Question

question 9 (multiple choice worth 10 points)
(01.02h mc)

a survey was taken of students in math classes to find out how many hours per day students spend on social media. the survey results for the first-, second-, and third-period classes are as follows:

first period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0
second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2
third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3

which is the best measure of center for second period, and why?

• mean, because there are no outliers that affect the center
• median, because there is 1 outlier that affects the center
• interquartile range, because there is 1 outlier that affects the center
• standard deviation, because there are no outliers that affect the center

Explanation:

Analyze the data set for the second period

$$ \text{Data: } \{3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2\} $$

$$ \text{Sorted Data: } \{0, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4\} $$

Determine outliers using the IQR method

$$ Q_1 = 1,\quad Q_3 = 3 $$

$$ \text{IQR} = Q_3 - Q_1 = 3 - 1 = 2 $$

$$ \text{Lower Bound} = Q_1 - 1.5 \times \text{IQR} = 1 - 3 = -2 $$

$$ \text{Upper Bound} = Q_3 + 1.5 \times \text{IQR} = 3 + 3 = 6 $$

Since all values in the second period data set lie within \([-2, 6]\), there are no outliers.

Choose the best measure of center

Since there are no outliers in the data set, the mean is the most appropriate and reliable measure of center.

Answer:

• (A) Mean, because there are no outliers that affect the center (Correct answer)
• (B) Median, because there is 1 outlier that affects the center
• (C) Interquartile range, because there is 1 outlier that affects the center
• (D) Standard deviation, because there are no outliers that affect the center