Solutions manual for essential statistics 1st edition william navidi

When a data set is skewed to the right, the mean is greater than the median.. The data is skewed to the right, so the mean is greater than the median.. Since there are 8 data values, t

Solutions Manual for Essential Statistics 1st Edition William Navidi

Essential Statistics 1st Edition Test Bank Navidi Monk

SECTION 3.1 EXERCISES

Understanding the Concepts

Exercises 1-6 are the Check Your Understanding exercises located within the section Their answers are found on page 104

7 mean

8 median

9 extreme values

10 mode

11 False If no data value appears more than once, then there is no mode

12 False The median is resistant

13 False When a data set is skewed to the right, the mean is greater than the median

14 True

Practicing the Skills

15 mean = 23.4, median = 26, and the mode = 27

16 mean = 3, median = 15, and the mode = −20

17 mean = 5.5, median = 14, and the mode = 28

18 mean = 81.5, median = 93.5, and the mode = 98

Chapter 3: Numerical Summaries of Data

19 mean = (5 13) (15 7) (25 10) (35 9) (45 11)

50

24.6

50 

20 mean = (8 2) (24 14) (40 6) (56 13) (72 15)

50

48

50 

21 mean = (25 17) (75 26) (125 14) (175 34) (225 26) (275 8)

125

145.0

125 

22 mean = (10 18) (30 11) (50 6) (70 6) (90 10) (110 5)

56

47.9

56 

23 The correct choice is option (ii) The data is skewed to the right, so the mean is greater

than the median This narrows the choices to (ii) and (iv) Since the data set “balances” around 4.5, (ii) is the answer

24 The correct choice is option (iv) The data is closely symmetrical, so the mean should be

close to the median Also, more of the data is clearly to the right of the largest rectangle

over 3.5, so the mean and median should be higher than 3.5 These observations yield

choice (iv)

25 The correct choice is (i) The data set is roughly symmetric, so the mean and median are

approximately equal This eliminates choices (ii) and (iii) Since the center of the graph is

around 8, the mean and median need to be around 8 This eliminates choice (iv)

26 The correct choice is (iii) When a data set is skewed to the left, the mean is less than the

median This eliminates (ii) and (iv) Since half of the data values lie below the median,

and the other half lie above, and the height of the last relative frequency “bar” is at 5, the

median needs to be somewhere between 5 and 6 This eliminates choice (i)

27 The data are 12, 20, 27, 27, 29, 37, 38, 41, and 43 The mean = 274

9  30.4 Since there are 9 data values, the median is the 5th number when sorted from low to high That number is 29 Since 27 is the only data value that repeats, the mode is 27

28 The data are 8, 11, 16, 23, 25, 25, 29, and 30 The mean = 167

8 20.9 Since there are 8 data values, the median is the average between the 4th and 5th numbers when sorted from low to

high The median is therefore the average between 23 and 25, which is 24 Since 25 is the

only data value that repeats, the mode is 25

Working with the Concepts

29 Since the mean is much greater than the median, we should expect that the data is skewed to

the right

30 Since the mean and median are basically equal, we should expect the histogram to be

approximately symmetric

31 (A) mean = 1740

66 290 calories per restaurant

(B) Since there are six numbers, the median is the average between the 3rd and 4th numbers when sorted from low to high The median is therefore the average between 290 and

310, which is 300 calories

32 (A) mean = 12.3

8  1.538 seconds per event

(B) Since there were 8 data values, the median is the average between the 4th and 5th sorted values Those numbers are 0.67 and 1.89, making the median 1.28 seconds

(C) The mean would increase more 18.9 is an extreme value in this particular data set, and

extreme values affect the mean The median is much more resistant

33 (A) mean = 699

20  35.0 hours per model

(B) Since there are 20 numbers, the median is the average between the 10th and 11th numbers when sorted from low to high Both of these values are 35, so the median is 35 hours

(C) 23, 25, 29, 35, 38, and 47 (all in hours)

(D) Since the mean and median are both basically 35, we should expect the data to be

approximately symmetric

34 (A) 6.65 million users per application

(B) Since there are 25 data values, the median is the 13th data value Noting that the data is already sorted from highest to lowest, the median is 4.7 million users

(C) Since the mean is greater than the median, we should expect that the data is skewed to

the right

35 (A) mean = 13.39 rating per program

The median is the average between the 10th and 11th data values Since both of these values are 12, the median is a 12 rating

(B) mean = 10.76 rating per program

The median is the average between the 10th and 11th data values The average between 10.4 and 9.7 is 10.05 The median is a 10.05 rating

