys or not
  4. Height of adults

Correct Answer: B

Explanation: Poisson models random counts over time/space.

8.2 Question 42

Learning Objective: Poisson Distribution

For X ~ Poisson(λ=5), what is E(X)?

  1. 2.24
  2. 5
  3. 25
  4. Cannot determine

Correct Answer: B

Explanation: For Poisson, E(X) = λ = 5.

8.3 Question 43

Learning Objective: Poisson Distribution

For a Poisson distribution, the variance equals:

  1. λ²
  2. λ
  3. √λ
  4. 1/λ

Correct Answer: B

Explanation: Var(X) = λ (same as mean).

9 Section 9: Expected Values & Variance

9.1 Question 44

Learning Objective: Expected Value Functions

If E(X) = 10 and Y = 3X + 7, what is E(Y)?

  1. 30
  2. 37
  3. 40
  4. 310

Correct Answer: B

Explanation: E(Y) = 3(10) + 7 = 37.

9.2 Question 45

Learning Objective: Variance Functions

If Var(X) = 16 and Y = 2X + 5, what is Var(Y)?

  1. 16
  2. 32
  3. 37
  4. 64

Correct Answer: D

Explanation: Var(Y) = 2²(16) = 64.

10 Section 10: Continuous Distributions

10.1 Question 46

Learning Objective: Continuous PDF

For a continuous distribution, the total area under the probability density curve equals:

  1. 0
  2. 0.5
  3. 1
  4. Infinity

Correct Answer: C

Explanation: Total probability = 1.

10.2 Question 47

Learning Objective: Discrete vs Continuous

Which is a continuous random variable?

  1. Number of customers
  2. Exact time to run a marathon
  3. Number of defects
  4. Count of emails

Correct Answer: B

Explanation: Time can take any value (continuous).

10.3 Question 48

Learning Objective: Uniform Distribution

For a uniform distribution on [0, 10], what is P(X ≤ 5)?

  1. 0.25
  2. 0.5
  3. 0.75
  4. 1.0

Correct Answer: B

Explanation: P = 5/10 = 0.5.

11 Section 11: Normal Distribution

11.1 Question 49

Learning Objective: Z-scores

A z-score of 2.5 means the value is:

  1. 2.5 below the mean
  2. 2.5 above the mean
  3. 2.5 times the mean
  4. At the 2.5th percentile

Correct Answer: B

Explanation: Positive z = above mean by that many SDs.

11.2 Question 50

Learning Objective: Empirical Rule

For a normal distribution, approximately 68% of values fall within:

  1. 1 SD of mean
  2. 2 SD of mean
  3. 3 SD of mean
  4. The median

Correct Answer: A

Explanation: 68-95-99.7 rule: 68% within 1 SD.

12 Calculating and Interpreting Probabilities

12.1 Question 51

Scenario: A telemarketing company knows that 15% of people they call will agree to a survey. They plan to call 200 people today.

Questions:

  1. What is the expected number of people who will agree to the survey?

  2. What is the standard deviation of the number of people who agree?

  3. Interpret what the expected value means in practical terms.

  4. If they need at least 25 people to complete the survey for their research to be valid, does the expected value suggest they’ll likely meet this goal? Explain.

Show Solution

Solution:

Part a) Expected Value:

For a binomial distribution: \(E(X) = np\)

\(E(X) = 200 \times 0.15 = 30\)

Answer: E(X) = 30 people

Part b) Standard Deviation:

\(SD(X) = \sqrt{np(1-p)} = \sqrt{200 \times 0.15 \times 0.85}\)

\(SD(X) = \sqrt{25.5} = 5.05\)

Answer: SD(X) ≈ 5.05 people

Part c) Interpretation:

The expected value of 30 means that on average, if the company calls 200 people repeatedly over many days, they can expect about 30 people per day to agree to the survey. It doesn’t mean they will get exactly 30 every day, but rather that 30 is the long-run average.

