Math 1530

Unit 1 Describing Data

Review

Determine whether the data are qualitative or quantitative:
1. the colors of automobiles on a used car lot qualitative
2. the numbers on the shirts of a soccer team qualitative
3. the number of seats in a movie theater quantitative
4. a list of house numbers on your street qualitative
5. the ages of a sample of 350 employees of a large hospital quantitative
Identify the data set’s level of measurement (nominal, ordinal, interval, ratio):
1. hair color of players on a high school tennis team nominal
2. numbers on the shirts of a soccer team nominal
3. ages of students in a statistics class ratio
4. temperatures of 22 selected refrigerators interval
5. number of milligrams of tar in 28 cigarettes ratio
6. number of pages in your statistics book ratio
7. marriage status of the faculty at the local community college nominal
8. the rank of a winning Super Bowl team in their division ordinal
9. the ratings of a movie ranging from “poor” to “good” to “excellent” ordinal
10. the final grades (A,B,C,D, and F) for students in a chemistry class ordinal
11. the annual salaries for all teachers in Utah ratio
12. list of zip codes for Chicago nominal
13. the nationalities listed in a recent survey nominal
14. the amount of fat (in grams) in 44 cookies ratio

Here we list the 20 countries that emitted the most carbon dioxide in 2015. Construct the following using the data: Frequency Distribution, Relative Frequency Distribution, Cumulative Frequency Distribution, Histogram, Dot Plot, Stem and Leaf. For the frequency distribution use a class width of 3 and 0 - 2.9 as the first class.

Rank and Country	2015 Per Capita Carbon Dioxide Emissions from Fuel Combustion (metric tons)
1 China	6.6
2 United States	15.5
3 India	1.6
4 Russia	10.2
5 Japan	9.0
6 Germany	8.9
7 South Korea	11.6
8 Iran	7.0
9 Canada	15.3
10 Saudia Arabia	16.9
11 Brazil	2.2
12 Mexico	3.7
13 Indonesia	1.7
14 south Africa	7.8
15 United Kingdom	6.0
16 Australia	15.8
17 Italy	5.5
18 Turkey	4.1
19 France	4.4
20 Poland	7.3

Frequency
EMISSIONS	FREQUENCY
0 – 2.9	3
3 - 5.9	4
6 - 8.9	6
9 - 11.9	3
12 - 14.9	0
15 - 17.9	4

Relative Frequency
EMISSIONS	RELATIVE FREQUENCY
0 - 2.9	15%
3 – 5.9	20%
6 - 8.9	30%
9 - 11.9	15%
12 - 14.9	0%
15 - 17.9	20%

Cumulative Frequency
EMISSIONS	CUMULATIVE FREQUENCY
0 - 2.9	3
3 - 5.9	7
6 - 8.9	13
9 - 11.9	16
12 - 14.9	16
15 - 17.9	20

Stem	Leaves
1	6 7
2	2
3	7
4	1 4
5
6	0 6
7	0 3 8
8	9
9	0
10	2
11	6
12
13
14
15	3 5 8
16	9
Legend 16	9 = 16.9

Below is a random sample of life expectancies from 20 countries:

70.5	65	70	51.5	57.5	61	78.5	61	72	64.5
56.5	73	69	52.5	78.5	54	74.5	76	70	68.5

Make a frequency table of the life expectancies.

Use a starting lower class limit of 50.0 and a class width of 5.0.

Class	Frequency
50.0 – 54.9	3
55.0 – 59.9	2
60.0 – 64.9	3
65.0 – 69.9	3
70.0 – 74.9	6
75.0 – 79.9	3

Answer the following questions based on your histogram:

What are the class midpoints?
52.45, 57.45, 62.45, 67.45, 72.45, 77.45
What are your lower class limits?
50.0, 55.0, 60.0, 65.0, 70.0, 75.0
What are your upper class limits?
54.9, 59.9, 64.9, 69.9, 74.9, 79.9
Draw a histogram using the class midpoints:

Use the same data to create a relative frequency distribution:

Classes	Relative Frequency
50.0 - 54.9	3/20 = 15%
55.0 - 59.9	2/20 = 10%
60.0 - 64.9	3/20 = 15%
65.0 - 69.9	3/20 = 15%
70.0 - 74.9	6/20 = 30%
75.0 - 79.9	3/20 = 15%

Use the following data to complete a-e:

AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest):

3	4	8	8	10	11	12	13	14	15
15	16	16	17	17	18	21	22	22	24
24	25	26	26	27	27	29	29	31	32
33	33	34	34	35	37	40	44	44	47

Calculate the measures of center from the given list of numbers.
Mean: 23.6

Median: 24

Mode: multi-modal

Midrange: 25

Create a frequency table using 2 as the lower limit of the first class and a class width of 8.

