Module 9 Assessment Readiness
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| Original Dot Plot | New Dot Plot |
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Now, we can find the median of the original 16 courses and then the new median when there are 17 courses. To determine the median, we need to find the middle value, or the mean of the two middle values, of the data when it is ordered from least to greatest.
Since we have 16 courses, the median M will be the mean of the 8^\text{th} and 9^\text{th} class sizes. 80,80,80,80,85,85,85, 85 85,90,90,90,95,95,100,105 Let's calculate it!
Thus, the median was 85 students.
Next, let's calculate the median of 17 courses. This time, because there is an odd number of values, the median will just be the middle value. Therefore, it is the 9^\text{th} value. 65,80,80,80,80,85,85,85, 85,85,90,90,90,95,95,100,105 The median number of students including the newly added course is 85. Since both of the medians are 85, the given statement is true.
First, let's determine Q_1 and Q_3 for the 17 courses. In Part A, we found that the median is 85. We find Q_1 and Q_3 by looking for the median values of the lower and upper halves of the data.
65,80,80, 80, 80,85,85,85, 85,
85,90,90, 90, 95,95,100,105
Finally, we can determine if the new course size is an outlier. Notice that, since it is below the least original value, we only need to check if it is an outlier on the lower end of things. x< Q_1-1.5(IQR) Let's substitute the given values into the inequality.
Substitute values
Multiply
Subtract term
Since 65 is not less than 61.25, it is not an outlier and the statement is false.