Part d) Meeting the Goal:

Yes, the expected value of 30 is above their minimum requirement of 25. The standard deviation of about 5 tells us that typical variation is around 5 people above or below 30. Since 30 - 5 = 25, even on a below-average day (one standard deviation below the mean), they would likely meet their goal. However, there’s still some risk of falling short on particularly unlucky days.

12.2 Question 52

Scenario: A coffee shop receives online orders at an average rate of 3.5 orders per hour during the afternoon.

Questions:

  1. Calculate the probability that the coffee shop receives exactly 3 orders in an hour.

  2. Calculate the probability that the coffee shop receives exactly 5 orders in an hour.

  3. Which is more likely: exactly 3 orders or exactly 5 orders? Why does this make intuitive sense?

  4. What is the expected number of orders during a 3-hour afternoon shift? What distribution would model this?

Show Solution

Solution:

Part a) P(X = 3) with λ = 3.5:

\(P(X = 3) = \frac{3.5^3 e^{-3.5}}{3!} = \frac{42.875 \times 0.0302}{6} = \frac{1.295}{6} = 0.2158\)

Answer: P(X = 3) ≈ 0.216 or 21.6%

Part b) P(X = 5) with λ = 3.5:

\(P(X = 5) = \frac{3.5^5 e^{-3.5}}{5!} = \frac{525.219 \times 0.0302}{120} = \frac{15.862}{120} = 0.1322\)

Answer: P(X = 5) ≈ 0.132 or 13.2%

Part c) Comparison and Interpretation:

Receiving exactly 3 orders is more likely (21.6% vs 13.2%).

Intuitive explanation: The coffee shop averages 3.5 orders per hour, so getting exactly 3 is very close to their average and thus quite likely. Getting exactly 5 orders would be considerably above average (about 1.5 orders more than expected), which is less probable though still possible.

Part d) Three-hour shift:

For a Poisson process, if the rate is λ per hour, then for t hours, the rate becomes λt.

\(E(X_{3 \text{ hours}}) = 3.5 \times 3 = 10.5\) orders

Answer: Expected value = 10.5 orders

Distribution: Poisson with λ = 10.5 (still Poisson, just with a different rate parameter)

12.3 Question 53

Scenario: A manufacturing machine’s calibration time is uniformly distributed between 15 and 45 minutes.

Questions:

  1. What is the probability that calibration takes less than 25 minutes?

  2. What calibration time represents the 60th percentile? Interpret this value.

  3. If you randomly select a calibration session, what is the expected duration? (Hint: For uniform [a,b], \(E(X) = \frac{a+b}{2}\))

  4. The operations manager claims: “Our machine calibration times are unpredictable - we’re just as likely to finish in 20 minutes as in 40 minutes.” Is this statement correct? Explain using the properties of the uniform distribution.

Show Solution

Solution:

Part a) P(X < 25):

\(P(X < 25) = \frac{25 - 15}{45 - 15} = \frac{10}{30} = 0.333\)

Answer: P(X < 25) ≈ 0.333 or 33.3%

Interpretation: About one-third of calibration sessions are completed in less than 25 minutes.

Part b) 60th Percentile:

For uniform distribution, the 60th percentile is located 60% of the way from a to b:

\(P_{60} = a + 0.60(b - a) = 15 + 0.60(45 - 15) = 15 + 18 = 33\) minutes

Answer: 60th percentile = 33 minutes

Interpretation: 60% of calibration sessions take 33 minutes or less, and 40% take 33 minutes or longer.

Part c) Expected Duration:

\(E(X) = \frac{a + b}{2} = \frac{15 + 45}{2} = \frac{60}{2} = 30\) minutes

Answer: E(X) = 30 minutes

Interpretation: The average calibration time is 30 minutes. For a uniform distribution, the expected value is always at the midpoint of the range.

Part d) Evaluating the Manager’s Claim:

The manager’s statement is CORRECT.

For a uniform distribution on [15, 45], every value in this range is equally likely. Since both 20 and 40 fall within this range, they have the exact same probability density.

