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AP Statistics 4.7 Introduction to Random Variables and Probability Distributions- MCQs - Exam Style Questions

Question

A veterinarian collected data on the weights of 1,000 cats and dogs treated at a veterinary clinic. The weight of each animal was classified as either healthy, underweight, or overweight. The data are summarized in the table.
 HealthyUnderweightOverweightTotal
Cat38654105545
Dog2998373455
Total6851371781,000
Based on the data in the table, which of the following is the most appropriate type of graph to visually show whether a relationship exists between the type of animal and the weight classification?
(A) Back-to-back stemplots
(B) Scatterplot
(C) Side-by-side boxplots
(D) Segmented bar chart
(E) Dotplot
▶️ Answer/Explanation
Detailed solution

1. Identify Variable Types:
– The variable “type of animal” (Cat, Dog) is categorical.
– The variable “weight classification” (Healthy, Underweight, Overweight) is also categorical.

2. Select the Appropriate Graph:
– We need a graph that shows a relationship between two categorical variables.
– Stemplots, scatterplots, boxplots, and dotplots are used for quantitative (numerical) data.
– A segmented bar chart is specifically used to compare the proportions of a categorical variable across different groups.

This makes it the best choice to see if the proportion of healthy, underweight, and overweight animals differs between cats and dogs.
Answer: (D)

Question

Events D and E are independent, with \(P(D)=0.6\) and \(P(D \text{ and } E)=0.18\). Which of the following is true?
(A) \(P(E)=0.12\)
(B) \(P(E)=0.4\)
(C) \(P(D \text{ or } E)=0.28\)
(D) \(P(D \text{ or } E)=0.72\)
(E) \(P(D \text{ or } E)=0.9\)
▶️ Answer/Explanation
Detailed solution

1. Find \(P(E)\) using the independence rule:
For independent events, \(P(D \text{ and } E) = P(D) \times P(E)\).
\(0.18 = 0.6 \times P(E)\)
\(P(E) = \frac{0.18}{0.6} = 0.3\)

2. Calculate \(P(D \text{ or } E)\) using the general addition rule:
\(P(D \text{ or } E) = P(D) + P(E) – P(D \text{ and } E)\)
\(P(D \text{ or } E) = 0.6 + 0.3 – 0.18\)
\(P(D \text{ or } E) = 0.9 – 0.18 = 0.72\)
Answer: (D)

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