One-Variable Data Analysis Rapid Review for AP Statistics
Review the following concepts if necessary:
- Graphical Analysis for AP Statistics
- Histogram for AP Statistics
- Measures of Center for AP Statistics
- Measures of Spread for AP Statistics
- Position of a Term in a Distribution for AP Statistics
- Normal Distribution for AP Statistics
- Describe the shape of the histogram below:
- For the graph of problem #1, would you expect the mean to be larger than the median or the median to be larger than the mean? Why?
- The first quartile (Q1) of a dataset is 12 and the third quartile (Q3) is 18. What is the largest value above Q3 in the dataset that would not be an outlier?
- A distribution of quiz scores has = 35 and s = 4. Sara got 40. What was her z-score? What information does that give you?
Answer: Bi-modal, somewhat skewed to the left.
Answer: The graph is slightly skewed to the left, so we would expect the mean, which is not resistant, to be pulled slightly in that direction. Hence, we might expect to have the median be larger than the mean.
Answer: Outliers lie more than 1.5 IQRs below Q1 or above Q3. Q3 + 1.5(IQR) = 18 + 1.5(18 - 12) = 27. Any value greater than 27 would be an outlier. 27 is the largest value that would not be an outlier.
This means that Sara's score was 1.25 standard deviations above the mean, which puts it at the 89.4th percentile (normalcdf (-100,1.25)).
(By calculator: normalcdf (-100,20,25,7)= 0.2375.)
Answer: Mean = median = mode = 0. Standard deviation = 1.
Answer: The five-number summary is [12, 13, 15, 18.5, 28]. 28 is an outlier (anything larger than 18.5 + 1.5(18.5 – 13) = 26.75 is an outlier by the 1.5(IQR) rule). Since 20 is the largest nonoutlier in the dataset, it is the end of the upper whisker, as shown in the following diagram:
Answer: Because the mean is not resistant and is pulled toward the tail of the skewed distribution, you would prefer to use the median and IQR.
Answer: Since the distribution is skewed to the left, we must use Chebyshev's rule. We note that the interval given is the same distance (8) above and below = 48. Solving 48 + k(6) = 56 gives k = 1.33. Hence, there are at least % = 43.5% of the terms between 40 and 56.
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