CLASS	FREQUENCY
2 - 9	4
10 - 17	11
18 - 25	7
26 - 33	10
34 - 41	5
42 - 49	3

ESTIMATE the mean of the data using the frequency table. Mean = 23.5
ESTIMATE the median of the data using the frequency table. First, identify the position of the median. Which CLASS in the frequency table contains the median?
The position of the median is 41/2 = 20.5, so the 21st term, which is in the third CLASS 18 - 25.

These are the volumes (in ounces) of randomly selected cans of Coke:

12.3

12.0

12.1

12.3

12.2

12.3

12.2

Find the Mean, median, mode, midrange. Mean = 12.2 Median = 12.2 Mode = 12.3 Midrange = 2.15

Find the mean of the following frequency distribution:

GPA	FREQUENCY	CLASS MIDPOINT	Frequency x Midpoint
0 - 0.9	4	0.45	1.8
1 - 1.9	7	1.45	10.15
2 - 2.9	12	2.45	29.4
3 - 3.9	15	3.45	51.75
4 - 4.9	6	4.45	26.7
	SUM = 44		SUM = 119.8

119.8/44 = 2.72

MEAN = 2.72

What is the shape of the data represented in the frequency distribution?
Skewed to the left

The ages of the employees at a local newspaper are given. Use the data to complete a-d:

Calculate the measures of variation from the given list of numbers. Range = 43 Variance = 232.6 Standard Deviation = 15.3

Create a frequency table using 20 as the lower limit of the first class and a class width of 10.

CLASS	FREQUENCY	MIDPOINT
20 -29	3	24.5
30 - 39	3	34.5
40 - 49	1	44.5
50 - 59	4	54.5
60 - 69	2	64.5

ESTIMATE the mean of the ages using the frequency table. Mean = 43.7
ESTIMATE the standard deviation of the ages using the frequency table. Standard Deviation = 15.0

Use the frequency table to estimate the mean and standard deviation of ticketed speeds:

Speed in mph of Driver Ticketed in 30 mph Zone	Frequency of Speed Reported on the Ticket	Midpoint
42 - 45	10	43.5
46 - 49	14	47.5
50 - 53	7	51.5
54 - 57	3	55.5
58 - 61	1	59.5

Estimate the mean of the data: Mean = 48.2
Estimate the standard deviation of the data: Standard Deviation = 4.2

FIVE-NUMBER SUMMARIES AND PERCENTILES

The circumference measurements (in cm) of a sample of randomly selected trees on a farmer’s property is given below. Use the data to answer the following questions.

18	18	19	24	31	34	37	37	38	39
40	41	49	51	51	52	53	55	83	112

CALCULATE THE FOLLOWING

Mean: 44.1

Median: 39.5

Mode: 18, 37, 51

MidRange: 65

Range: 94

Variance: 492.8

Standard Deviation: 22.2

Q₁: 32.5

Q₃: 51.5

IQR: 51.5 - 32.5 = 19

Are there any outliers in the data?

Yes, 83 and 112 are outliers

Lower Outlier Limit: Q1 – 1.5*IQR 32 – 1.5*19 = 4

Any data point less than 4 is an outlier.

Upper Outlier Limit: Q3 + 1.5*IQR 51.5 + 1.5*19 = 80

Any data point greater than 80 is an outlier.
CREATE A BOXPLOT OF THE DATA: Mark outliers clearly on the boxplot.
What is the Percentile of 24? 15^th percentile
$\frac{3}{20} \times 100=15$

The average life expectance of a specific brand of tires is 40,000 miles and has a mound shaped distribution (bell-shaped). The standard deviation is 7,500 miles.
1. Construct and label a bell curve representing the distribution.
2. What percentage of all tires have a life expectancy that is:
  Below 32,500 miles? 16%     0.15% + 2.35% + 13.5%
  
