Exercise 2 Page 379

a A new member who is 30 joins the club which is in the previous exercise. Thus, the new data plot can be shown as the following.

Let's order the entries including the new one.

30,32,33,34,34,35, 35,35,35,36,36,37,37 37,37,38,38,39,40 Let's find the new range. 30,32,33,34,34,35, 35,35,35,36,36,37,37 37,37,38,38,39, 40 We will find the difference between the extremes of the data set. Range: 40- 30=10 The range in the previous exercise was 8. The new range is 10 making the first statement true.

b To determine the median, we determine the middle entry as this list has an odd number of elements.

30,32,33,34,34,35, 35,35,35, 36,36,37,37 37,37,38,38,39,40 In this case the median will be the 10^\text{th} value. Median: 36 Since the median is the same as in the previous exercise, the second statement is false.

c We say that an entry of the data set x is an outlier if it satisfies the following inequality. x < Q_1-1.5(IQR) or x > Q_3+1.5(IQR) Now, let's find the IQR so we can use it in the inequality above.

&30,32,33,34, 34,35,35,35,35, 36, &36,37,37,37, 37,38,38,39,40 The IQR will be the difference between the lower quartile and the upper quartile.

IQR: 37-34=3 Next, let's determine inequalities for the outliers. Inequality I:& x< 34-1.5*3 Inequality II:& x> 37+1.5*3 Thus, the compound inequality can be written as the following. Inequality I:& x< 29.5 Inequality II:& x> 41.5 Compund:& x< 29.5 or x>41.5 Recall that the age of the new member is 30. We put this value into the inequality used to determine outliers. 30 ≮ 29.5 or 30 ≯ 41.5 Since the inequality is not satisfied, the new age is not considered at outlier making the last statement false.

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Exercises p.379-382

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Exercise 2 Page 379

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