Explanation: In a uniform distribution, the probability density is constant across the entire range. Whether it’s 20, 25, 30, 35, or 40 minutes, each specific value has the same “likelihood.” The word “unpredictable” captures the essence of the uniform distribution - we have no reason to expect any particular time within the range over another. The manager’s statement accurately reflects this property.

12.4 Question 54

Scenario: A quality control inspector finds that 12% of items produced on a particular assembly line have minor cosmetic defects. She inspects a random sample of 40 items.

Questions:

  1. Calculate the expected number of defective items and the standard deviation.

  2. Calculate P(X ≤ 3), the probability of 3 or fewer defective items.

  3. Calculate P(X ≥ 8), the probability of 8 or more defective items.

  4. If the inspector finds 10 defective items in this sample of 40, should she be concerned about the assembly line? Use probability and standard deviations to support your answer.

Show Solution

Solution:

Given: \(n = 40\), \(p = 0.12\)

Part a) Expected Value and Standard Deviation:

\(E(X) = np = 40 \times 0.12 = 4.8\) defective items

\(SD(X) = \sqrt{np(1-p)} = \sqrt{40 \times 0.12 \times 0.88} = \sqrt{4.224} = 2.06\) items

Answer: E(X) = 4.8 defective items, SD(X) ≈ 2.06 items

Part b) P(X ≤ 3):

\(P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)\)

\(P(X=0) = \binom{40}{0}(0.12)^0(0.88)^{40} = 0.006\)

\(P(X=1) = \binom{40}{1}(0.12)^1(0.88)^{39} = 40 \times 0.12 \times 0.0076 = 0.0328\)

\(P(X=2) = \binom{40}{2}(0.12)^2(0.88)^{38} = 780 \times 0.0144 \times 0.0086 = 0.0873\)

\(P(X=3) = \binom{40}{3}(0.12)^3(0.88)^{37} = 9880 \times 0.001728 \times 0.0098 = 0.1507\)

\(P(X \leq 3) = 0.006 + 0.0328 + 0.0873 + 0.1507 = 0.2768\)

Answer: P(X ≤ 3) ≈ 0.2768 or 27.7%

Interpretation: There’s about a 27.7% chance of finding 3 or fewer defective items, which is fairly common.

Part c) P(X ≥ 8):

This is best calculated using complement: \(P(X \geq 8) = 1 - P(X \leq 7)\)

Using binomial cumulative distribution: \(P(X \leq 7) \approx 0.9004\)

\(P(X \geq 8) = 1 - 0.9004 = 0.0996\)

Answer: P(X ≥ 8) ≈ 0.0996 or 9.96%

Interpretation: There’s about an 9.96% chance of finding 8 or more defective items, which is somewhat unusual but not extremely rare.

Part d) Should She Be Concerned About 10 Defective Items?

To assess this, we can:

  1. Compare to expected value: 10 defects is more than double the expected value of 4.8, which seems quite high.

  2. Use standard deviations: \(z = \frac{10 - 4.8}{2.06} = 2.52\)

    This is more than 2.5 standard deviations above the mean!

  3. Calculate probability: P(X ≥ 10) would be approximately 1% or less

Answer: Yes, the inspector should be concerned. With an expected value of 4.8 and standard deviation of 2.06, observing 10 defective items is more than 2.5 standard deviations above the mean. This would occur by chance less than 2% of the time under normal operating conditions. The inspector should investigate whether: - The assembly line is malfunctioning - Workers need additional training - Raw materials have changed in quality - The defect rate has increased beyond the historical 12%

This result is statistically unusual enough to warrant a thorough investigation of the production process.

12.5 Question 55

Scenario: A tech support call center receives technical issue calls at an average rate of 2.4 calls per hour during the evening shift.

