1. REVIEW OF UNIT 1 1) The table displays the number of videos rented. Number of Videos Rented Number of Families 316 414 611 82 104 a. How many families were surveyed? 16 + 14 + 11 + 2 + 4 = 47

2. REVIEW OF UNIT 1 1) The table displays the number of videos rented. Number of Videos Rented Number of Families 316 414 611 82 104 b. Find the median number of videos rented by these families. Median # of families: 47+1 / 2 = 48 / 2 = 24 Since 16 families rented 3 videos, the 24 families would fall into the 4- video category. 16 + 14 = 30

3. REVIEW OF UNIT 1 1) The table displays the number of videos rented. Number of Videos Rented Number of Families 316 414 611 82 104 c. Find the mean number of videos rented by these families.

4. 2) The histogram below shows the ages of the people attending the showing of a movie. a. How many people attended this move? 5 60 55 20 10 15 10 5 2 5 + 60 + 55 + 20 + 10 + 15 + 10 + 5 + 2 = 182

5. 2) The histogram below shows the ages of the people attending the showing of a movie. b. Circle the interval below that contains the mean age of the people attending this movie. Explain how you can make this choice without actually computing the mean.5-10 years old15-20 years old 30-35 years old 5 60 55 20 10 15 10 5 2

6. 2) HOW TO COMPUTE MEAN FROM HISTOGRAM: 5 60 55 20 10 15 10 5 2 1) Find the central value of each bin (interval). 2.5 7.5 12.5 17.5 32.537.5 42.5 67.5 72.5

7. 2) HOW TO COMPUTE MEAN FROM HISTOGRAM: 5 60 55 20 10 15 10 5 2 2) Multiply central value times the entries in that bin and add the products: 2.5 7.5 12.5 17.5 32.537.5 42.5 67.5 72.5 2.5(5) + 7.5(60) + 12.5(55) + 17.5(20) + 32.5(10) + 37.5(15) +42.5(10) + 67.5(5) + 72.5(2) = 2945

8. 2) HOW TO COMPUTE MEAN FROM HISTOGRAM: 5 60 55 20 10 15 10 5 2 3) Divide this sum by total entries: 2.5 7.5 12.5 17.5 32.537.5 42.5 67.5 72.5 This means that the mean age is 16.18 which falls beteen 15-20. Mean age: 16.18

9. 2) The histogram below shows the ages of the people attending the showing of a movie. c. Approximately, what is the median? 5 60 55 20 10 15 10 5 2 Since 182 people attended, the median number of people is (182+1) / 2 = 183 / 2 = 91.5. Since 5 people are in the first bin, 60 people are in the second bin (5 + 60 = 65), then the 91.5 person has to lie in the third bin about ½ way through. This is the 10-15 age range. 12.5 is the age about ½ way into this 10-15 interval.

10. 2) The histogram below shows the ages of the people attending the showing of a movie. d. Approximately, what is the mean? 5 60 55 20 10 15 10 5 2

11. 2) The histogram below shows the ages of the people attending the showing of a movie. e. Draw an approximate box-whiskers plot for the histogram. 5 60 55 20 10 15 10 5 2 12.5 Median age: 12.5 26.5 28.5 Compute Q 1 : Find the number of people halfway from the minimum to the median (91.5 + 1) / 2 = 46.25 41.25 Counting over from the minium, Q 1 should fall around 7 or 8. Q1:7

12. 2) The histogram below shows the ages of the people attending the showing of a movie. e. Draw an approximate box-whiskers plot for the histogram. 5 60 55 20 10 15 10 5 2 12.5 Median age: 12.5 26.5 28.5 Compute Q 3 : Find the number of people halfway from the median to the maximum (91.5 + 1) / 2 = 46.25 41.25 Counting over from the median, Q 3 should fall around 19 or 20. Q 1 :7Q 3 :7

