Advanced Placement: Can You Solve These Statistics Questions? Trivia Quiz

1. Which of the following is a true statement about experimental design?

Randomization is a key component in experimental design. Randomization is used to spread uncontrolled variability among treatment groups.

Blocking eliminates the effects of all lurking variables.

The placebo effect is a concern for all experiments.

Randomization is a key component in experimental design because it helps to spread uncontrolled variability among treatment groups. By randomly assigning participants to different groups, any potential confounding variables are more likely to be evenly distributed across the groups. This increases the internal validity of the experiment and allows for more accurate conclusions to be drawn about the effects of the treatment.

Explanation

Randomization is a key component in experimental design because it helps to spread uncontrolled variability among treatment groups. By randomly assigning participants to different groups, any potential confounding variables are more likely to be evenly distributed across the groups. This increases the internal validity of the experiment and allows for more accurate conclusions to be drawn about the effects of the treatment.

2. Summary statistics are calculated for a data set that includes an outlier. If the outlier is removed, which summary statistic would be least affected?

Mean

Median

Range

Standard deviation

Variance

When an outlier is present in a dataset, it can significantly affect the mean since it is sensitive to extreme values. However, the median is not influenced by outliers as it only considers the middle value of the dataset. Therefore, if the outlier is removed, the median would be the least affected summary statistic.

Explanation

When an outlier is present in a dataset, it can significantly affect the mean since it is sensitive to extreme values. However, the median is not influenced by outliers as it only considers the middle value of the dataset. Therefore, if the outlier is removed, the median would be the least affected summary statistic.

3. Two friends, Tom and Janice, have cars in desperate need of repair. On any given day, the probability that Tom's car will break down is 0.5, the probability that Janice's car will break down is 0.5, and the probability that both cars will break down is 0.3. What is the probability that Tom and Janice's car will break down?

1.3

1.0

0.7

0.4

0.2

The probability that Tom's car will break down is 0.5, and the probability that Janice's car will break down is also 0.5. The probability that both cars will break down is 0.3. To find the probability that either Tom or Janice's car will break down, we can add the individual probabilities and subtract the probability of both cars breaking down. Therefore, the probability that either Tom or Janice's car will break down is 0.5 + 0.5 - 0.3 = 0.7.

Explanation

The probability that Tom's car will break down is 0.5, and the probability that Janice's car will break down is also 0.5. The probability that both cars will break down is 0.3. To find the probability that either Tom or Janice's car will break down, we can add the individual probabilities and subtract the probability of both cars breaking down. Therefore, the probability that either Tom or Janice's car will break down is 0.5 + 0.5 - 0.3 = 0.7.

4. Mr. B teaches two periods of English. He has 38 seniors in one period and 24 juniors in the other period. The overall mean for both periods combined on their Fall final grades was 87. If the junior period had a mean of 92, what was the approximate mean for the senior period?

82.6

83.8

89.5

87.0

90.4

The approximate mean for the senior period can be calculated by using the overall mean and the mean of the junior period. Since the overall mean for both periods combined is 87, and the mean of the junior period is 92, the mean of the senior period can be found by subtracting the junior period mean from the overall mean and then adding it to the overall mean. This can be represented as: (87 - 92) + 87 = 83.8. Therefore, the approximate mean for the senior period is 83.8.

Explanation

The approximate mean for the senior period can be calculated by using the overall mean and the mean of the junior period. Since the overall mean for both periods combined is 87, and the mean of the junior period is 92, the mean of the senior period can be found by subtracting the junior period mean from the overall mean and then adding it to the overall mean. This can be represented as: (87 - 92) + 87 = 83.8. Therefore, the approximate mean for the senior period is 83.8.

5. In a game of chance, three fair coins are tossed simultaneously. If all three coins are heads, then the player wins $15. If all three coins show tails, then the player wins $10. If it costs $5 to play the game, what are the players expected gain or loss at the end of two games?

The player can expect to gain $15 after two games.

The player can expect to gain $1.88 after two games.

The player can expect to gain $3.75 after two games.

The player can expect to lose $1.88 after two games.

The player can expect to lose $3.75 after two games.

Since it costs $5 to play the game and the player can win either $15 or $10, the player's expected gain or loss can be calculated by multiplying the probability of winning each amount by the respective amount and subtracting the cost of playing the game. The probability of getting all three heads is 1/8, and the probability of getting all three tails is also 1/8. Therefore, the player's expected gain from winning $15 is (1/8) * $15 = $1.88, and the expected gain from winning $10 is (1/8) * $10 = $1.25. The player's total expected gain is $1.88 + $1.25 - $5 = -$1.87. Since the player plays two games, the total expected gain or loss is -$1.87 * 2 = -$3.75. Therefore, the player can expect to lose $3.75 after two games.

Explanation

Since it costs $5 to play the game and the player can win either $15 or $10, the player's expected gain or loss can be calculated by multiplying the probability of winning each amount by the respective amount and subtracting the cost of playing the game. The probability of getting all three heads is 1/8, and the probability of getting all three tails is also 1/8. Therefore, the player's expected gain from winning $15 is (1/8) * $15 = $1.88, and the expected gain from winning $10 is (1/8) * $10 = $1.25. The player's total expected gain is $1.88 + $1.25 - $5 = -$1.87. Since the player plays two games, the total expected gain or loss is -$1.87 * 2 = -$3.75. Therefore, the player can expect to lose $3.75 after two games.