Ch. 3 HW

IQ scores are measured with a test designed so that the mean is 104 and the standard deviation is 11. Consider the group of IQ scores that are unusual. What are the z scores that separate the unusual IQ scores from those that are​ usual? What are the IQ scores that separate the unusual IQ scores from those that are​ usual? (Consider a value to be unusual if its z score is less than −2 or greater than​ 2.)

***The range rule of thumb states that the vast majority​ (such as​ 95%) of sample values lie within 2 standard deviations of the mean. Recall that a value is considered to be unusual if its z score is less than −2 or greater than 2. The lower z score boundary is −2. The higher z score boundary is 2. z = (x - µ)/σ Rearrange the formula to solve for x. z = μ + σ • z compute x, lower boundary IQ score use lower z score x = 104 + 11 • -2 x = 82 compute x, higher boundary IQ score x = 104 + 11 • 2 x = 126 the IQ scores that separate the unusual IQ scores from those that are usual are 82 and 126

The following are the interval times​ (minutes) between eruptions of a geyser. Detemine the five number summary and construct a box plot from the data below. 81, 83, 84, 86, 87, 90, 90, 91, 92, 93, 95, 97, 98, 100, 100, 107

For a set of​ data, the​ 5-number summary consists of the five values listed below. 1. Minimum 2. First​ quartile, Q1 3. Second​ quartile, Q2​ (same as the​ median) 4. Third​ quartile, Q3 5. Maximum minimum = 21 Q1 = (25/100) x 16 = 4; so, in between 4th and 5th value Q1 = (86+87)/2 = 86.5 median = 0.5 x 16 = 8; so, in between the 8th and 9th value median = (91+92)/2 = 91.5 Q3 = 0.75 x 16 = 12; so, in between the 12th and 13th value Q3 = (97+98)/2 = 87.5 maximum = 107 The​ 5-number summary is 81​, 86.5​, 91.5​, 97.5​, 107.

In an​ editorial, the Poughkeepsie Journal printed this​ statement: "The median price − the price exactly in between the highest and lowest−​..." Does this statement correctly describe the​ median? Why or why​ not?

No. It describes the​ midrange, not the median.

Below are 36 sorted ages of an acting award winner. Find P75 using the method presented in the textbook 17, 18, 19, 20, 22, 22, 23, 26, 26, 28, 29, 32, 34, 42, 43, 46, 48, 49, 50, 50, 51, 56, 60, 62, 64, 65, 65, 67, 68, 69, 70, 71, 71, 74, 79, 80

The first step in converting from a percentile to the corresponding data value is to arrange the data in order of lowest to highest. L = (k/100) •​ n k = 75 n = 36 L = (75/100) x 36 L = 27 Since L is a whole #, it is between the 27th and 28th value P75 = (65+67)/2 P75 = 66

Below are 36 sorted ages of an acting award winner. Find P80 using the method presented in the textbook. 16, 17, 17, 20, 20, 22, 23, 24, 25, 26, 30, 30, 32, 34, 37, 38, 39, 41, 41, 42, 42, 49, 52, 52, 53, 65, 66, 66, 69, 69, 69, 70, 72, 79, 79, 80

The first step in converting from a percentile to the corresponding data value is to arrange the data in order of lowest to highest. Next compute L = (k/100) • n n = the total number of values in the data set k = the percentile being used. L = locator that gives the position of a value.​ (For example, for the 12th value in the sorted​ list, L=​12.) n = 36 k = 80 L = (k/100) • n L = (80/100) • 36 L = 28.8 L = 30th value **always round L up to the next larger whole number ex: 14.4 round to 15 P80 = 30th value = 69

Find the third quartile Q3 of the list of 24 sorted values shown below. 31 33 34 36 41 46 48 48 50 50 52 54 55 55 57 59 59 59 61 69 71 75 77 78

To find the kth​ percentile, Pk​, first sort the data​(arrange the data in order of lowest to​ highest). compute L = (k/100) •​ n, where n is the number of values. k = Q3 = 75 n = 24 L = (75/100) x 24 L = 18th value ***If L is not a whole​ number, change L by rounding it up to the next larger whole number. The value of Pk is the Lth value. If L is a whole​ number, Pk is midway between the Lth value and the next value in the sorted set of data. Q3 or P75 = between the 18th and 19th value Q3 = (59 + 61)/2 Q3 = 60

Find the standard​ deviation, s, of sample data summarized in the frequency distribution table below by using the formula​ below, where x represents the class​ midpoint, f represents the class​ frequency, and n represents the total number of sample values.​ Also, compare the computed standard deviation to the standard deviation obtained from the original list of data​ values, 11.1.