13. 2) The histogram below shows the ages of the people attending the showing of a movie. e. Draw an approximate box-whiskers plot for the histogram. 5 60 55 20 10 15 10 5 2 12.5 Median age: 12.5 26.5 28.5 All values for the box and whisker plot are now labeled in the histogram. 41.25 Q 1 :7Q 3 :7 min max Q 2 : 12.5 Q1:7Q1:7 Q 3 : 20 Min: 0 Max: 75

14. 2) The histogram below shows the ages of the people attending the showing of a movie. f. Describe the spread in the data. Skewed right

15. 3) (a.) Are there any outliers? Explain how you know..4 1.72.64 20.1 HOW TO DETERMINE IF THERE IS(ARE) OUTLIERS: If any value lies beyond either of the following, then these are outliers. If no value lies beyond either of the following, then there are no outliers. Q 1 – IQR(1.5) = 1.7 – 2.3(1.5) = 1.7 – 3.45 = - 1.75 IQR = Q 3 – Q 1 = 4 – 1.7 = 2.3 No outlier to the left of Q 1 Q 3 + IQR(1.5) = 4 + 2.3(1.5) = 4 + 3.45 = 7.45 Since 20.1 lies to the right of 7.45, 20.1 is an outlier.

16. 3) The box plot below shows the number of licensed drivers (in millions) for 25 states in the United States. The minimum value is 0.4 million, the lower quartile is 1.7 million, the median is 2.6 million, the upper quartile is 4.0 million, and the maximum is 20.1 million. (b.) Describe the shape of the distribution. Explain your reasoning. Skewed right

17. 3) The box plot below shows the number of licensed drivers (in millions) for 25 states in the United States. The minimum value is 0.4 million, the lower quartile is 1.7 million, the median is 2.6 million, the upper quartile is 4.0 million, and the maximum is 20.1 million. (c.) How many of these states have more than 4 million licensed drivers? Explain how you know. 25 % of the values lie above Q3. So 25(.25) = 6.25 states

18. (d.) Draw an appropriate histogram that matches this box and whisker plot. 3)

19. 4)Tamika kept track of the number of minutes that she exercised each day for two months. The distribution was approximately normal. She calculated the following summary statistics. Mean = 35 minutes, median = 34 minutes, standard deviation = 8 minutes, lower quartile = 24 minutes, and upper quartile = 41 minutes (a.) Find the interquartile range and write a sentence that describes what it tells you about Tamika’s exercise times. IQR = Q 3 – Q 1 = 41 – 24 = 17 50 % of her exercise minutes fell within this range

20. 4)Tamika kept track of the number of minutes that she exercised each day for two months. The distribution was approximately normal. She calculated the following summary statistics. Mean = 35 minutes, median = 34 minutes, standard deviation = 8 minutes, lower quartile = 24 minutes, and upper quartile = 41 minutes (b.) Suppose that she ran five more minutes each day for the next two months. What would the new quartiles be? Mean:Median: Standard Deviation:IQR: 40 39 8 17

21. 5)Jack’s homework grades are 75, 78, 96, 94, 93, 100, 101, 93, 60, 65, 64, 62, 73, 85, 87. Find the: a.) mean b.) median c.) mode d.) range e.) IQR 60, 62, 64, 65, 73, 75, 78, 85, 87, 93, 93, 94, 96, 100, 101 81.73 85 93 101 – 60 = 41 Q3 – Q1 = 94 – 65 = 29

22. 5)Jack’s homework grades are 75, 78, 96, 94, 93, 100, 101, 93, 60, 65, 64, 62, 73, 85, 87. Find the: f.) standard deviation 60, 62, 64, 65, 73, 75, 78, 85, 87, 93, 94, 96, 100, 101 14.49 or 14.006

23. 5)Jack’s homework grades are 75, 78, 96, 94, 93, 100, 101, 93, 60, 65, 64, 62, 73, 85, 87. Make a histogram and a box plot that displays his homework grades. 60708090100110 Q 1 65 Q 2 85 Q 3 94 min 60 max 101

