Probability and Statistics Study Guidance – Ch. 3

Q1. What is the predicted exam score for a student who studied for 8 hours?

Background

Topic: Simple Linear Regression

This question tests your ability to use a regression equation to predict an outcome (exam score) based on an explanatory variable (hours studied).

Key Terms and Formulas

• Regression Equation: $\text{ExamScore} = 50.263 + 3.427 \times \text{HoursStudied}$

• Predicted Value ($\hat{Y}$): The value calculated from the regression equation for a given input.

Step-by-Step Guidance

1. Identify the regression equation and the value for HoursStudied: $\text{ExamScore} = 50.263 + 3.427 \times \text{HoursStudied}$, with $\text{HoursStudied} = 8$.

2. Substitute 8 for HoursStudied in the equation: $\text{ExamScore} = 50.263 + 3.427 \times 8$.

3. Calculate $3.427 \times 8$ to find the contribution from hours studied.

4. Add this value to 50.263 to get the predicted exam score.

Try solving on your own before revealing the answer!

Final Answer: 77.679

$\text{ExamScore} = 50.263 + 3.427 \times 8 = 77.679$

The regression equation predicts an exam score of 77.679 for a student who studied 8 hours.

Q2. What is the sample space and number of possible outcomes for a computer program that randomly generates a digit from 0 to 9?

Background

Topic: Basic Probability – Sample Space

This question tests your understanding of sample spaces and counting possible outcomes in a random experiment.

Key Terms and Formulas

• Sample Space ($S$): The set of all possible outcomes.

• Number of Outcomes: The count of elements in the sample space.

Step-by-Step Guidance

1. List all possible digits the program can generate: 0, 1, 2, ..., 9.

2. Write the sample space as a set: $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

3. Count the number of elements in the set to determine the number of possible outcomes.

Try solving on your own before revealing the answer!

Final Answer: Sample space is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; Number of possible outcomes is 10.

There are 10 possible outcomes, one for each digit from 0 to 9.

Q3. Are the events "being a nurse" and "having completed advanced training" independent?

Background

Topic: Probability – Independence of Events

This question tests your ability to determine whether two events are independent using probability concepts.

Key Terms and Formulas

• Independent Events: Two events A and B are independent if $P(A \cap B) = P(A) \times P(B)$.

• Conditional Probability: $P(B|A) = \frac{P(A \cap B)}{P(A)}$

Step-by-Step Guidance

1. Find the total number of nurses and doctors: 3,200 nurses and 1,800 doctors.

2. Find the number of nurses and doctors who completed advanced training: 480 nurses and 720 doctors.

3. Calculate the probability of being a nurse: $P(\text{Nurse}) = \frac{3200}{3200 + 1800}$.

4. Calculate the probability of having completed advanced training: $P(\text{Advanced}) = \frac{480 + 720}{3200 + 1800}$.

5. Calculate the probability of being a nurse and having completed advanced training: $P(\text{Nurse} \cap \text{Advanced}) = \frac{480}{3200 + 1800}$.

6. Compare $P(\text{Nurse} \cap \text{Advanced})$ to $P(\text{Nurse}) \times P(\text{Advanced})$ to check independence.

Try solving on your own before revealing the answer!

Final Answer: No, the events are not independent.

The calculated probabilities do not satisfy the independence condition.

Q4. What is the probability that all four items are non-defective if the probability that an item is defective is 0.474?

Background

Topic: Probability – Complements and Multiplication Rule

This question tests your ability to use the complement rule and multiplication rule for independent events.

Key Terms and Formulas

• Complement Rule: $P(\text{Non-defective}) = 1 - P(\text{Defective})$

• Multiplication Rule for Independent Events: $P(\text{All Non-defective}) = [P(\text{Non-defective})]^4$

Step-by-Step Guidance

1. Find the probability that an item is non-defective: $P(\text{Non-defective}) = 1 - 0.474$.

2. Since the items are selected randomly and independently, raise the probability to the fourth power: $[P(\text{Non-defective})]^4$.

3. Calculate the value of $P(\text{Non-defective})$.

4. Set up the expression for $[P(\text{Non-defective})]^4$.

Try solving on your own before revealing the answer!