Questions:

  1. What is the probability of receiving exactly 3 technical calls in a 1-hour period?

  2. What is the expected number of technical calls in a 6-hour evening shift?

  3. What is the probability of receiving no technical calls during a 6-hour shift? (Hint: scale λ appropriately)

  4. The call center has capacity to handle up to 4 calls per hour without significant wait times. What’s the probability that they’ll exceed this capacity in a given hour?

Show Solution

Solution:

Part a) P(X = 3) in 1 hour with λ = 2.4:

\(P(X = 3) = \frac{2.4^3 e^{-2.4}}{3!} = \frac{13.824 \times 0.0907}{6} = \frac{1.254}{6} = 0.2090\)

Answer: P(X = 3) ≈ 0.209 or 20.9%

Part b) Expected Value in 6 Hours:

For a Poisson process over time period t with rate λ per hour: \(E(X) = \lambda t = 2.4 \times 6 = 14.4\) calls

Answer: E(X) = 14.4 technical calls

Interpretation: On average, during a 6-hour evening shift, the call center expects about 14-15 technical calls.

Part c) P(X = 0) in 6 hours:

For 6 hours, the rate parameter is \(\lambda = 2.4 \times 6 = 14.4\)

\(P(X = 0) = \frac{14.4^0 e^{-14.4}}{0!} = e^{-14.4} = 0.00000055\)

Answer: P(X = 0) ≈ 0.00000055 or 0.000055%

Interpretation: It’s extremely unlikely (less than 0.0001% chance) that the call center would go an entire 6-hour evening shift without receiving any technical calls. This is virtually impossible.

Part d) P(X ≥ 5) in 1 hour (exceeding capacity):

Using complement: \(P(X \geq 5) = 1 - P(X \leq 4)\)

With λ = 2.4:

\(P(X=0) = e^{-2.4} = 0.0907\)

\(P(X=1) = \frac{2.4 \times 0.0907}{1} = 0.2177\)

\(P(X=2) = \frac{5.76 \times 0.0907}{2} = 0.2613\)

\(P(X=3) = \frac{13.824 \times 0.0907}{6} = 0.2090\)

\(P(X=4) = \frac{33.178 \times 0.0907}{24} = 0.1254\)

\(P(X \leq 4) = 0.0907 + 0.2177 + 0.2613 + 0.2090 + 0.1254 = 0.9041\)

\(P(X \geq 5) = 1 - 0.9041 = 0.0959\)

Answer: P(X ≥ 5) ≈ 0.096 or 9.6%

Interpretation: The call center has about a 10% chance of exceeding their comfortable capacity in any given hour. This means that roughly 1 out of every 10 hours will experience longer wait times. Over a 6-hour shift, they might expect this to happen about once per shift on average. Management should consider this when planning staffing levels.

12.6 Question 56

Scenario: A streaming service analyzes user behavior and creates the following contingency table showing the relationship between subscription type and whether users watch more than 10 hours per week:

Watch >10 hrs/week Watch ≤10 hrs/week Total
Premium Subscription 180 120 300
Basic Subscription 90 210 300
Total 270 330 600

Questions:

  1. Calculate P(Premium), P(Watch >10 hrs/week), and P(Premium AND Watch >10 hrs/week).

  2. Are “having a Premium subscription” and “watching more than 10 hours per week” independent events? Show your work.

  3. Can a user simultaneously have both a Premium subscription and a Basic subscription? Are these events mutually exclusive?

  4. Calculate P(Watch >10 hrs/week | Premium) and interpret this in context. How does this compare to P(Watch >10 hrs/week | Basic)?

Show Solution

Solution:

Part a) Basic Probabilities:

\(P(\text{Premium}) = \frac{300}{600} = 0.50\)

\(P(\text{Watch >10 hrs/week}) = \frac{270}{600} = 0.45\)

\(P(\text{Premium AND Watch >10 hrs/week}) = \frac{180}{600} = 0.30\)

Answers: - P(Premium) = 0.50 or 50% - P(Watch >10 hrs/week) = 0.45 or 45% - P(Premium AND Watch >10 hrs/week) = 0.30 or 30%

Part b) Testing for Independence:

For independence, we need: \(P(A \cap B) = P(A) \times P(B)\)

\(P(\text{Premium}) \times P(\text{Watch >10 hrs/week}) = 0.50 \times 0.45 = 0.225\)

But: \(P(\text{Premium AND Watch >10 hrs/week}) = 0.30\)

Since \(0.30 \neq 0.225\), the events are NOT independent.

