Chapter 1 Exploring Data
Back to Lesson 1-4
Chapter 1 Exploring Data Lesson 1-4
(pp. 31–37)
1. A; The five-number summary includes the maximum, minimum, and median values, not the mean. 2. a. minimum = 15,199; Q1 = 15,253; median =
15,457 + 16,263 2
= 15, 860 ;
Q3 = 16,919; maximum = 18,721 b. IQR = 16,919 – 15,253 = 1,666 c. 1.5(1,666) = 2,499 so Q1 – 2,499 = 12,754 and Q3 + 2,499 = 19,418. All charges are between these values so there are no outliers using this criterion. 3.
4. a. About 25% of the data in a distribution will fall below the Q1 value. b. About 50% of the data in a distribution will fall between the Q1 and Q3 values. c. About 25% of the data in a distribution will fall above the Q3 value. 5. B; Because the total count is 100 the median resides in the 5-10 bin. The only plot whose median is in this range is Plot B. 6. minimum 37 mm.; Q1 43.5 mm; median 44.5 mm; Q3 54.6 mm; maximum 72.5; 72.5 is an outlier 7. a. Answers vary. Sample: the frequency of scores in each interval b. Answers vary. Sample: the values of the median, Q1 , and Q3
must be natural numbers, the highest possible Q3 is 36. b. Q3 must always be greater than or equal to the mean so the least possible Q3 is 34. 9. Answers vary. Sample: {0, 0, 0, 0, 2, 2, 4, 4, 4, 8} 10. Answers vary. Sample: {0, 0, 2, 2, 4, 4, 6, 6, 8, 8} 11. a. about 20 + 15 + 7 + 4 + 3 + 1 + 2 + 1 = 53 b. about 3 + 26 + 71 = 100 people and = 15 . total people = 500 so about 100 500 c. about 85 + 89 + 81 + 46 + 44 = 345 345 so about 500 = 69% . d. Answers vary. Sample: The median wait time is between 5 and 6 minutes. The range of wait times is about 19 minutes. The data is skewed with a tail on the right. 12. Total spent on stock = $16,550. Total number of shares purchased = 2,200. = $7.52. Average price = $16,550 2,200 13. a. b.
5(4) + 8(3.7) 3.815 3.82 13 5(4) + 8(3.7) + 3(2.3) 3.531 16
c. 3.815 – 3.531 0.28
14. a. all of the words in English literature b. the first 250 words of a random Ernest Hemmingway novel c. the average number of letters per word 15. For n 6 , when n is of the form 4k or 4k + 1, for any k , then Q1 does not have to me a member of the set.
8. a. Q3 + 1.5(Q3 Q1 ) < 50 so 2.5Q3 < 50 + 1.5(28) and Q3 < 36.8 . Assuming the scores 8
Functions, Statistics and Trigonometry Solution Manual
3.53
Chapter 1, Lesson 1-4
Chapter 1 Exploring Data Lesson 1-4
(pp. 31–37)
1. A; The five-number summary includes the maximum, minimum, and median values, not the mean. 2. a. minimum = 15,199; Q1 = 15,253; median =
15,457 + 16,263 2
= 15, 860 ;
Q3 = 16,919; maximum = 18,721 b. IQR = 16,919 – 15,253 = 1,666 c. 1.5(1,666) = 2,499 so Q1 – 2,499 = 12,754 and Q3 + 2,499 = 19,418. All charges are between these values so there are no outliers using this criterion. 3.
4. a. About 25% of the data in a distribution will fall below the Q1 value. b. About 50% of the data in a distribution will fall between the Q1 and Q3 values. c. About 25% of the data in a distribution will fall above the Q3 value. 5. B; Because the total count is 100 the median resides in the 5-10 bin. The only plot whose median is in this range is Plot B. 6. minimum 37 mm.; Q1 43.5 mm; median 44.5 mm; Q3 54.6 mm; maximum 72.5; 72.5 is an outlier 7. a. Answers vary. Sample: the frequency of scores in each interval b. Answers vary. Sample: the values of the median, Q1 , and Q3
must be natural numbers, the highest possible Q3 is 36. b. Q3 must always be greater than or equal to the mean so the least possible Q3 is 34. 9. Answers vary. Sample: {0, 0, 0, 0, 2, 2, 4, 4, 4, 8} 10. Answers vary. Sample: {0, 0, 2, 2, 4, 4, 6, 6, 8, 8} 11. a. about 20 + 15 + 7 + 4 + 3 + 1 + 2 + 1 = 53 b. about 3 + 26 + 71 = 100 people and = 15 . total people = 500 so about 100 500 c. about 85 + 89 + 81 + 46 + 44 = 345 345 so about 500 = 69% . d. Answers vary. Sample: The median wait time is between 5 and 6 minutes. The range of wait times is about 19 minutes. The data is skewed with a tail on the right. 12. Total spent on stock = $16,550. Total number of shares purchased = 2,200. = $7.52. Average price = $16,550 2,200 13. a. b.
5(4) + 8(3.7) 3.815 3.82 13 5(4) + 8(3.7) + 3(2.3) 3.531 16
c. 3.815 – 3.531 0.28
14. a. all of the words in English literature b. the first 250 words of a random Ernest Hemmingway novel c. the average number of letters per word 15. For n 6 , when n is of the form 4k or 4k + 1, for any k , then Q1 does not have to me a member of the set.
8. a. Q3 + 1.5(Q3 Q1 ) < 50 so 2.5Q3 < 50 + 1.5(28) and Q3 < 36.8 . Assuming the scores 8
Functions, Statistics and Trigonometry Solution Manual
3.53
Chapter 1, Lesson 1-4
