Exercise 1 Page 348

a To find if the new course will affect the median class size, let's first look at how the new student will change the dot plot. Remember, the new course has only 65 enrolled students.

Original Dot Plot	New Dot Plot

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.

Median of 16 courses

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!

M=85+85/2

AddTerms

Add terms

M=170/2

CalcQuot

Calculate quotient

M=85

Thus, the median was 85 students.

Median of 17 courses

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.

b Now we want to know if the new course is an outlier. A data point is an outlier if one of two things is true.

The value x is less than 1.5 times the IQR below the first quartile, x
The value x is greater than 1.5 times the IQR above the third quartile, x>Q_3+1.5(IQR).

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, 95,95,100,105Because there are an even number of values in each half of the data, we can calculate Q_1 and Q_3 by finding the mean of the middle two values. &Q_1: 80+80/2=80 &Q_3: 90+95/2=92.5 Next, we will find the interquartile range IQR.

IQR=Q_3-Q_1

SubstituteValues

Substitute values

IQR=92.5-80

SubTerm

Subtract term

IQR=12.5

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.

x< Q_1-1.5(IQR)

SubstituteValues

Substitute values

65? <80-1.5(12.5)

Multiply

65? <80-18.75

SubTerm

Subtract term

65≮ 61.25

Since 65 is not less than 61.25, it is not an outlier and the statement is false.

c Finally, we need to check whether or not the range increases by 40 when the 17^\text{th} course is added. The range R is the difference between the least and the greatest data values.

16 courses:& 80,80,80,80,85,85,85,85,85, &90,90,90,95,95,100, 105 17 courses:& 65,80,80,80,80,85,85,85,85, &85,90,90,90,95,95,100, 105 Let's calculate the range for both data sets to see how they differ. &16 courses: R=105-80=25 &17 courses: R=105-65=40 Since the difference between the two ranges is only 15, the statement is false.

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Exercises p.348

Exercise 1 Page 348

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Median of 16 courses

Median of 17 courses