Final Answer: 0.0765

The probability that all four items are non-defective is 0.0765.

Q5. What is the probability that a customer did not tip, given that the customer paid by card?

Background

Topic: Conditional Probability

This question tests your ability to calculate conditional probabilities using a contingency table.

Key Terms and Formulas

• Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

• "Did Not Tip" given "Paid by Card": $P(\text{Did Not Tip} | \text{Paid by Card})$

Step-by-Step Guidance

1. Identify the number of customers who paid by card: $18 + 42 = 60$.

2. Identify the number of customers who did not tip and paid by card: $42$.

3. Set up the conditional probability formula: $P(\text{Did Not Tip} | \text{Paid by Card}) = \frac{42}{60}$.

4. Simplify the fraction to get the probability.

Try solving on your own before revealing the answer!

Final Answer: 0.70

70% of customers who paid by card did not tip.

Q6. If the probability that event E will not occur is 0.86, what is the probability that event E will occur?

Background

Topic: Probability – Complements

This question tests your understanding of the complement rule in probability.

Key Terms and Formulas

• Complement Rule: $P(E) = 1 - P(\overline{E})$

Step-by-Step Guidance

1. Identify the probability that event E will not occur: $P(\overline{E}) = 0.86$.

2. Use the complement rule: $P(E) = 1 - 0.86$.

3. Set up the calculation for $P(E)$.

Try solving on your own before revealing the answer!

Final Answer: 0.14

The probability that event E will occur is 0.14.

Q7. What is the probability that a randomly selected student either sped regularly or used a phone while biking, but not both?

Background

Topic: Probability – Addition Rule (Excluding Overlap)

This question tests your ability to use the addition rule for probabilities, excluding the overlap (students who did both).

Key Terms and Formulas

• Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

• Probability of "either but not both": $P(A) + P(B) - 2P(A \cap B)$

Step-by-Step Guidance

1. Identify the total number of students: $800$.

2. Identify the number who sped regularly: $280$; used phone while biking: $190$; did both: $60$.

3. Calculate the probability for each group: $P(\text{Speed}) = \frac{280}{800}$, $P(\text{Phone}) = \frac{190}{800}$, $P(\text{Both}) = \frac{60}{800}$.

4. Apply the formula for "either but not both": $P(\text{Speed or Phone but not both}) = P(\text{Speed}) + P(\text{Phone}) - 2P(\text{Both})$.

5. Set up the calculation for the probability.

Try solving on your own before revealing the answer!

Final Answer: 0.4375

The probability is 0.4375 for a student who either sped regularly or used a phone while biking, but not both.

Q8. If the probability of event A is 0.2 and the probability of event B is 0.3, what is the probability of either event A or event B occurring if they are mutually exclusive?

Background

Topic: Probability – Addition Rule for Mutually Exclusive Events

This question tests your understanding of the addition rule for mutually exclusive events.

Key Terms and Formulas

• Mutually Exclusive Events: Events that cannot occur at the same time.

• Addition Rule: $P(A \cup B) = P(A) + P(B)$ for mutually exclusive events.

Step-by-Step Guidance

1. Identify the probabilities: $P(A) = 0.2$, $P(B) = 0.3$.

2. Since the events are mutually exclusive, use the addition rule: $P(A \cup B) = P(A) + P(B)$.

3. Set up the calculation for $P(A \cup B)$.

Try solving on your own before revealing the answer!

Final Answer: 0.5

The probability of either event A or event B occurring is 0.5.

Q9. What is the probability of selecting a shopper who believes the policy is extremely unfair?

Background

Topic: Probability – Empirical Probability from Frequency Data

This question tests your ability to calculate probability from frequency data (Pareto chart).

Key Terms and Formulas

• Empirical Probability: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Step-by-Step Guidance

1. Identify the number of shoppers who believe the policy is extremely unfair: $345$.

2. Identify the total number of shoppers surveyed: $1115$.

3. Set up the probability formula: $P = \frac{345}{1115}$.

4. Simplify the fraction to get the probability.

Try solving on your own before revealing the answer!

Final Answer: 0.31

The probability is 0.31 for selecting a shopper who believes the policy is extremely unfair.