2.7 Measures of the Spread of the Data

Step 1 — Find the mean first. The 20 distances total 1,681 miles, so

Step 2 — Enter the data and read the standard deviation. These 20 stores are a sample, so read \(Sx\) rather than \(\sigma x\) from 1-VarStats:

Note on the graphing calculator. Put all 20 distances in L1, leave the frequency argument blank, and run STAT CALC 1:1-VarStats. The screen reports \(\bar{x} = 84.05\), \(Sx = 34.5169\ldots\), and \(\sigma x = 33.6429\ldots\), with \(n = 20\).

Answer: \(s \approx 34.5\) miles. (Had the 20 stores been the entire population, the answer would be \(\sigma \approx 33.6\) miles.)

Step 1 — Recall the equation. A value one standard deviation below the mean is \(\text{value} = \bar{x} + (-1)(s)\).

Step 2 — Substitute the mean and standard deviation from the previous problem.

Answer: About 49.5 miles. A store 49.5 miles from the distribution center is one standard deviation closer than average.

Step 1 — Recognize that the two averages are not comparable directly. The two players bat in different leagues with different team means and different team spreads, so compare z-scores instead of raw averages.

Step 2 — Fredo's z-score.

Step 3 — Karl's z-score.

Step 4 — Decide which direction is better. For a batting average, higher is better, so the larger z-score wins. Both players are below their own team's average, but \(-0.67\) is larger than \(-0.80\).

Answer: Fredo had the higher batting average when compared to his team. His raw average of 0.158 is the lower of the two, so comparing raw averages would give the wrong answer — Karl simply plays on a team that hits better overall.

Step 1 — Set up the equation for each team. Three standard deviations from a team mean is \(\text{value} = \text{mean} + (\pm 3)(s)\), applied with that team's own mean and standard deviation.

Step 2 — Fredo's team (mean 0.166, standard deviation 0.012), part a and part b.

Step 3 — Karl's team (mean 0.189, standard deviation 0.015), part a and part b.

Answer: a) Three standard deviations above the mean is 0.202 for Fredo's team and 0.234 for Karl's team. b) Three standard deviations below the mean is 0.130 for Fredo's team and 0.144 for Karl's team. By Chebyshev's Rule at least 89% of each team's batting averages fall inside its own interval.

Step 1 — Find each class midpoint. Average the two endpoints of every class. For all three tables the midpoints are 54.5, 64.5, 74.5, 84.5, and 94.5.

Step 2 — Table 2.7.8, the grades (part a). With frequencies 2, 3, 8, 12, 5 the total is \(n = 30\) and

The squared-deviation column totals \(\Sigma f(m - \bar{x})^2 = 3{,}550\), so

Step 3 — Table 2.7.9, the daily low temperatures (part b). With frequencies 53, 32, 15, 1, 0 the total is \(n = 101\) and

Here \(\Sigma f(m - \bar{x})^2 \approx 5{,}916.9\), so

Step 4 — Table 2.7.10, the points per game (part c). With frequencies 14, 32, 15, 23, 2 the total is \(n = 86\) and

Here \(\Sigma f(m - \bar{x})^2 \approx 10{,}633.8\), so

Note on the graphing calculator. For each table, put the five midpoints in L1 and that table's frequencies in L2, then run 1-VarStats L1, L2. The \(Sx\) value on the screen matches the hand calculation in each case.

Answer: a) \(s \approx 11.1\); b) \(s \approx 7.7\); c) \(s \approx 11.2\). The temperature data are the most tightly clustered of the three, even though they have by far the largest \(n\) — a bigger sample does not mean a bigger spread.

Step 1 — Notice which statistic 1,014 is. The problem gives the median as 1,014 FTES.

Step 2 — Use what the median means. The median splits the data in half: 50% of the 29 years are at or below 1,014 FTES and 50% are at or above it.

Step 3 — Apply that fraction to the sample of 11.

Answer: About 6 years (5 or 6 in practice, since you cannot have half a year). Because 1,014 FTES is the median, half of all years sit at or above it, so half of any sample of 11 years is the expected count.

Step 1 — Translate 75% at or below into a quartile. The value with 75% of the data at or below it is the third quartile, \(Q_3\).

Step 2 — Translate 75% at or above into a quartile. The value with 75% of the data at or above it has 25% at or below it, which is the first quartile, \(Q_1\).