  Above 25,000 miles? 97.5%     13.5% + 34% + 34% 13.5% + 2.35% + 0.15%
  
  Between 40,000 and 55,000 miles? 47.5%     34% + 13.5%
According to dairymoos.com, the mean weight of an adult cow is 1500 lb. Assuming the weights are normally distributed with a standard deviation of 180lb,
1. Construct and label a bell curve representing the distribution.
2. What percentage of adult cows weigh:
  Between 1320 pounds and 1860 pounds? 81.5% 34% + 34% + 13.5%
  
  Less than 1500 pounds? 50%
  
  More than 1680 pounds? 16%
Heights of women in the population have a bell-shaped distribution with a mean of 161cm and a standard deviation of 7cm. If one woman is randomly selected, what is the probability her height will be:
1. less than 168 cm? 84%
2. greater than 147 cm? 97.5%
3. between 154 cm and 168 cm? 68%
4. between 147 cm and 175 cm? 95%
5. Find a z score for a woman that is 156 cm tall: $z_{156}=\frac{156-161}{7}=-0.71$
6. What does this z-score mean in words:
  The woman’s height of 156 cm is 0.71 standard deviations below the mean of 161 cm.
7. Find a z score for a woman that is 170 cm tall: $z_{170}=\frac{170-161}{7}=1.29$
8. What does this z-score mean in words:
  The woman’s height of 170 cm is 1.29 standard deviations above the mean of 161 cm.
9. Find a z score for a woman that is 161 cm tall: $z_{161}=\frac{161-161}{7}=0$
10. What does this z-score mean in words:
  The woman’s height of 161 cm is 0 standard deviations above or below the mean of 161 cm.
According to MarathonGuide.com, in 2010 the average time it took to run a marathon was about 4.5 hours with a standard deviation of about 1 hour.
1. Construct and label a bell curve representing the distribution.
2. Find the z-score for a runner who finishes in 3.2 hours $z_{3.2}=\frac{3.2-4.5}{1}=-1.3$
3. What does this z-score mean in words:
  A runner’s time of 3.2 hours is 1.3 standard deviations below the mean time of 4.5 hours. (faster than average)
4. Find the z-score for a runner who finishes in 5.2 hours: $z_{5.2}=\frac{5.2-4.5}{1}=0.7$
5. What does this z-score mean in words:
  A runner’s time of 5.2 hours is 0.7 standard deviations above the mean time of 4.5 hours. (slower than average)
6. Would you consider someone who finishes a marathon in 2 hours and 15 minutes to be very fast? What would their z-score be and what does that mean? hours and 15 minutes = 2.25 hours (because 15 minutes is ¼ of an hour)
  $z_{2.25}=\frac{2.25-4.5}{1}=-2.25$
  
  A runner’s time of 2.25 hours is 2.25 standard deviations below the mean time of 4.5 hours. This runner’s time would be more than 2 standard deviations below the mean and would be considered unusually fast.

Review of Frequency Distributions

Use the data to construct a Frequency Distribution Table: Begin with a lower class limit of 30 and a class width of 15.

32 49 53 57 61 64 66 68 68 68 71 72 72 75 79 80 83 85 90 93

CLASS	FREQUENCY	RELATIVE FREQUENCY
30- 44	1	0.05
45-59	3	0.15
60-74	9	0.45
75-89	5	0.25
90-104	2	0.10

Cumulative Class	Cumulative Frequency
Less than 45	1
Less than 60	4
Less than 75	13
Less than 90	18
Less than 105	20

Find the following using the Frequency Table (not the Relative or Cumulative summary information):
- •Lower Class Limit of the 3rd Class 60
- Lower Class Boundary of the 3rd Class 59.5
- Midpoint of the 3rd Class 67
Use the Frequency Distribution (not Relative or Cumulative) to draw a Histogram of the data:

3	4	8	8	10	11	12	13	14	15
15	16	16	17	17	18	21	22	22	24
24	25	26	26	27	27	29	29	31	32
33	33	34	34	35	37	40	44	44	47

3	4	8	8	10	11	12	13	14	15
15	16	16	17	17	18	21	22	22	24
24	25	26	26	27	27	29	29	31	32
33	33	34	34	35	37	40	44	44	47

3	4	8	8	10	11	12	13	14	15
15	16	16	17	17	18	21	22	22	24
24	25	26	26	27	27	29	29	31	32
33	33	34	34	35	37	40	44	44	47