Solutions manual for essential statistics 1st edition william navidi

Solutions Manual for Essential Statistics 1st Edition William Navidi Essential Statistics 1st Edition Test Bank Navidi Monk Chapter 3: Numerical Summaries of Data 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 mean median 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 41 19 mean = (5  13)  (15  7)  (25  10)  (35  9)  (45  11) 1230   24.6 50 50 20 mean = (8  2)  (24 14)  (40  6)  (56 13)  (72 15) 2400   48 50 50 21 mean = (25 17)  (75  26)  (125  14)  (175  34)  (225  26)  (275  8) 18125  145.0  125 125 22 mean = (10  18)  (30  11)  (50  6)  (70  6)  (90  10)  (110  5) 2680   47.9 56 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 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 and This eliminates choice (i) 274  30.4 Since there are 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 27 The data are 12, 20, 27, 27, 29, 37, 38, 41, and 43 The mean = 167  20.9 Since there are 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 28 The data are 8, 11, 16, 23, 25, 25, 29, and 30 The mean = 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 42 30 Since the mean and median are basically equal, we should expect the histogram to be approximately symmetric 31 (A) mean = 1740  290 calories per restaurant 66 (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  1.538 seconds per event (B) Since there were 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  35.0 hours per model 20 (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 43 used to be 36 (A) mean = 474  29.6 breweries per state 16 (B) mean = 619  56.3 breweries per state 11 (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 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 data values, the median is the 5th value when sorted That value is 3.59 dollars per gallon (C) The mean did, by cents 7580.3  315.85 thousands of dollars per city 24 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) 38 (A) mean = 6225.9  259.41 thousands of dollars per city 24 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) (B) mean = (C) The mean decreased more than the median did 44 3552  296 pounds per man 12 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 39 (A) mean = (B) mean = 3426  285.5 pounds per man 12 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 10658.38  484.472 stock price per day 22 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 40 (A) mean = 9932.29  472.966 stock price per day 21 Because there are 21 data values, the median is the 11th value when sorted The median is 481.59 stock price (B) mean = (C) Comparing both means and both medians, we see that the prices are about the same 41 (A) mean = 17.3  1.73 billion dollars per company 10 (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  20.117 dollars per magazine 22 (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 45 (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 (16  339)  (26 1544)  (36  844)  (46  686)  (56  413)  (66 132) 139348  35.2  3958 3958 Therefore, the mean is 35.2 years of age per victim 43 (A) mean = (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 (5  283)  (15  203)  (25  217)  (35  256)  (45 176)  1286 36220  (55  92)  (65  21)  (75  23)  (85  12)  (95  3)   28.2 1286 44 (A) mean = 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  37.5 thousands of dollars per state 51 (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): 1622.54  37.733 stock price per company Since there are 43 data values, the median is 43 the 22nd sorted value The median = 28.42 stock price 46 (A) mean = 46 (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 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 numbers above the median 50 (A) 17 Eight numbers are below the median, there is only one 10, so numbers will also be above the median, making n = 17, which is indeed odd (B) 18 There will be the numbers below the median, the two tens at the median, and the numbers above the median of 10 51 208 This answers comes from solving the equation: 202  52 88 This answer comes from solving the equation: 85  802  x 422  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 