Step 3 — Read both from the given parameters. \(Q_3 = 1{,}447.5\) FTES and \(Q_1 = 528.5\) FTES.

Answer: a) At or below 1,447.5 FTES. b) At or above 528.5 FTES.

Step 1 — Find the symbol for a population standard deviation. A population standard deviation is written \(\sigma\), not \(s\), and the problem states these parameters describe the full 29-year population.

Step 2 — Read it off the given list.

Answer: \(\sigma = 474\) FTES. Because these are population parameters, the correct symbol is \(\sigma\); using \(s\) would claim the 29 years were only a sample.

Step 1 — Identify the two endpoints. 528.5 FTES is \(Q_1\) and 1,447.5 FTES is \(Q_3\).

Step 2 — Use what the quartiles mean. \(Q_1\) has 25% of the data at or below it and \(Q_3\) has 75% at or below it, so the stretch between them holds \(75\% - 25\% = 50\%\) of the data.

Answer: 50% of the years had an FTES between 528.5 and 1,447.5. We know it because those two numbers are the first and third quartiles, and by definition the quartiles cut off the middle half of the data.

Step 1 — Recall the definition. The interquartile range is \(\text{IQR} = Q_3 - Q_1\).

Step 2 — Substitute.

Answer: \(\text{IQR} = 919\) FTES. It represents the spread of the middle 50% of the years — the width of the range that the typical half of the data occupies, with the smallest quarter and the largest quarter of the years both excluded.

Step 1 — Set up the z-score. The question asks for \(\#\text{ofSTDEVs}\), which is the z-score of the median treated as a value in the population.

Step 2 — Substitute the median for \(x\).

Answer: About 0.03 standard deviations above the mean — essentially zero. The mean and the median are so close together that, on the scale of this data's spread, they are the same number. From Section 2.6, that is what a roughly symmetrical distribution looks like.

Step 1 — The mean. The six values total 10,856 FTES, so

Step 2 — The median. Ordered, the values are 1,585; 1,690; 1,735; 1,890; 1,935; 2,021. With six values the median is the average of the 3rd and 4th:

Step 3 — The standard deviation. The squared deviations from 1,809.3 total about 137,193. These six years are described as the population, so divide by \(N = 6\):

Step 4 — The quartiles. Split the ordered list at the median. The lower half is 1,585; 1,690; 1,735, whose median is \(Q_1 = 1{,}690\). The upper half is 1,890; 1,935; 2,021, whose median is \(Q_3 = 1{,}935\).

Step 5 — The IQR.

Answer: \(\bar{x} \approx 1{,}809.3\); median \(= 1{,}812.5\); \(\sigma \approx 151.2\); \(Q_1 = 1{,}690\); \(Q_3 = 1{,}935\); \(\text{IQR} = 245\), all in FTES. (Treating the six years as a sample instead would give \(s \approx 165.6\). The interpolation formula from Section 2.3 gives slightly different quartiles, \(Q_1 \approx 1{,}663.8\) and \(Q_3 \approx 1{,}956.5\); a TI-84 reports the median-of-halves values above.)

Step 1 — List what a box plot needs. A box plot is drawn from the five-number summary: the smallest value, \(Q_1\), the median, \(Q_3\), and the largest value.

Step 2 — Check the six-year data against that list. We have the full list of six values, so the smallest (1,585) and largest (2,021) are already known along with the three middle numbers. Nothing further is needed.

Step 3 — Check the 29-year data. For that period we were given only \(\mu\), the median, \(\sigma\), \(Q_1\), and \(Q_3\). The two ends of the whiskers are missing.

Answer: For the 29-year period you need the smallest and the largest FTES values — the minimum and maximum that complete the five-number summary. For Years 1 to 6 no additional information is needed, because the raw data list supplies the minimum and maximum directly.

Step 1 — Put the two figures side by side. The 29-year period has \(\sigma = 474\) FTES. The six-year period has \(\sigma \approx 151.2\) FTES.

Step 2 — Compare them. The 29-year standard deviation is more than three times the six-year one.

Step 3 — Explain why the window matters. Over 29 years the college grew, so the FTES values run from small early-year enrollments up to much larger recent ones, and the standard deviation has to cover that whole climb. Six consecutive years catch only a short stretch of the same trend, where every value sits within a few hundred FTES of the others.

Answer: The 29-year standard deviation is about three times larger. A longer time window captures more of the college's growth, so the values it contains are spread over a much wider range. The six-year window is short enough that enrollment barely changed across it, which is why its spread is small.