Interpretation: Premium subscribers are more likely to watch more than 10 hours per week compared to what we’d expect if subscription type and viewing habits were unrelated. The fact that 30% of all users are premium subscribers who watch heavily, rather than the expected 22.5%, suggests subscription type and viewing behavior are related.

Part c) Mutual Exclusivity:

No, a user cannot simultaneously have both Premium and Basic subscriptions - each user has exactly one subscription type.

These events ARE mutually exclusive because: - P(Premium AND Basic) = 0 (there are no users in both categories) - A user must be in one category or the other, never both - The subscription types are separate, non-overlapping categories

Note: This is different from independence! Mutually exclusive events cannot be independent (except in the trivial case where one has probability 0).

Part d) Conditional Probabilities:

\(P(\text{Watch >10 hrs/week | Premium}) = \frac{180}{300} = 0.60\) or 60%

Interpretation: Among Premium subscribers, 60% watch more than 10 hours per week.

\(P(\text{Watch >10 hrs/week | Basic}) = \frac{90}{300} = 0.30\) or 30%

Comparison: Premium subscribers are twice as likely to watch more than 10 hours per week compared to Basic subscribers (60% vs 30%). This makes intuitive sense - users who pay for premium features may be more engaged with the service and use it more heavily. This relationship further confirms that subscription type and viewing behavior are not independent - knowing someone has a Premium subscription significantly changes the probability they’re a heavy viewer.

12.7 Question 57

Scenario: An airport parking lot has the following distribution of parking duration (uniformly distributed) and payment method:

  • Short-term parking: Duration uniformly distributed between 1 and 4 hours
  • Long-term parking: 35% of customers use credit cards, 65% use mobile payment

A customer parks in the short-term lot for a random duration.

Questions:

  1. What is the probability that a short-term customer parks for more than 3 hours?

  2. What is the expected parking duration for short-term customers?

  3. Among long-term customers, what is the expected number who use credit cards if 80 customers park long-term today? What distribution models this?

  4. If a short-term customer has already been parked for 2.5 hours, what’s the probability they’ll park for more than 3 hours total? Explain why this differs from part (a).

Show Solution

Solution:

Part a) P(X > 3) for Uniform [1, 4]:

\(P(X > 3) = \frac{4 - 3}{4 - 1} = \frac{1}{3} = 0.333\)

Answer: P(X > 3) ≈ 0.333 or 33.3%

Interpretation: About one-third of short-term parking customers stay longer than 3 hours.

Part b) Expected Duration:

\(E(X) = \frac{a + b}{2} = \frac{1 + 4}{2} = 2.5\) hours

Answer: E(X) = 2.5 hours

Interpretation: The average short-term parking duration is 2.5 hours, which is the midpoint of the range since the distribution is uniform.

Part c) Expected Credit Card Users:

This follows a binomial distribution with \(n = 80\), \(p = 0.35\)

\(E(X) = np = 80 \times 0.35 = 28\) customers

Answer: 28 customers expected to use credit cards

Distribution: Binomial with n = 80 and p = 0.35

Interpretation: On average, if 80 customers park long-term, about 28 will pay with credit cards. The actual number will vary from day to day around this average.

Part d) Conditional Probability P(X > 3 | X > 2.5):

For uniform distributions, the conditional distribution given X > 2.5 is uniform on [2.5, 4].

