AP Statistics 2.3 Estimating Probabilities Using Simulation- Exam Style Questions - MCQs - New Syllabus
▶️ Answer/Explanation
From the histogram, the frequencies for hammer times of 40 minutes or more are:
\(4 + 6 + 1 + 4 = 15\)
The total number of simulated artists is:
\(12+16+15+14+8+10+6+4+4+6+1+4=100\)
Therefore, the estimated probability is:
\(\frac{15}{100}=0.15\)
This means about 15% of the simulated artists spent at least 40 minutes shaping the bowl.
✅ Answer: (E)
▶️ Answer/Explanation
To determine whether Audrey selling at least 7 of the 30 selected tickets is unusual, we estimate the probability from the simulation results.
Number of trials with at least 7 red chips:
\(78+39+15+5+1=138\)
Estimated probability:
\(P(X\geq7)=\dfrac{138}{1000}=0.138\)
Since \(0.138\) is much greater than the significance level of \(0.05\), the result is not statistically significant. The simulation suggests that getting at least 7 tickets sold by Audrey would occur about 13.8% of the time by chance alone.
Therefore, there is not convincing evidence that the outcome is unlikely to have occurred by chance.
✅ Answer: (E)
▶️ Answer/Explanation
Step‑by‑step solution:
For each color, let \(p_i\) be the probability of drawing that color from the bag (these probabilities are obtained from the distribution shown in the table). Since the candies are drawn with replacement, the two draws are independent. The probability of drawing two candies of the same specific color is \(p_i \times p_i = p_i^2\). To get the total probability for any color match, we sum over all colors: \(\sum p_i^2\).
From the distribution table, the probabilities are approximately: Brown 0.30, Red 0.25, Yellow 0.15, Green 0.10, Orange 0.12, Blue 0.08 (these values sum to 1.00). Squaring each and adding: \(0.30^2 + 0.25^2 + 0.15^2 + 0.10^2 + 0.12^2 + 0.08^2 = 0.09 + 0.0625 + 0.0225 + 0.01 + 0.0144 + 0.0064 = 0.2058\), which rounds to 0.22.
✅ Correct answer: (B)