One-Variable Data Analysis Multiple Choice Practice Problems for AP Statistics (page 2)

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By — McGraw-Hill Professional
Updated on Feb 5, 2011


  1. The correct answer is (a). I is correct since the mean is pulled in the direction of the large maximum value, 150 (well, large compared to the rest of the numbers in the set). II is not correct because the mode is y—there are three ys and only two 26s. III is not correct because 150 is an outlier (you can't actually compute the upper boundary for an outlier since the third quartile is y , but even if you use a larger value, 33, in place of y, 150 is still an outlier).
  2. The correct answer is (e).
  3. One-Variable Data Analysis Solutions to Practice Problems

    (On the TI-83/84, normalcdf (-100, 1.92) = normalcdf (-1000,70,65,206) = 0.9726 up to rounding error.)

  4. The correct answer is (d). The effect on the mean of a dataset of subtracting the same value is to reduce the old mean by that amount (that is, μx-k = μx-k). Because the original mean was 19, and 19 has been subtracted from every term, the new mean is 0. The effect on the standard deviation of a dataset of dividing each term by the same value is to divide the standard deviation by that value, that is,
  5. One-Variable Data Analysis Solutions to Practice Problems.

    Because the old standard deviation was 4, dividing every term by 4 yields a new standard deviation of 1. Note that the process of subtracting the mean from each term and dividing by the standard deviation creates a set of z-scores

    One-Variable Data Analysis Solutions to Practice Problems

    so that any complete set of z-scores has a mean of 0 and a standard deviation of 1. The shape is normal since any linear transformation of a normal distribution will still be normal.

  6. The correct answer is (b). The maximum length of a "whisker" in a modified boxplot is 1.5(IQR) = 1.5(40 - 18) = 33.
  7. The correct (best) answer is (d). Using Table A, the area under a normal curve between 63 and 75 is 0.6247 (z63 = –1.5 A1 = 0.0668, z75 = 0.5 A2 = 0.6915 A2 – A1 = 0.6247). Then (0.6247)(5,000) = 3123.5. Using the TI-83/84, normalcdf (63,75,72,6) × 5000 = 3123.3.
  8. The correct answer is (b). Since we do not know that the empirical rule applies, we must use Chebyshev's rule. Since 72–k (6) = 58, we find k = 2.333. Hence, there are at most % = 18.37% of the scores less than 58. Since there are 5000 scores, there are at most (0.1837)(5,000) = 919 scores less than 58. Note that it is unlikely that there are this many scores below 58 (since some of the 919 scores could be more than 2.333 standard deviation above the mean)—it's just the strongest statement we can make.
  9. The correct answer is (e). The graph is clearly not symmetric, bi-modal, or uniform. It is skewed to the right since that's the direction of the "tail" of the graph.
  10. The correct answer is (a). The median is resistant to extreme values, and the mean is not (that is, extreme values will exert a strong influence on the numerical value of the mean but not on the median). II and III involve statistics equal to or dependent upon the mean, so neither of them is resistant.
  11. The correct answer is (d). One-Variable Data Analysis Solutions to Practice Problems
  12. The correct answer is (b). A score at the 90th percentile has a z-score of 1.28.
  13. Thus, One-Variable Data Analysis Solutions to Practice Problems

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