Section 1.2 Solutions

1.42

a. There are 50 observations, so the median is in the (50+1)/2 = 25.5^th position. Hence, the median is (12.7 + 12.8)/2 = 12.75%. Now the data set is divided into two groups of 25. The median of a group of 25 observations is in the (25+1)/2 = 13^th position. Hence, the first quartile is 11.7% and the third quartile is 13.5%. Finally, the minimum value is 5.7% and the maximum value is 17.6%. Hence, the five-number summary is (5.7%, 11.7%, 12.75%, 13.5%, 17.6%).

b. The IQR is 13.5% - 11.7% = 1.8%. Then 1.5(IQR) = 1.5(1.8%) = 2.7%. So an observation is a suspected outlier if it lies below 11.7% - 2.7% = 9% or if it lies above 13.5% + 2.7% = 16.2%. Hence, both Alaska and Hawaii are flagged as suspected outliers; the value of 8.5% is also flagged.

1.44

a. The ordered data are shown below. There are 48 countries listed, so the median is in the (48+1)/2 = 24.5^th position. So the median is (2.8 + 3.6)/2 = 3.2 metric tons per person. Now the data set is divided into two groups of 24. The median of a group of 24 observations is in the (24+1)/2 = 12.5^th position. Hence, the first quartile is (0.7 + 0.8)/2 = 0.75 metric tons per person, and the third quartile is (7.6 + 8.0)/2 = 7.8 metric tons per person. Finally, the minimum value is 0.0 metric tons per person, and the maximum is 19.9 metric tons per person. Hence, the five-number summary is (0.0, 0.75, 3.2, 7.8, 19.9), where each value is measured in metric tons per person. Note that the distance between the median and the first quartile is 2.45 while the distance between the third quartile and the median is 4.6. Because the same number of observations fall within these ranges, the observations are more densely packed at the low end of the distribution and more spread out at the high end of the distribution; hence the distribution is skewed right.

Ordered Data:

0.0 0.0 0.1 0.1 0.2 0.2 0.2 0.2 0.3 0.3 0.5

0.7 0.8 0.9 0.9 1.0 1.2 1.4 1.7 1.8 2.3 2.5

2.5 2.8 3.6 3.7 3.8 3.9 3.9 4.6 4.8 5.1 6.1

6.8 7.3 7.6 8.0 8.1 8.8 9.0 9.1 9.7 10.0 10.2

11.0 16.0 17.0 19.9

b. The IQR is 7.8 – 0.75 = 7.05. Then 1.5(IQR) = 1.5(7.05) = 10.575. So an observation is a suspected outlier if it lies below 0.75 – 10.575 = -9.825 or if it lies above 7.8 + 10.575 = 18.375. Hence, the United States value of 19.9 metric tons per person is flagged as an outlier.

A histogram of the CO2 emissions is shown below. Graphically, the values of 16.0, 17.0, and 19.9 all appear to be outliers (i.e., they seem to lie outside the overall pattern of the data).

1.47

Salaries

$25,000, $25,000, $25,000, $25,000, $25,000, $60,000, $60,000, $255,000

The sample mean is .

Seven of the 8 employees earn less than the mean. The median salary is $25,000. The median is a better measure of a typical value (the mean is pulled up by the extreme value, $255,000).

1.49

The new mean is .

The median remains the same at $25,000.

1.55

This applet helps the user understand the relationship between mean and median (for distributions of different shapes). When there are only two data points, the mean and median are exactly the same (hence, only one arrow).

1.56

a. Because outliers don’t affect the median, it stays the same. The mean is affected by the extreme point, and is pulled toward the right.

b. The mean is now pulled to the left. The median switches to the middle-most point.

1.58

The ordered lengths are shown below.

Bihai - Ordered

46.34 46.44 46.64 46.67 46.75 46.81 46.94 47.12 47.12

47.43 48.07 48.15 48.34 48.36 50.12 50.26

Red - Ordered

37.40 37.78 37.87 37.97 38.01 38.07 38.10 38.20 38.23

38.79 38.87 39.16 39.63 39.78 40.57 40.66 41.47 41.69

41.90 41.93 42.01 42.18 43.09

Yellow - Ordered

34.57 34.63 35.17 35.45 35.68 36.03 36.03 36.11 36.52

36.66 36.78 36.82 37.02 37.10 38.13

For the H. bihai, the five-number summary is (46.34 mm, 46.71 mm, 47.12 mm, 48.245 mm, 50.26 mm).

For the H. caribaea red, the five-number summary is (37.40 mm, 38.07 mm, 39.16 mm, 41.69 mm, 43.09 mm).

For the H. caribaea yellow, the five-number summary is (34.57 mm, 35.45 mm, 36.11 mm, 36.82 mm, 38.13 mm)

The boxplots are shown below:

As measured by the median, the H. bihai variety is the longest flower, followed by the H. caribaea red and then the H. caribaea yellow. Furthermore, the H. caribaea red variety has the most variability in length.

1.66

In order for a data set of 5 numbers to have 10 as a median, the value 10 must be included in the data set; then there must be two numbers at or below 10 and two numbers at or above 10. For the mean to be 7, the sum of the five numbers must be 35. Furthermore, there must be a left skew to the distribution (since the mean is less than the median). One possible set of numbers is (2, 3, 10, 10, 10).

1.68

Here is the derivation not using summation notation:

Note:

The first equality holds simply by rearranging terms. The second equality holds by the definition of the sample mean.

1.69

a. The smallest possible standard deviation is 0, and that only happens if all the observations are exactly the same. For example, choosing (0, 0, 0, 0).

b. In this case, we want the largest possible spread, so pick the extremes: (0, 0, 10, 10).

c. There are 11 possible answers to part a, as any repeated number will work. The answer to part b is unique.

1.72

To determine the appropriate linear transformation, set up the following ratio and solve for : . Both the mean (measure of center) and the standard deviation (measure of spread) are affected by the multiplier 2.2. Hence, the new mean is 2.2(2.39) = 5.258 pounds, and the new standard deviation is 2.2(1.14) = 2.508 pounds.

1.76

Recall the definition of sample variance is

If we multiply each observation by 2.54 (note the sample mean will then be multiplied by 2.54) then the new sample variance is

Hence, the new variance is multiplied by . In general (for a transformation ), the new variance will be multiplied by .