Asample of blood pressure measurements is taken for a group of adults, and those values (mm Hg) are listed below. The values are matched so that 10 subjects each have a systolic and diastolic measurement. Find the coefcient of variation for each of the two samples; then compare the variation. Systolic 11TIIr 13E] 157!Ir 98 155 121 116 136 125 122% Diastolic 1"8 1"? T4 53 88 90 58 66 TU 34 The coefcient of variation for the systolic measurements is I: 'lia. (Type an integer or decimal rounded to one decimal place as needed.) The coefcient of variation for the diastolic measurements is Bill}. (Type an integer or decimal rounded to one decimal place as needed.) Compare the variation. The coefcients of variation for each data set are Y Therefore, the systolic measurements vary 7 the diastolic measurements. Use the same scales to construct modified boxplots for the pulse rates of males and females from the accompanying data sets. Identify any outliers. Use the boxplots to compare the two data sets. Click the icon to view the data sets. Determine the boxplot for the men's pulse rate data. Choose the correct graph below. - O B. Pulse rates for men and women O A. HH X xx - Men's Pulse Rates 40 50 60 70 80 90 100110 C Full data set 40 50 60 70 80 90 100110 61 62 68 70 59 86 85 74 79 74 OD. 53 103 67 66 80 64 72 79 O C. 85 56 54 61 65 66 66 50 66 56 70 57 x 68 80 68 63 72 74 75 62 50 58 40 50 60 70 80 90 100 110 40 50 60 70 80 90 100 110 Women's Pulse Rates What points are outliers? 80 68 67 71 76 76 97 74 62 79 70 57 75 79 62 99 74 62 83 61 78 73 90 89 73 O A. 103 83 70 77 105 73 86 73 80 O B. 86, 103 O c. 50 O D. 50, 86, 103 Print Done Determine the boxplot for the women's boxplot data. Choose the correct graph below. O B. A. HX C 40 50 60 70 80 90 100 110 C 40 50 60 70 80 90 100 110 OD. O C. xx C 10 50 60 70 80 90 100 110 C 10 50 60 70 80 90 100 110What points are outliers? o A. 5161.62 0 B. 99105 O c. 57 o D. 105 Compare the two boxplots. Choose The coned answer below. 0 A. In general it appears That males have higher pulse rates than females The variation ameng The male pulse rates is similar to the variation ameng The female pulse rates 0 B. In generalr it appears That males have higher pulse rata than temala. The variation among The male pulse rates is much greater than the variation among the female pulse rates. 0 C. In generalr it appears That males have lower pulse rats than femala The variation among the male pulse rates is similar to the varlation among The female pulse rats. 0 D. In general it appears That males have lower pulse rates than females The variation among the male pulse rates is much greater than the variation among the female pulse rates, The brain volumes (cm3} o12l] brains have a mean of 1116?.8 cm3 and a standard deviation of 127.3 cma'. Use the given standard deviation and the range rule of thumb to identify the limits separating values that are signicantly low or signicantly high. For such data, would a brain volume of 13424 cm3 be signicantly high? E) Signicantly low values are El cm3 or lower. (Type an integer or a decimal. Do not round.) Signicantly high values are El cm3 or higher. (Type an integer or a decimal. Do not round.) ls 1342.4 cm3 signicantly high? Q A. Yes, because it is below the lower limit separating value. C) B. Yes, because it is above the upper limit separating value. C) C. No, because it is between the limits separating values. 0 D. No, because it is above the upper limit separating value. 0 E. No, because it is below the lower limit separating value. C) F. Yes, because it is between the limits separating values. Find the (a) mean, (b) median, {c} mode, and (d) midrange for the data and then (e) answer the given questions. Listed below are the highest amounts of net worth (in millions of dollars] of all celebrities. What do the results tell us about the population of all celebrities? Based on the nature of the amounts. what can be inferred about their precision? 231] 190 185 171] 151] 160 155 155 155 1553 I a. Find the mean. The mean is $D million. (Type an integer or a decimal rounded to one decimal place as needed.) b. Find the median. The median is $ million. (Type an integer or a decimal rounded to one decimal place as needed.) c. Find the mode. Select The correct choice below and, if necessary, ll in The answer box to complete your choice. O A_ The mode(s] is[are} $ million. (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed} 0 B. There is no mode. d. Find the midrange. The midrange is $ll million. ' d. Find the midrange. The midrange is $D million. (Type an integer or a decimal rounded to one decimal place as needed.) e. What do the results tell us about the population of all celebrities? O A. The results tell us that The most common celebrity net worth is the mode, but all other celebrities are expected to have net worths approximately equal to the mean, median, or midrange. O B. The results tell us that all celebrities are expected to have amounts of net worth approximately equal to one of the measures of center found in parts (a) through (d). C) C. Apart from the tact that all other celebrities have amounts of net worth lower than those given, the results in parts [a], (b), and {d} do not given meaningful results. However, the result from part {c} shows that The most common celebrity net worth is equal to The mode. 0 D. Apart from the tact that all other celebrities have amounts of net worth lower than those given, nothing meaningtul can be known about the population. Based on the nature of the amounts, what can be inferred about their precision? C) A. Since no information is given, nothing can be said about the precision of the given values. 