\(P(X > 3 | X > 2.5) = \frac{4 - 3}{4 - 2.5} = \frac{1}{1.5} = 0.667\)

Answer: P(X > 3 | X > 2.5) ≈ 0.667 or 66.7%

Interpretation and Comparison: If we know a customer has already been parked for 2.5 hours and hasn’t left yet, there’s now a 66.7% chance they’ll stay beyond 3 hours. This is double the unconditional probability of 33.3% from part (a).

Why the difference? The conditioning changes everything: - Originally, the customer could park anywhere from 1 to 4 hours (3-hour range) - After learning they’ve been there 2.5+ hours, we know they must park between 2.5 and 4 hours (1.5-hour range) - We’ve eliminated all durations less than 2.5 hours from consideration - Within the remaining 1.5-hour window, 1 hour is above 3, making it more likely

This demonstrates the memoryless-like property of uniform distributions under conditioning - the conditional distribution is still uniform, just on a shifted range.

12.8 Question 58

Scenario: A hospital emergency department tracks patient outcomes and creates the following contingency table:

Admitted to Hospital Discharged Home Total
Critical Cases 85 15 100
Non-Critical 45 355 400
Total 130 370 500

Questions:

  1. Calculate P(Critical), P(Admitted), and P(Critical AND Admitted).

  2. Are “being a critical case” and “being admitted to the hospital” independent events? Provide statistical evidence.

  3. Is it possible for a patient to be both admitted to the hospital and discharged home? Are these events mutually exclusive? Explain the practical meaning.

  4. Calculate and interpret P(Admitted | Critical) and P(Admitted | Non-Critical). What does the comparison tell us about the emergency department’s triage system?

Show Solution

Solution:

Part a) Basic Probabilities:

\(P(\text{Critical}) = \frac{100}{500} = 0.20\)

\(P(\text{Admitted}) = \frac{130}{500} = 0.26\)

\(P(\text{Critical AND Admitted}) = \frac{85}{500} = 0.17\)

Answers: - P(Critical) = 0.20 or 20% - P(Admitted) = 0.26 or 26% - P(Critical AND Admitted) = 0.17 or 17%

Part b) Testing for Independence:

For independence: \(P(A \cap B) = P(A) \times P(B)\)

\(P(\text{Critical}) \times P(\text{Admitted}) = 0.20 \times 0.26 = 0.052\)

But: \(P(\text{Critical AND Admitted}) = 0.17\)

Since \(0.17 \neq 0.052\), the events are NOT independent.

Statistical Evidence: The probability of being both critical and admitted (17%) is much higher than what we’d expect if these were independent events (5.2%). This difference is substantial and meaningful.

Interpretation: Patient severity (critical vs non-critical) and hospital admission are strongly related. Critical patients are much more likely to be admitted than we’d expect by chance alone. This makes medical sense - the triage system appropriately identifies critical patients and admits them at higher rates.

Part c) Mutual Exclusivity:

No, a patient cannot be both admitted to the hospital AND discharged home - these are mutually exclusive outcomes.

These events ARE mutually exclusive because: - P(Admitted AND Discharged) = 0 (no patients appear in both columns) - Each patient receives exactly one disposition: either admission or discharge - These are the two possible outcomes with no overlap

Practical meaning: In emergency medicine, admission and discharge are the two distinct final dispositions for a patient. Once the medical team makes a decision, the patient follows only one path. A patient cannot simultaneously be admitted to an inpatient bed and sent home.

Part d) Conditional Probabilities and Triage Analysis:

\(P(\text{Admitted | Critical}) = \frac{85}{100} = 0.85\) or 85%

Interpretation: Among critical cases, 85% are admitted to the hospital.

\(P(\text{Admitted | Non-Critical}) = \frac{45}{400} = 0.1125\) or 11.25%

Interpretation: Among non-critical cases, only about 11% are admitted to the hospital.