47 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, 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 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 100 20 = Also, since = 20 10, the median is the average between the 10th and 11th data values (when sorted in order) These two numbers are and 8, respectively, which average to Five is therefore the median 63 (A) Since the sum of the 20 numbers is 100, the mean equals (B) The median is affected more It rises to 8, whereas the mean only goes to (C) The mean is affected more It rises to 9.5, whereas the median only goes to 48 (D) This is a summary of parts (B) and (C) 64 (A) mean = 0.76, and the median = (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 10 variance 11 68% 12  K2 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 and 25 16 true Practicing the Skills 49 17 variance = x 18 variance = x  nx 3045  5(23)2 3045  2645 400   100  n 1 4 The standard deviation = 100  10 =  nx n 1 The standard deviation = x x x x  nx = 7276  9(24.667)2 7276  5476.148 1799.853   224.98  8 224.98  15 1913  9(13)2 1913  1521 392   49  n 1 8 The standard deviation = 49  19 variance =  nx = 1734  6(15)2 1734  1350 384    64 n 6 The standard deviation = 64  20 variance =  nx = 6544  8(23)2 6544  4232 2312    289 n 8 The standard deviation = 289  17 21 variance =  nx = 6780  12(23)2 6780  6348 432    36 n 12 12 12 The standard deviation = 36  22 variance = = 23 s  (5  24.6)2  13  (15  24.6)2   (25  24.6)2  10  (35  24.6)2   (45  24.6)2  11 11192  228.41 49 49 The standard deviation = 228.41  15.11 (8  48)   (24  48)  14  (40  48)   (56  48) 13  (72  48) 15 21120 =  431.02 49 49 The standard deviation = 431.02  20.76 24 s  (25  145)2  17  (75  145)  26  (125  145)  14  125 710000  (175  145)2  34  (225  145)  26  (275  145)    5680 125 The standard deviation = 5680  75.37 25 σ = 50 (10  47.86)2  18  (30  47.86)  11  (50  47.86)   56 69342.8576  (70  47.86)   (90  47.86)2  10  (110  47.86)2    1238.27 56 The standard deviation = 1238.27  35.19 26 σ = 27 (A) By the Empirical Rule, roughly 68% of all data values lie within one standard deviation of the mean (B) 20 and 44 By the Empirical Rule, roughly 95% of all data values lie within two standard deviations of the mean (C) 75%, by Chebyshev’s Inequality 28 (A) 95% By the Empirical Rule, roughly 95% of all data values lie within two standard deviations of the mean (B) 129 and 139 By the Empirical Rule, roughly 68% of all data values lie within one standard deviation of the mean (C) 88.9%, by Chebyshev’s Inequality 29 (A) 95% By the Empirical Rule, roughly 95% of all data values lie within two standard deviations of the mean (B) 140 and 212 By the Empirical Rule, roughly 68% of all data values lie within one standard deviation of the mean (C) 88.9%, by Chebyshev’s Inequality 30 (A) 68% By the Empirical Rule, roughly 68% of all data values lie within one standard deviation of the mean (B) 65 and 93 By the Empirical Rule, roughly 95% of all data values lie within two standard deviations of the mean (C) 75%, by Chebyshev’s Inequality Working with the Concepts 31 (A) s  (B) x  nx n 1 = 54.8134  9(2.30667)2 54.8134  47.8865 6.9269    8659 8 s= 8659  931 dollars per gallon s2  x  nx n 1 162.8533  9(3.9678)2 162.8533  141.6909 21.1624  =   2.6453 8 51 s= 2.6453  1.626 dollars per gallon (C) yes 32 (A) s  x  nx 2 n 1 = 3549  7(22.4286)2 3549  3521.2947 27.7053   4.6176  6 4.6176  2.1 MPG s= x (B) s   nx 2 n 1 3665  7(22.7143)2 3665  3611.5760 53.424  =   8.904 6 8.904  3.0 MPG S= (C) sports cars 33 (A) s x   nx n 1 Therefore, s = (B) s  s= x  nx n 1 = 1085524  12(299.667)2 1085524  1077603.731 7920.269   720.0245  11 11 11 720.0245  26.8 pounds = 1150470  12(308.833)2 1150470  1144533.863 5936.137   539.6488  11 11 11 539.6488  23.2 pounds (C) offensive lineman Links for completed download: essential statistics navidi solutions manual download essential statistics navidi test bank pdf essential statistics navidi access code essential statistics pdf elementary statistics navidi and monk pdf essential statistics navidi ebook essential statistics access code essential statistics connect plus access code 52