CHAPTER 2 Descriptive Statistics I: Elementary Data Presentation and Description Level of Difficulty (Moderate to Challenging) Learning Objectives 1 2 3 1.Compute and interpret the three main measures of central tendency. 1,2,3,4 5,6,7,11,12,13 ,14 2.Compute and interpret the four main measures of dispersion. 8,9,10,50 ,53 11,12,13,14,51 ,52,54,55,56, 57,58 3. Summarize data in a frequency distribution. 15,16,17 18,19,20,21,22 ,23,24,25,26,5 9,60,61 62,63 4. Summarize data in a relative frequency distribution. 27,28,29, 32,33,64, 30,31,34,35,36 ,65 5. Build and interpret a cumulative frequency distribution. 37,38,66 39,40,41,67,68 ,69 70 6. Analyze grouped data and show that data in a histogram. 42,43,44, 45 46,47,48,49,71 ,72,73,74,75,7 6 77,78
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CHAPTER 2Descriptive Statistics I:
Elementary Data Presentation and Description
Level of Difficulty (Moderate to Challenging)
Learning Objectives 1 2 3
1.Compute and interpret the three main measures of central tendency. 1,2,3,4 5,6,7,11,12,13,14
2.Compute and interpret the four main measures of dispersion. 8,9,10,50,53 11,12,13,14,51,52,
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CHAPTER 2
1.Mean() =
2+6+2+0+2+3+5+9+1+510 = 3.5. This is the center of the data set
insofar as it represents the balance point for the data.
The ordered list is 0, 1, 2, 2, 2, 3, 5, 5, 6, 9.
The median is the middle value; in this case,it’s 2.5, the value in position (10+1
2 )=5.5—or halfway between the 5th and 6th values— in the ordered list.) At least half the values in the data set are at or above 2.5; at least half of the values are at or below 2.5.
Mode = 2. This is the most frequently occurring value.
c) mode =2 median = 2.5 22501235 69
The high value, 9, exerts a great deal of leverage in setting the balance point (that is, the mean) because it sits so far to the right.
2.
Mean()=
471+300+.. .+26712 = $533.25 million. This is the center of the data set
insofar as it represents the balance point for the data.
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The median is the middle value.It’s the value in position (12+1
2 )= 6.5 in the ordered list, halfway between the 6th and the 7th value: 479 and 495. We can calculate
its value as (479+495
2 )= $487. Interpretation: At least half the movies had revenues of $487 million or more; at least half had revenues of $487 million or less.
There’s no mode, since no value appears more than once.
3. a) Mean() =15.09 This is the center of the data set insofar as it represents the balance point for the data.
The ordered list is 11, 11, 12, 12, 14, 15, 17, 17, 17, 20, 20.
The median is the middle value; in this case, 15 (the value in position (11+1
2 )= 6 in the ordered list) At least half the months had 15 or more bankruptcies; at least half had 15 or less.
The mode (or modal value) is 17, the most frequently occurring value.
c) median =15 mode = 171711 12 17 20 11 12 14 15 17 20
4. Mean(x̄ ) =142. This is the center of the data set insofar as it represents the balance point for the data.
The ordered list is 40, 99, 115, 130, 130, 190, 290.
The median is the middle value—here, 130 (the value in position (7+1
2 )= 4 in the ordered list) At least half the scores were 130 or more; at least half the scores were 130 or less.
mean = 15.09
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The mode is 130—the most frequent score.
b) A “typical” or “representative” score is about 130.There’s a chance for an occasional great game and an occasional dismal performance.(For purposes of comparison, a professional bowler will average above 230.)
5. Mean () =1.411. This is the center of the data set insofar as it represents the balance point for the data.
The ordered list is: 0 0.21 0.23 0.60 1.43 1.64 2.24 2.52 2.69 3.06
The median is the middle value—here, 1.535 (the value in position (10+1
2 )= 5.5 in
the ordered list, or halfway between the 5th and 6th value: (1. 43+1.64
2 )= 1.535. ) At least half the values were 1.535% or more; at least half the scores were 1.535% or less.
There is nomode here.
6.Mean () =$60,010. This is the center of the data set insofar as it represents the balance point for the data.
The ordered list is:
55,430 57,118 60,055 60,451 60,946 63,011 63,057
The median is the middle value—here, 60,451 (the value in position (7+1
2 )= 4 in the ordered list, 60,451. At least half the values were $60,451 or more; at least half the scores were $60,451 or less.
There is no mode.
7. This suggests that there are a relatively few very high volume texters, pulling the average well to the right of (that is, above) the median. The mean is generally more sensitive to extreme values than is the median.
8. Range: 22 – 2 = 20
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Interpretation: The difference between the largest value and the smallest value is 20.