0 B. The values all end in E] or 5, so they appearto be rounded estimates. 0 C. Since celebrity information is public, these values can be assumed to be unrounded. C) D. The values are all whole numbers, so they appear to be accurate to the nearest whole number. The following are the ratings of males by females in an experiment involving speed dating. Use the given data to construct a boxplot and identify the 5-number summary. 1.0 1.5 2.5 2.5 2.5 2.5 3.5 3.5 3.5 3.5 3.5 4.5 5.0 5.5 5.5 5.5 6.0 6.0 6.5 7.5 The 5-number summary is. . . and. (Use ascending order. Type integers or decimals. Do not round.) Which boxplot below represents the data? O A. OB. CO- 2 4 6 10 0 2 4 6 10 Ratings Ratings OC. OD. 2 4 6 10 2 4 6 10 Ratings RatingsLet a population consist of the values 11 cigarettes, 21 cigarettes, and 22 cigarettes smoked in a day. Shov.r that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. 11I'|.that does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? E) It Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {11,11} {11,21} {11.22} {21.11} {21,21} {21,22} {22,11} {22,21} {22,22} (Type integers or decimals rounded to one decimal place as needed.) Calculate the mean oi the sample mean absolute deviations. The mean of the sample mean absolute deviations is (Type an integer or decimal rounded to one decimal p ace as needed.) Calculate the population mean absolute deviation. The population mean absolute deviation is '1. (Type an integer or decimal rounded to one decimal place as needed.) The population mean absolute deviation is :|. (Type an integer or decimal rounded to one decimal place as needed} What do these values indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? 0 A. The sample mean absolute deviation is an unbiased estimator of the population mean absolute deviation because the sample statistic is the same value regardless of which sample is chosen. 0 B. The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around a different value than the population parameter. Q C. The sample mean absolute deviation is an unbiased estimator of the population mean absolute deviation because the sample statistic centers around the mean value of the population. 0 D. The sample mean absolute deviation is a biased estimator of the population mean absolute deviation because the sample statistic centers around the mean value ot the population. \fFind the range, variance, and standard deviation for the given sample data, if possible. If the measures of valiation can be obtained for these values, do the results make sense? Biologists conducted experiments to determine whether a deciency of carbon dioxide in the soil aects the phenotypes oi peas. Listed below are the phenotype codes, where '1 = smoothyellow, 2 = smoothgreen, 3 =wrinkled yellow, and 4 = wrinkledgreen. 11224332222222221222% Can the range of the sample data be obtained for these values? Choose the correct answer below and, ifnecessary, ll in the answer box within your choice. 0 A- The range of the sample data is E. (Type an integer or a decimal. Do not round.) G) B. The range of the sample data cannot be calculated. Can the standard deviation of the sample data be obtained for these values? Choose the correct answer below and, it necessary, ll in the answer box within your choice. 0 A- The standard deviation of the sample data is D. (Round to one decimal place as needed.) 0 B. The standard deviation of the sample data cannot be calculated. Can the variance of the sample data be obtained for these values? Choose the correct answer below and, it necessary, ll in the answer box within your choice. 0 A- The variance of the sample data is El. (Round to one decimal place as needed.) 0 B. The variance of the sample data cannot be calculated. Do the results make sense? 0 A. The measures of variation make sense because the data is numeric, so the spread between the values is meaningful. O B. While the measures of variation can be found, they do not make sense because the data are nominal; they don't measure or count anything. 0 C. It makes sense that the measures of variation cannot be calculated because there is not a large enough sample size to calculate the measures of variation. 0 D. The measures of variation do not make sense because the standard deviation cannot be greater than the variance. Researchers measured the data speeds for a particular smartphcne carrier at 5!] airports. The highest speed measured was 75.3 Mbps. The complete list of 50 data speeds has a mean of x=15.35 Mbps and a standard deviation of s = 38.46 Mbps. a. What is the di'erence between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part {at}? c. lConvert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scc res between 2 and 2 to be neither significantlyr low nor signicantly high, is the carrier's highest data speed signicant? (... a. The difference is I: Mbps. (Type an integer or a decimal. Do not round.) b. The difference is I: standard deviations. (Round to two decimal places as needed.) c. The z score is z= :l. (Round to two decimal places as needed.) d. The canier's highest data speed is T