To use the given​ formula, first find the midpoint of each class. Interval Frequency Midpoint (x) 20-26 2 23 27-33 4 30 34-40 2 37 41-47 4 44 48-54 17 51 55-61 37 58 62-68 32 65 n = ∑f n = 2 + 4 + 2 + 4 + 17 + 37 +32 n = 98 find ∑(f • x^2) ∑(f • x^2) = (f1 • x1^2) + (f2 • x2^2) + ..... ∑(f • x^2) = (2 • 23^2) + (4 • 30^2) + (2 • 37^2) + (4 • 44^2) + (17 • 51^2) + (37 • 58^2) + (32 • 65^2) ∑(f • x^2) = 319,025 find [∑(f • x)]^2 ∑(f • x)^2 = [(f1 • x1) x (f2 • x2) ....]^2 ∑(f • x)^2 = [(2 x 23) + (4 x 30) + (2 x 37) + (4 x 44) + (17 x 51) + (37 x 58) + (32 x 65)]^2 ∑(f • x)^2 = 30,349,801 solve for standard deviation s =√n[∑(f • x^2)] - [∑(f • x)]^2 / n(n-1) s = √98(319,025) - 30,349,801 / 98(98-1) s = 9.8 ***Consider a difference of​ 20% between two values of a standard deviation to be significant. 11.1/9.8 = 1.13:1; 13% difference The computed value is not statistically different from the given value

Listed below are the top 10 annual salaries​ (in millions of​ dollars) of TV personalities. Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the given sample data in millions of dollars. Given that these are the top 10​ salaries, do we know anything about the salaries of TV personalities in​ general? Are such top 10 lists valuable for gaining insight into the larger​ population? 37, 35, 34, 26, 14, 12, 11, 9, 8.6, 7.4

a. mean = (37 + 35 + 34 + 26 + 14 + 12 + 11 + 9 + 8.6 + 7.4)/10 mean = 19.4 b. median is between 14 and 12. median = (14+12)/2 median = 13 c. There is no mode d. midrange = (maximum value + minimum value)/2 midrange = (37 + 7.4)/2 midrange = 22.2 Given that these are the top 10​ salaries, do we know anything about the salaries of TV personalities in​ general? Since the sample values are the 10​ highest, they give almost no information about the salaries of TV personalities in general.

With a height of 69 ​in, William was the shortest president of a particular club in the past century. The club presidents of the past century have a mean height of 74.7 in and a standard deviation of 1.5 in. a. What is the positive difference between William​'s height and the​ mean? b. How many standard deviations is that​ [the difference found in part​ (a)]? c. Convert William​'s height to a z score. d. If we consider​ "usual" heights to be those that convert to z scores between −2 and​ 2, is William​'s height usual or​ unusual?

a. |69 - 74.7| = 5.7 b. compare the difference and the standard deviation 5.7/1.5 = 3.8 c. A z score is the number of standard deviations that a given value x is above or below the mean. It is found using the following expressions. sample population z = (x - x̄)/s z = (x - µ)/σ µ/x̄ = mean s/σ = standard deviation the club is a population z = (x - µ)/σ z = (69 - 74.7)/1.5 z = -3.8 **** "usual" heights are considered to be those that convert to z scores between −2 and​ 2 William's height is unusual

Listed below are the annual tuition amounts of the 10 most expensive colleges in a country for a recent year. What does this​ "Top 10" list tell us about the population of all of that​ country's college​ tuitions? $51,823, $52,673, $52,406, $50,747, $5,1050, $52,085, $52,340, $51,792, $50,747, $52,559

mean = $51,822.20 midrange = (maximum value + minimum value)/2 midrange = $52,673 + $50,747/2 midrange = $51,710 median = (51823 +52085)/2 median = 51,954 mode(s)= $50,747 Nothing meaningful can be concluded from this information except that these are the largest tuitions of colleges in the country for a recent year.

Listed below are the lead concentrations​ (in mg/g) measured in different samples of a medicine. What do the results suggest about the safety of this​ medicine? What do the decimal values of the listed amounts suggest about the precision of the​ measurements? 18.5, 5, 17.5, 9, 10.5, 13.5, 3.5, 13.5, 16, 6

mean = 11.3 midrange = (18.5 + 3.5)/2 midrange = 11 median = (13.5 + 10.5)/2 median = 12 mode(s) = 13.5 There is not enough information for any meaningful conclusion. They are rounded to one half unit measurements.

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.5 degrees.

mean from freq. dist. = ∑(f • x) / ∑f First, find class midpoints (X) Temperature Frequency f Class Midpoint x f • x 40-44 1 42 42 45-49 4 47 188 50-54 9 52 468 55-59 4 57 228 60-64 1 62 62 Totals: ∑f = 19 ∑(f • x) = 988 mean from freq. dist. = ∑(f • x) / ∑f mean from freq. dist. = 988/19 mean from freq. dist. = 52 degrees ***The computed mean and the actual mean are considered close if the difference is less than​ 5% of the actual mean. Otherwise the means are said to be substantially different. (computed mean - actual mean)/actual mean x 100 (52-51.5)/51.5 x 100 = 0.97% difference The computed mean is close to the actual mean because the difference between the means is less than​ 5% of the actual mean.