Step 1 — Recognize the three grading systems are different scales. A 2.7, an 87, and an 8.6 cannot be compared directly, so compute each student's z-score against their own school.

Step 2 — Thuy's z-score.

Step 3 — Vichet's z-score.

Step 4 — Kamala's z-score.

Step 5 — Compare. For a GPA, higher is better, so the largest z-score wins: \(1.5 > 0.6 > -0.625\).

Answer: Kamala had the best GPA compared to other students at her school — she is 1.5 standard deviations above her school's average, while Vichet is 0.6 above his and Thuy is 0.625 below hers. Note that Vichet's raw 87 is the largest number on the list and he still does not win.

Step 1 — Set up a z-score for each instrument against its own market.

Step 2 — The piano.

Step 3 — The guitar.

Step 4 — The drum set.

Step 5 — Compare, remembering which direction is which. The most negative z-score is the cheapest relative to its own market; the largest is the most expensive.

Answer: The drum set is the lowest cost when compared to other drum sets, at 1.0 standard deviation below the mean drum price. The guitar is the highest, at 0.25 standard deviations above the mean guitar price. The piano is the most expensive item in raw dollars, at $3,000, yet it is a comparative bargain — 0.4 standard deviations below the mean piano price.

Step 1 — Compute a z-score for each runner against their own class.

Step 2 — Yasmin's z-score.

Step 3 — Kenji's z-score.

Step 4 — Nedda's z-score.

Step 5 — Read the signs (part a). These are times, so a negative z-score means faster than the class average. Kenji's \(-0.25\) puts him below his class mean; Nedda's \(+0.25\) puts her above hers.

Step 6 — Find the most negative z-score (part b). Comparing \(-1.0\), \(-0.25\), and \(0.25\), the most negative is Yasmin's.

Answer: a) Kenji ran a quarter of a standard deviation faster than his class average, while Nedda ran a quarter of a standard deviation slower than hers. Nedda's raw time of 8 minutes beats Kenji's 8.5 minutes, but she was racing a much faster class, so relative to her own classmates she is the weaker runner. b) Yasmin is the fastest with respect to her class, at 1.0 standard deviation below her class mean.

Step 1 — Find the class midpoints. Averaging each class's endpoints gives 15.925, 24.95, 33.95, 42.95, 51.95, 60.95, 69.95, and 78.95 percent.

Step 2 — Estimate the mean. The frequencies are 29, 13, 4, 0, 2, 1, 0, 1, so \(n = 50\) and \(\Sigma f m = 1{,}165.775\):

Step 3 — Estimate the standard deviation. The squared-deviation column totals \(\Sigma f(m - \bar{x})^2 \approx 8{,}222.5\), so

Step 4 — Place the United States on that scale.

Step 5 — Judge how unusual that is. The rule of thumb calls a value unusual once it is more than two standard deviations from the mean. This one is less than one.

Answer: The best estimate of the average poverty percentage is about 23.3%, with a standard deviation of about 13.0%. At 33.9% the United States is above average, but only about 0.82 standard deviations above — well inside the two-standard-deviation line, so it is not an unusual rate for this group of countries.

Step 1 — Find the class midpoints. Averaging each class's endpoints gives 18.725, 24.175, 29.625, 35.075, 40.525, and 45.975 percent.

Step 2 — Estimate the mean. The frequencies are 23, 4, 9, 7, 6, 1, so \(n = 50\) and \(\Sigma f m = 1{,}328.65\):

Step 3 — Estimate the standard deviation. The squared-deviation column totals \(\Sigma f(m - \bar{x})^2 \approx 3{,}573.8\), so

Step 4 — Mark the unusual boundary. More than two standard deviations from the mean counts as unusual, which here means below \(26.573 - 2(8.54) \approx 9.5\%\) or above \(26.573 + 2(8.54) \approx 43.7\%\).

Step 5 — Check each interval against that boundary. Only the top class, 43.25-48.7, has a midpoint (45.975) beyond 43.7%. Its z-score is

Answer: The best estimate of the mean is about 26.6% with a standard deviation of about 8.5%. The 43.25-48.7 interval could be considered unusual: its midpoint sits about 2.3 standard deviations above the mean, past the two-standard-deviation rule of thumb, and only one country falls in it. No interval reaches the unusual boundary on the low side, because the data stop at 16%.