Comparison and Triage System Analysis:

Critical patients are 7.6 times more likely to be admitted compared to non-critical patients (85% vs 11.25%). This substantial difference demonstrates that:

  1. The triage system is working effectively - there’s a strong relationship between patient severity and admission decisions
  2. Critical patients receive appropriate intensive care - the vast majority (85%) are kept for further treatment
  3. Non-critical patients are efficiently managed - most (89%) are safely discharged, avoiding unnecessary hospital resource use
  4. The emergency department appropriately differentiates between patients who need inpatient care and those who can be managed as outpatients

The strong dependence between critical status and admission (proven in part b) reflects sound medical decision-making rather than random chance. This is exactly what we want to see in a well-functioning emergency department.

12.9 Question 59

Scenario: A social media marketing campaign targets young adults, and historical data shows that 18% of people who see the ad will click on it. The campaign will display the ad to 150 people.

Questions:

  1. What is the expected number of clicks and the standard deviation?

  2. Calculate the probability that exactly 25 people click the ad.

  3. The marketing team considers the campaign successful if at least 30 people click. What’s the probability they’ll meet this goal?

  4. After running the campaign, they observe 35 clicks. The team leader says: “This proves our campaign performed better than average!” Evaluate this claim using appropriate statistical reasoning. Calculate how many standard deviations above the mean this result is.

Show Solution

Solution:

Given: n = 150, p = 0.18

Part a) Expected Value and Standard Deviation:

\(E(X) = np = 150 \times 0.18 = 27\) clicks

\(SD(X) = \sqrt{np(1-p)} = \sqrt{150 \times 0.18 \times 0.82} = \sqrt{22.14} = 4.71\) clicks

Answer: E(X) = 27 clicks, SD(X) ≈ 4.71 clicks

Part b) P(X = 25):

Using binomial with n = 150, p = 0.18:

\(P(X = 25) = \binom{150}{25}(0.18)^{25}(0.82)^{125}\)

Answer: P(X = 25) ≈ 0.073 or 7.3%

Interpretation: There’s about a 7% chance of getting exactly 25 clicks, which is close to but slightly below the expected value of 27.

Part c) P(X ≥ 30):

Answer: P(X ≥ 30) ≈ 0.30 or 30%

Interpretation: There’s about a 30% chance the campaign will achieve at least 30 clicks. This is a reasonable probability - it’s not highly likely, but it happens roughly 3 out of 10 times, so it’s a feasible goal.

Part d) Evaluating the Claim of 35 Clicks:

Statistical Analysis:

  1. Compare to expected value: 35 clicks exceeds the expected value of 27 by 8 clicks

  2. Calculate standard deviations from mean: \(z = \frac{35 - 27}{4.71} = 1.70\)

    This is 1.70 standard deviations above the mean.

  3. Probability assessment: P(X ≥ 35) with z = 1.70 corresponds to approximately 4.5% probability

Evaluation of the Team Leader’s Claim:

The claim is somewhat supported, but should be made cautiously:

Arguments supporting the claim: - 35 clicks is 1.70 standard deviations above the expected value, which is moderately unusual - This result would occur by chance only about 4-5% of the time - It does represent above-average performance

Arguments for caution: - While 1.70 standard deviations is above average, it’s not dramatically unusual (typically we consider 2+ standard deviations as strong evidence) - There’s still a roughly 4-5% chance this could happen due to random variation alone - With 95% confidence intervals, we’d expect results within about 2 standard deviations (roughly 18-36 clicks), and 35 falls within this range - One successful campaign doesn’t prove the strategy is superior - it could be random variation

Better statement: “Our campaign performed above average with 35 clicks compared to the expected 27. While this is encouraging (occurring only 4-5% of the time by chance), we should run additional campaigns to confirm whether this represents a

genuinely improved strategy or just favorable random variation.”

Conclusion: The result is moderately impressive (1.70 SD above mean) and suggests better-than-expected performance, but falls short of the strong statistical evidence (2+ SD) typically needed to definitively claim superior performance. The team should be pleased but not overconfident based on this single result.