A certain group of test subjects had pulse rates with a mean of 72.7 beats per minute and a standard deviation of 11.2 beats per minute. Would it be unusual for one of the test subjects to have a pulse rate of 85.1 beats per​ minute?

minimum "usual" value = (mean) − 2 × (standard deviation) maximum "usual" value = (mean) + 2 × (standard deviation) minimum​ "usual" value = 72.7 - 2(11.2) = 50.3 maximum​ "usual" value = 72.7 + 2(11.2) = 95.1 Is 85.1 beats per minute an unusual pulse​ rate? No, because it is between the minimum and maximum "usual" values

Below are 36 sorted ages of an acting award winner. Find the percentile corresponding to age 76 using the method presented in the textbook 18, 19, 21, 29, 30, 31, 31, 31, 31, 34, 37, 37, 37, 39, 43, 43, 46, 47, 48, 52, 54, 55, 57, 59, 60, 61, 66, 66, 67, 71, 76, 77, 77, 77, 80, 80

percentile value of x = (# of values less than x / total # of values) x 100 percentile value of 76 = (30/36) x 100 = 83

Listed below are the top 10 annual salaries​ (in millions of​ dollars) of TV personalities. Find the​ range, variance, and standard deviation for the sample data. Given that these are the top 10​ salaries, do we know anything about the variation of salaries of TV personalities in​ general? 38, 37, 36, 30, 22, 21, 20, 19, 18.9, 18.4

range = (maximum data ​value)−​(minimum data​ value) range = 38 - 18.4 range = 19.6 variance is also s^2, x̄ is the mean of the​ sample, and n is the size of the sample. s^2 = ∑(x - x̄ )^2 / n - 1 find the sample mean, x̄, first x̄ = ∑x / n ∑x is the sum of all data values n is the number of data values x̄ = (38+37+36+30+22+21+20+19+18.9+18.4)/10 x̄= 26.03 calculate (x - x̄) and (x - x̄)^2 x (x - x̄) (x - x̄)^2 38 11.97 143.28 37 10.97 120.34 36 9.97 99.4 30 3.97 15.76 22 -4.03 16.24 21 -5.03 25.3 20 -6.03 36.36 19 -7.03 49.42 18.9 -7.13 50.84 18.4 -7.63 58.22 ∑(x - x̄ )^2 = 615.16 find variance and substitute variables into the equation s^2 = 615.16 / 10 - 1 s^2 = 615.16 / 9 s^2 = 615.16 / 10 -1 s^2 = 68.35 find standard deviation s = √variance s = √68.35 s = 8.27 Is the standard deviation of the sample a good estimate of the variation of salaries of TV personalities in​ general? No, because the sample is not representative of the whole population. *** samples can never represent population must be all

Six different​ second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic readings​ (in mmHg) are listed below. Find the​ range, variance, and standard deviation for the given sample data. If the​ subject's blood pressure remains constant and the 120, 120, 120, 120, 120, 120

range = 0 sample variance = 0 sample standard deviation =. 0 ​Ideally, the standard deviation would be zero because all the measurements should be the same.

Listed below are the amounts of mercury​ (in parts per​ million, or​ ppm) found in tuna sushi sampled at different stores. Find the​ range, variance, and standard deviation for the set of data. What would be the values of the measures of variation if the tuna sushi contained no​ mercury? 1.01 1.03 0.38 0.24 0.48 0.51 1.08

range = 0.84 sample variance = 0.124 sample standard deviation = 0.352 What would be the values of the measures of variation if the tuna sushi contained no​ mercury? The measures of variation would all be 0.

Six different​ second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic readings​ (in mmHg) are listed below. Find the​ range, variance, and standard deviation for the given sample data. If the​ subject's blood pressure remains constant and the medical students correctly apply the same measurement​ technique, what should be the value of the standard​ deviation? 149 136 125 128 121 133

range = 189 -121 = 28 mmHG s = √[ n (∑x^2) - (∑x)^2 / n(n-1) ] n = 6 ∑x = 149 + 136 + 125 + 128 + 121 + 133 = 792 ∑x^2 = 149^2 + 136^2 + 125^2 + 128^2 + 121^2 + 133^2 = 105,036 Substitute ∑x​, ∑x2​, and n into the formula to find the sample​ variance, s^2 s^2 = n(∑x^2) - (∑x)^2 / n(n-1) s^2 = 6(105,036) - (792)^2 / 6(6-1) s^2 = 98.4 mmHg^2 find the square root of variance, s^2, to get the standard deviation s = √98.4 s = 9.9 mmHg