Question
1. The following list is ordered from smallest to largest: 25, 26, 26, 30, y, y, y, 33, 150. Which of the following statements is (are) true?
I. The mean is greater than the median
II. The mode is 26
III. There are no outliers in the data
a. I only
b. I and II only
c. III only
d. I and III only
e. II and III only
Answer/Explanation
Markscheme: 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).
Question
2. Megan wonders how the size of her beagle Herbie compares with other beagles. Herbie is 40.6 cm tall. Megan learned on the internet that beagles heights are approximately normally distributed with a mean of 38.5 cm and a standard deviation of 1.25 cm. What is the percentile rank of Herbie’s height?
a. 59
b. 65
c. 74
d. 92
e. 95
Answer/Explanation
Markscheme: e
\(z=\frac{40.6-38.5}{1.25}=1.68\)\(\rightarrow \) percentile = 0.9726 (see drawing below):
Question
3. The mean and standard deviation of a normally distributed dataset are 19 and 4, respectively. 19 is subtracted from every term in the dataset and then the result is divided by 4. Which of the following best describes the resulting distribution?
a. It has a mean of 0 and a standard deviation of 1.
b. It has a mean of 0, a standard deviation of 4, and its shape is normal.
c. It has a mean of 1 and a standard deviation of 0.
d. It has a mean of 0, a standard deviation of 1, and its shape is normal.
e. It has a mean of 0, a standard deviation of 4, and its shape is unknown.
Answer/Explanation
Markscheme: d
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,
\(\sigma_{\frac{x}{k}} = \frac{\sigma_x}{k}\)
Because the old standard deviation was 4, dividing every team 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
\(z_x = \frac{x-\bar{x}}{s}\)
so that 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 normal distribution will still be normal.
Question
4. The five-number summary for a one-variable dataset is {5, 18, 20, 40, 75}. If you wanted to construct a box-and-whiskers plot for the dataset, what would be the maximum possible length of the right side “whisker”?
a. 35
b. 33
c. 5
d. 55
e. 53
Answer/Explanation
Markscheme: b
The maximum length of a “whisker” in a box-and-whiskers plot is 1.5(IQR) = 1.5(40-18) = 33.
Question
5. A set of 5000 scores on a college readiness exam are known to be approximately normally distributed with mean 72 and standard deviation 6. To the nearest integer value, how many scores are there between 63 and 75?
a. 0.6247
b. 4,115
c. 3,650
d. 3,123
e. 3,227
Answer/Explanation
Markscheme: d
Using Table A, the area under a normal curve between 63 and 75 is 0.6247 (\(z_{53}\) = -1.5 \(\Rightarrow A_1\) = 0.0668, \(z_{75}\) = 0.5 \(\Rightarrow A_2 = 0.6915 \Rightarrow A_2 – A_1 = 0.6247\)). Then (0.6247)(5,000) = 3123.5. Using the TI-83/84, normalcdf(63,75,72,6) \(\times \) 5000 = 3123.3