MAD: Given that the mean of the data is 9,
MAD =
|2-9|+|12-9|+|6-9|+|4-9|+|8-9|+|12-9| 6 =
326 ≈ 5.33
Interpretation: The average difference between the individual values and the overall average for the data is 5.33.
Variance: 2 =
(2-9)2 +(12-9 )2 + (6-9)2 + .. . + (12-9)2
6
=
2626 ≈43.667
Standard Deviation: = √43 . 667 ≈ 6.61
Interpretation: Roughly speaking, the individual values are, on average, about 6.61 units away from the overall mean for the data. As is typically the case, the standard deviation here, 6.61, is greater than the MAD, 5.33.
9. Range: 116 – 111 = 5 yen
Interpretation: The difference between the highest exchange rate and the lowest exchange rate was 5 yen.
MAD: Given that the mean of the data is 114,
MAD =
|112-114|+|115-114|+|111-114|+. ..+|112-114| 7 =
147 = 2 yen
Interpretation: The average difference between the daily exchange rate and the overall average exchange rate for the week was 2 yen.
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=
307 = 4.286
Interpretation: The average squared difference between the daily exchange rate and the overall average exchange rate for the week was 4.286 (yen2).
Standard Deviation: = √4 .286 =2.07
Interpretation: Roughly speaking, the daily exchange rate is, on average, about 2.07 yen away from the overall mean exchange rate for the week (114 yen/dollar). As is typically the case, the standard deviation here, 2.07, is slightly greater than the MAD, 2.0.
10. Range: 146 – 122 = $24
Interpretation: The difference between the highest price and the lowest priceis $24.
Interpretation: The average squared difference between the individual competitor prices and the overall average competitor price is 44.33(squared dollars).
Standard Deviation: = √44 .33 ≈$6.66
Interpretation: Roughly speaking, the price charged by individual competitors is, on average, about $6.66 away from the overall average price charged by the group.
11.
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Interpretation: The average difference between individual year sales and the overall average sales level for the 10-year period is approximately .753 million cars.
Interpretation: The average squared difference between individual year sales and the overall average sales level is .715 million cars2.
Standard Deviation: = √ . 7148 ≈.845 million cars. (845,000 cars)
Interpretation: : Roughly speaking, individual year sales are, on average, about 845,000 cars away from overall average car sales for the 10-year period.
12. a) Range: 290 – 40 = 250 pins
Interpretation: The difference between the highest score and the lowest score rate is 250.
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MAD: Given that the mean score is 142 pins,
MAD =
|130-142|+|99-142|+|190-142|+. ..+|115-142| 7 =
3927 = 56 pins.
Interpretation: The average difference between the individual scores and the overall average score for the sample is 56 pins.
Interpretation: If we looked at the population being represented here, the average squared difference between the individual scores and the overall average score in that population would be about 6246 (pins2).
Standard Deviation: s = √6246 . 33 ≈79.0 pins.
Interpretation: Roughly speaking, the scores in the sample suggest that if we looked at the population being represented here, the individual scores in that population would be, on average, about 79 pins away from the overall population average score.
b) Your bowling looks to be pretty erratic.
13. a) Range: 3927 – 0 = 3927 megawatts
Interpretation: The difference between the highest capacity and the lowest is 3927 megawatts.
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=
165844968 = 2073062
Interpretation: The average squared difference between the individual values and the overall average value is about 2,073,062 (megawatts2).
Standard Deviation: = √2073062 ≈1439.8megawatts.
Interpretation: Roughly speaking, the individual state capacities are, on average, about 1440 megawatts away from the overall averagecapacity of 1234 megawatts.
14. a) Range: 40.5 – .6 = 39.9 $billion
Interpretation: The difference between the highest assets and the lowest is 39.9 $billion.
Interpretation: The average squared difference between the individual asset values and the overall average asset value is 158.852 $billion.
Standard Deviation: = √158 .852 ≈12.604$billion.
Interpretation: Roughly speaking, the individual asset values are, on average, about 12.604 $billion away from the overall average asset value of 9.07 $billion.
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15. Rating
xFrequency
f(x)3 84 165 10
16. a) errors
xfrequency
f(x)0 11 12 13 24 25 36 67 4
b)
Stock Rating
0
5
10
15
20
3 4 5 More
rating
Freq
uenc
y
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c) The distribution is negatively skewed and unimodal.
17. a) score
xfrequency
f(x)5 36 67 18 49 6
b)
c) The distribution is bi-modal.
