Multiple Choice Questions
Advanced Statistics and Data Analysis
Topic: Advanced Statistics and Data Analysis
Grade: 12
Question 1:
In a survey of 100 students, it was found that 40 students like apples, 30 students like oranges, and 20 students like both fruits. What is the probability that a randomly selected student likes neither apples nor oranges?
A) 0.1
B) 0.2
C) 0.3
D) 0.4
Answer: B) 0.2
Explanation: To find the probability that a student likes neither apples nor oranges, we need to subtract the probability of liking either apples or oranges from 1. The probability of liking apples is 40/100, the probability of liking oranges is 30/100, and the probability of liking both is 20/100. Therefore, the probability of liking either apples or oranges is (40/100) + (30/100) – (20/100) = 50/100 = 0.5. Subtracting this from 1 gives us the probability of liking neither, which is 1 – 0.5 = 0.5.
Example: Let\’s say we have another group of 100 students. Out of these 100 students, 60 students like apples, 40 students like oranges, and 10 students like both fruits. The probability of liking either apples or oranges can be found as (60/100) + (40/100) – (10/100) = 90/100 = 0.9. Therefore, the probability of liking neither apples nor oranges is 1 – 0.9 = 0.1.
Question 2:
The heights (in inches) of a group of 50 students are normally distributed with a mean of 65 and a standard deviation of 3. What is the probability that a randomly selected student has a height between 62 and 68 inches?
A) 0.3413
B) 0.6826
C) 0.9544
D) 0.9973
Answer: C) 0.9544
Explanation: Since the heights are normally distributed, we can use the empirical rule to find the probability. According to the empirical rule, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations. In this case, the range of heights between 62 and 68 inches is within two standard deviations of the mean, so the probability is approximately 95% or 0.9544.
Example: Let\’s say we have another group of 50 students with heights that are normally distributed with a mean of 70 and a standard deviation of 4. The probability of randomly selecting a student with a height between 66 and 74 inches can be found using the same logic as above. Since this range is also within two standard deviations of the mean, the probability is approximately 95% or 0.9544.
Question 3:
A company wants to estimate the average monthly sales of its product with 95% confidence. In a sample of 100 customers, the average monthly sales were $500 with a standard deviation of $50. What is the margin of error for the estimate?
A) $1.96
B) $2.58
C) $4.32
D) $6.42
Answer: B) $2.58
Explanation: The margin of error can be calculated using the formula: Margin of Error = Critical value * Standard deviation / Square root of sample size. Since we want a 95% confidence level, the critical value is 1.96. The standard deviation is $50 and the sample size is 100. Therefore, the margin of error is 1.96 * ($50 / √100) = $2.58.
Example: Let\’s say another company wants to estimate the average monthly sales of its product with 99% confidence. In a sample of 200 customers, the average monthly sales were $1000 with a standard deviation of $100. The critical value for a 99% confidence level is 2.58. Using the same formula as above, the margin of error would be 2.58 * ($100 / √200) = $5.77.
Question 4:
A group of 500 students took a math test. The scores were normally distributed with a mean of 75 and a standard deviation of 10. What is the minimum score needed to be in the top 10% of the test takers?
A) 79
B) 81
C) 85
D) 89
Answer: D) 89
Explanation: To find the minimum score needed to be in the top 10%, we need to find the z-score corresponding to the 90th percentile and then convert it back to the original scale. The z-score can be found using the z-table or a calculator, which gives us a value of 1.28. Converting this back to the original scale using the formula X = Z * σ + μ, where X is the score, Z is the z-score, σ is the standard deviation, and μ is the mean, we get X = 1.28 * 10 + 75 = 89.
Example: Let\’s say another group of 500 students took a different math test. The scores were normally distributed with a mean of 80 and a standard deviation of 15. To be in the top 5% of test takers, we need to find the z-score corresponding to the 95th percentile. Using the same logic as above, the z-score is 1.65. Converting this back to the original scale, the minimum score needed to be in the top 5% would be 1.65 * 15 + 80 = 105.75.