24. 6) The box plots below show the ages of the members of the 2005 U. S. Olympic Hockey teams. Answer the questions that follow using the information in these plots. Be sure to clearly explain each of your answers. a. Which team had the greater range in ages? Min 18 Max 31 Q 1 22 Q 2 24 Q 3 26 Min 24 Max 43 Q 1 27 Q 2 32 Q 3 34 Women: 31 – 18 = 13 Men: 43 – 24 = 19 Men

25. 6) The box plots below show the ages of the members of the 2005 U. S. Olympic Hockey teams. Answer the questions that follow using the information in these plots. Be sure to clearly explain each of your answers. b. What percentage of the female players were at least 22 years old? Min 18 Max 31 Q 1 22 Q 2 24 Q 3 26 Min 24 Max 43 Q 1 27 Q 2 32 Q 3 34 75 %

26. 6) The box plots below show the ages of the members of the 2005 U. S. Olympic Hockey teams. Answer the questions that follow using the information in these plots. Be sure to clearly explain each of your answers. c. True or False? More men were older than 34 than were younger than 27. Min 18 Max 31 Q 1 22 Q 2 24 Q 3 26 Min 24 Max 43 Q 1 27 Q 2 32 Q 3 34 Both are in the 25 % range.

27. 6) The box plots below show the ages of the members of the 2005 U. S. Olympic Hockey teams. Answer the questions that follow using the information in these plots. Be sure to clearly explain each of your answers. d. Which team had the higher standard deviation? Min 18 Max 31 Q 1 22 Q 2 24 Q 3 26 Min 24 Max 43 Q 1 27 Q 2 32 Q 3 34 Men

28. 6) e. Draw a histogram for the men. Min 18 Max 31 Q 1 22 Q 2 24 Q 3 26 Min 24 Max 43 Q 1 27 Q 2 32 Q 3 34

29. 7) a. Which of the following examples are most likely to have the largest interval ranges for a histogram and interquartile range for a box plot? I: The weight (in pounds) of 30 people at the mall. II: The ages of 30 people at the mall. III: The amount of money in the wallet of 30 people at the mall. IV: The shoe sizes of 30 people at the mall. b. Which of the above is most likely to have the smallest intervals and quartile range? I IV

30. 8) The following are attendance for home football games: 287, 263, 286, 286, 256, 70, 255, and 300. If the attendance for the 6 th game (70) wasn’t included, what would be the answer for each of the following? a.Which measure would increase the most: mean, mode, median, or standard deviation? b. Will the interquartile range increase or decrease? mean neither

31. 8) The following are attendance for home football games: 287, 263, 286, 286, 256, 70, 255, and 300. If the attendance for the 6 th game (70) wasn’t included, what would be the answer for each of the following? c. Will the range increase or decrease? d. Will the standard deviation increase or decrease? decrease

32. 9) A teacher interviewed 200 students and found the following results: MaleFemale Sophomore.2.1 Junior.1.25 Senior.25.1 a. How many senior males did the teacher interview? b. If the student was a sophomore, what gender would this student more likely be? c. How many more female juniors were there than male juniors?.2(200) = 40.1(200) = 20.25(200) = 50.1(200) = 20.25(200) = 50.1(200) = 20 50 male 50 – 20 = 30

33. 10) A van holds 15 people. Which of the following would you need to know if you wanted to calculate the approximate total weight of the people in the van? (median, mode, standard deviation, range, or mean) mean

34. 11) Solve the following equations: a. b. - 5 x < - 4 - 3 2 x = - 18 22 x = - 9

35. 11) Solve the following equations: c. d. e. 14 x = 7 / 14 = ½ =.5 - 2 x 3 = 3 x – 10 + 10 13 = 3 x 3 3 x = 13 / 3 = 4.3 - 5 - x = - 17 (- 1)(- x) = (- 17)(- 1) x = 17