(C) Seeing that the mean and median are both considerably higher in Part A than they are in Part B, the answer to Part C is Yes That is, the audience today is now spread out over

many more channels than they used to be, so that ratings are not as high today as they

used to be

36 (A) mean = 474

16  29.6 breweries per state

(B) mean = 619

11  56.3 breweries per state

(C) The median is the average between the 8th and 9th data values when the data is first sorted The average between 20 and 22 is 21 breweries

(D) Because there are 11 data values, the median is the 6th data value when sorted That value is 30 breweries

(E) The West has a lot more, since both its mean and median are considerably greater than

are the East’s

(F) mean = 31.4 breweries per state, and the median = 25.5 breweries

(G) The means and medians become much more similar This is because CA represented an

extreme data vale on the high end So its removal caused the statistics in that region to

go way down

37 (A) mean = 2.307 dollars per gallon per country

Because there are 9 data values, the median is the 5th value when sorted That value is 2.02 dollars per gallon

(B) mean = 3.968 dollars per gallon per country

Because there are 9 data values, the median is the 5th value when sorted That value is 3.59 dollars per gallon

(C) The mean did, by 9 cents

38 (A) mean = 7580.3

24  315.85 thousands of dollars per city Since there are 24 data values, the median is the average between the 12th and 13th values when sorted Those numbers are 255.5 and 286.4, making the median 270.95 dollars (in thousands)

(B) mean = 6225.9

24  259.41 thousands of dollars per city Since there are 24 data values, the median is the average between the 12th and 13th values when sorted Those numbers are 227.2 and 241.1, making the median 234.15 dollars (in thousands)

(C) The mean decreased more than the median did

39 (A) mean = 3552

12  296 pounds per man Since there are 12 data values, the median is the average between the 6th and 7th sorted values Those numbers are 302 and 303, making the median 302.5 pounds

(B) mean = 3426

12  285.5 pounds per man

Since there are 12 data values, the median is the average between the 6th and 7th sorted values Those numbers are 274 and 296, making the median 285 pounds

(C) Offensive lineman tend to be heavier

40 (A) mean = 10658.38

22  484.472 stock price per day Because there are 22 data values, the median is the average between the 11th and 12th data values when sorted Those numbers are 485.52 and 486.25 The median is 485.885 stock price

(B) mean = 9932.29

21  472.966 stock price per day Because there are 21 data values, the median is the 11th value when sorted The median

is 481.59 stock price

(C) Comparing both means and both medians, we see that the prices are about the same

41 (A) mean = 17.3

10  1.73 billion dollars per company

(B) The data is already sorted Since there are 10 data values, the median is the average

between the 5th and 6th values Both of these are 1.4, so the median is 1.4 billion dollars

(C) There are two modes They are 1.2 and 1.4 billion dollars

(D) Since 32 would now represent an extreme data value on the high end, the mean would be

pulled away from the median in the direction of that value Therefore, the mean would increase, but the median would be unaffected

42 (A) mean = 442.58

22 20.117 dollars per magazine

(B) There are 22 data values, therefore the median is the average between the 11th and 12th sorted values Both of these are 14.98, so the median is $14.98

(C) $14.98

(D) $116.99

(E) The mean would move much closer towards the median It would decrease to 15.504

dollars per magazine The median and the mode would remain unchanged

43 (A) mean = (16 339) (26 1544) (36 844) (46 686) (56 413) (66 132)

3958

35.2

3958  Therefore, the mean is 35.2 years of age per victim

(B) Too small, because we multiply the midpoint of the class interval by the number of

fatalities The midpoint of that first interval is 16 However, really that is the start of the

interval since in most states people cannot start driving until 16

44 (A) mean = (5 283) (15 203) (25 217) (35 256) (45 176)

1286

(55 92) (65 21) (75 23) (85 12) (95 3)          36220 28.2

1286

Therefore, the mean is 28.2 years of age per resident

(B) Too large, because we multiply the midpoint of the class interval by the number of

residents The midpoint of that last interval is 95, so if all three residents were just 90,

we are adding 15 extra years to the numerator of the mean

45 (A) mean = 1914

51  37.5 thousands of dollars per state

(B) Since there are 51 data values, the median is the 26th sorted value This number is 36

thousand dollars

(C) Skewed to the right, because the mean is greater than the median

(D) The histogram below does agree with our expectation in part (C):

46 (A) mean = 1622.54

43  37.733 stock price per company Since there are 43 data values, the median is the 22nd sorted value The median = 28.42 stock price