12.10 Question 60

Scenario: A city’s emergency services department analyzes response patterns:

  • Fire calls: Arrive at an average rate of 1.8 calls per day
  • Medical emergencies: 22% require advanced life support (ALS), while 78% need only basic life support (BLS). Today they respond to 50 medical emergencies.
  • Response time: For routine calls, response time is uniformly distributed between 4 and 12 minutes.

Questions:

  1. For each scenario, calculate the expected value and interpret what it means in context.

  2. For the fire calls scenario, what’s the probability of receiving exactly 2 fire calls in one day?

  3. For the medical emergencies scenario, what’s the probability that exactly 15 of the 50 calls require ALS?

  4. A dispatcher notices they’ve already had 1 fire call today. Does this change the expected number of additional fire calls for the rest of the day? Explain using properties of the Poisson distribution.

Show Solution

Solution:

Part a) Expected Values and Interpretations:

Poisson (Fire calls): \(E(X) = \lambda = 1.8\) calls per day

Interpretation: On average, the city receives 1.8 fire calls per day. Over many days, the long-run average will be about 1.8 fire calls daily, though on any particular day they might receive 0, 1, 2, 3, or more calls.

Binomial (Medical emergencies requiring ALS): \(E(X) = np = 50 \times 0.22 = 11\) ALS calls

Interpretation: Out of 50 medical emergency calls, the department expects about 11 will require advanced life support. This doesn’t mean exactly 11 every time they have 50 calls, but 11 is the average across many sets of 50 calls.

Uniform (Response time): \(E(X) = \frac{a + b}{2} = \frac{4 + 12}{2} = 8\) minutes

Interpretation: The average response time for routine calls is 8 minutes, which is the midpoint of the possible range. Response times are equally likely to be anywhere between 4 and 12 minutes.

Part b) P(X = 2) for Poisson with λ = 1.8:

\(P(X = 2) = \frac{1.8^2 e^{-1.8}}{2!} = \frac{3.24 \times 0.1653}{2} = \frac{0.5356}{2} = 0.2678\)

Answer: P(X = 2) ≈ 0.268 or 26.8%

Interpretation: There’s about a 27% chance of receiving exactly 2 fire calls on any given day. This is actually the most likely single outcome, since 2 is close to the mean of 1.8.

Part c) P(X = 15) for Binomial with n = 50, p = 0.22:

\(P(X = 15) = \binom{50}{15}(0.22)^{15}(0.78)^{35} = 0.0515\)

Answer: P(X = 15) ≈ 0.0515 or 5.15%

Interpretation: There’s only about a 4.5% chance of having exactly 15 ALS calls out of 50 medical emergencies. This is considerably higher than the expected 11, occurring about 1.4 standard deviations above the mean - somewhat unusual but not extremely rare.

Part d) Does One Fire Call Change Expectations? (Memoryless Property)

Answer: No, the expected number of additional fire calls remains 1.8.

Explanation using Poisson properties:

The Poisson process has the memoryless property - past events don’t affect future probabilities. Here’s why:

  1. Independent increments: Fire calls in one part of the day are independent of fire calls in another part
  2. The process “restarts”: After observing 1 call, the Poisson process doesn’t “remember” this happened
  3. Rate remains constant: The underlying rate of 1.8 calls per day hasn’t changed

Common misconception: “We’ve already had 1 fire call, so we should expect fewer additional calls to stay near the average of 1.8 total.”

Correct reasoning: “The rate is 1.8 calls per day regardless of what’s already happened. If we’ve had 1 call already, we still expect 1.8 more calls on average for a total expectation of 2.8 calls for the day.”

Important note: This seems counterintuitive, but remember: - The original expectation of 1.8 was for the entire day before any information - Once we know 1 call has occurred, that’s no longer a “future” event - The expectation of 1.8 applies to any time period of length “one day” going forward - The total expected calls for the day, given we’ve seen 1, is now 1 (observed) + 1.8 (still expected) = 2.8

Practical interpretation: The dispatcher should still prepare for an average of 1.8 more fire calls, regardless of whether they’ve already had 0, 1, or even 3 calls so far today. Each day period has the same expected rate.