(B) Since the mean is considerably greater than the median, we would expect a histogram of

the data to be skewed to the right

(C) The histogram below does agree with our expectation in part (B):

47 Since fiction occurs in the data the most often, it is the mode

48 Since television occurs in the data the most often, it is the mode

49 (A) 13 Six numbers are below the median, there is only one 17, so 6 numbers will also be

above the median, making n = 13, which is indeed odd

(B) 12 Since n is even, and the median value is not an actual value in the data set, there

must also be 6 numbers above the median

50 (A) 17 Eight numbers are below the median, there is only one 10, so 8 numbers will also be

above the median, making n = 17, which is indeed odd

(B) 18 There will be the 8 numbers below the median, the two tens at the median, and the 8

numbers above the median of 10

51 208 This answers comes from solving the equation: 202 802

5

x





52 88 This answer comes from solving the equation: 85 422

6

x





53 (A) $220,600 per year

(B) $20,000

(C) The mean would better depict their current immediate situation That is because the last year of

the data is when they won the lottery money, so currently, they have much more money than they are used to having

54 The median would Since the data is heavily skewed to the right, there are some very rich

families that are extreme data values, which pull the mean away from the median in their

direction However, the median is more indicative of the norm for the residents of the town Therefore, the estimate should be based on this statistic

55 Answers will vary One possible one is 1, 4, 5, 6, 9

56 Answers will vary The data set given in exercise 55 works here as well

57 Answers will vary One possible one is 10, 20, 30, 40, 50, 60

58 Answers will vary The data set given in exercise 57 works here as well

59 No, the midrange is not resistant This is because it is computed using the largest and

smallest data values Therefore, if either the largest or smallest data value is an extreme data value, the midrange will be highly affected, and thus, is not resistant

60 Yes, all three will be equal because the data set contains only 2 numbers In such an

instance, all three indicators will simply be the average of the two numbers

Extending the Concepts

61 (A) 68.4 inches per person

(B) 68 inches

(C) 5.417, 6, 5.667, 5.583, 5.833

(D) 5.7 feet per person Yes, the two are equal

(E) 5.667 feet Yes, the two are equal

62 (A) $45,000 per employee

(B) $40,000

(C) $46,000 per employee, yes

(D) $47,250 per employee, yes

63 (A) Since the sum of the 20 numbers is 100, the mean equals 100

20 = 5 Also, since

20

2 =

10, the median is the average between the 10th and 11th data values (when sorted in order) These two numbers are 2 and 8, respectively, which average to 5 Five is

therefore the median

(B) The median is affected more It rises to 8, whereas the mean only goes to 6

(C) The mean is affected more It rises to 9.5, whereas the median only goes to 8

(D) This is a summary of parts (B) and (C)

64 (A) mean = 0.76, and the median = 1

(B) When the mean is less than the median, the data set is generally skewed to the left

(C) However, as seen in the histogram of the data below, the data is skewed to the right

SECTION 3.2 EXERCISES

Understanding the Concepts

Exercises 1-8 are the Check Your Understanding exercises located within the section Their answers are found on page 123

9 0

10 variance

11 68%

12 1 12

K



13 False They are measures of variability or spread of the data (not measures of the center)

14 true

15 False 68% of the data will be between 10 and 20 95% of the data will be between 5 and

25

16 true

Practicing the Skills

17 variance =

1

x nx n





3045 5(23) 4



 3045 2645

4



 400

4 100

The standard deviation = 100 10

18 variance =

1

x nx n





7276 9(24.667)

8



 7276 5476.148

8



 1799.853

8 224.98

The standard deviation = 224.9815

19 variance =

1

x nx n





1913 9(13) 8



 1913 1521 8



 392 49

8  The standard deviation = 497

20 variance =

x nx n



1734 6(15) 6

  1734 1350 384

64

The standard deviation = 648

21 variance =

x nx n



6544 8(23) 8

  6544 4232 2312

289

The standard deviation = 289 17

22 variance =

x nx n



6780 12(23) 12

  6780 6348 432

36

The standard deviation = 36 6

23

s2 (5 24.6) 13 (15 24.6)2 2 7 (25 24.6) 102 (35 24.6)2 9 (45 24.6) 112

49

228.41

49  The standard deviation = 228.4115.11

24 s2 (8 48)2 2 (24 48) 142 (40 48)2 6 (56 48) 13 (722 48) 152

49

= 21120 431.02

49  The standard deviation = 431.0220.76

25 σ2=

(25 145) 17 (75 145) 26 (125 145) 14

125

(175 145)  234 (225 145)  226 (275 145)  28 710000 5680

125

The standard deviation = 568075.37