Audit Results
01234567
0 1 2 3 4 5 6 7
Mor
e
errors
Freq
uenc
y
Test Results
0
2
4
6
8
5 6 7 8 9 More
score
Freq
uenc
y
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32. median = 7. Summing down the relative frequency (proportion) column, we reach a sum of .50 part way through the 7s (.22 + part of .36). The median, then, is 7).
x p(x)
6 0.22
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7 0.36
8 0.20
9 0.12
10 0.101.00
33. a) = 2.25 days (See detailed calculations below.)b) median = 2 days (Summing down the relative frequency (proportion) column, we reach a sum of .50 part way through the 2s (.39 + part of .27). The median, then, is 2.
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x p(x) xp(x)
17 0.08 1.36
18 0.14 2.52
19 0.18 3.42
20 0.22 4.40
21 0.16 3.36
22 0.1 2.20
23 0.08 1.84
24 0.04 0.96 = 20.06
c) Median = 20 games (Summing down the relative frequency (proportion) column, we reach a sum of .50 part way through the 20s (.08 + .14 + +.18 + part of .22). The median, then, is 20.)
35. a) = 2.03 offers
b) Median = 2 offers (Summing down the relative frequency (proportion) column, we reach a sum of .50 part way through the 2s (.08 + .288 + part of .367). The median, then, is 2.)
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b) Median = 7 interviews (Summing down the relative frequency (proportion) column, we reach a sum of .50 part way through the 7s (.043 + .095 + .136 +.192 + part of .235). The median, then, is 7.)
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b) Estimated Mean = 13.5 hours Estimated Variance = 94.75
Estimated Standard Deviation = √Variance≈9.73 hours
47. a) To produce relative frequencies, divide the frequency column by the total projected population of 357.4 (million).
Age Group
mid f(x) p(x) xp(x)0 to under 20 10 94.3 0.264 2.6420 to under 40 30 92.9 0.260 7.80 40 to under 60 50 85 0.238 11.89 60 to under 80 70 70.3 0.197 13.7780 to under 100 90 14.8 0.041 3.73 100 to 120 110 0.1 0.000 0.03
Total = 357.4 mean = 39.85
b) Estimated Mean age = 39.85 Estimated Variance = 569.15
−6 to −3 4−3+ to 0 00+ to 3 123+ to 6 146+ to 9 159+ to 12 5
12+ to 15 1
02468
10121416
No.
of S
tate
s
% change
% change in food stamp participation (Jan 2011-Jan 2012)
-6 - 3 0 3 6 9 12 15
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50. a) = 100 strokes.100 here represents the central tendency of your golf scores insofar as it serves as a “balance point” forthe scoresb) The ordered list is 60, 80, 90, 100, 100, 110, 120, 140.
Median = 100 strokes, halfway between the 4th and 5th values (here, these values are both 100). It’s the middle value insofar as at least half the scores are at or above this value and at least half the scores are at or below this value.Mode=100 strokes. The most frequent score.
c) range = 140− 60 = 80 strokes. The difference between the highest and lowest score.MAD = 17.5 strokes. The average difference between the individual scores and the overall average score.2 = 525. The average squared difference between your individual scores and the overall average score.
= √525 = 22.9 strokes. Roughly the average difference between the individual scores and the overall average score.
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51. a) = $17.51 billion (See detailed calculation below.) b) The ordered list is 13.7,15, 16.3, 18.6, 19, 19.2, 20.8 Median = $18.6 billion, the 4th value
No Mode.c) range = 20.8−13.7 = $7.1 billion; MAD = $2.16 billion
52. a) = 7 hours (See detailed calculation below.) b) The ordered list is 3, 5, 5, 6, 7, 10, 13
Median = 6 hours, the 4th value ((7+1)/2 = 4)Mode =5 hours.
c) range = 13 − 3 = 10 hours; MAD = 2.57 hours
2 = 10; = √10 = 3.16 hours
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x x− lx−l (x−)2
5 −2 2 46 −1 1 13 −4 4 165 −2 2 47 0 0 0
10 3 3 913 6 6 3649 0 18 70
= 49/7= 7
MAD = 18/7 = 2.57
2 = 70/7 = 10
53. a) x̄ = 106. (See the detailed calculations below.)b) The ordered list is 89, 93, 95, 97, 99, 100, 100, 100, 106, 181Median = 99.5, a value halfway between the 5th and 6th values in the ordered list.Mode = 100.c) Range = 181−89 = 92; MAD = 15
Variance= 715.8; Standard deviation = √715 .8 = 26.75. d) The median or mode would better represent the “typical” time since these are measures less influenced by the one extreme of 181.
x x− x̄ lx− x̄ l (x−x̄ )2
106 0 0 0
100 −6 6 36
100 −6 6 36
97 −9 9 81
89 −17 17 289
95 −11 11 121
93 −13 13 169
181 75 75 5625
99 −7 7 49
100 −6 6 361060 0 150 6442
x̄ =1060/10=106
MAD = 150/10 = 15
s2 = 6442/9= 715.8
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54. a) = $40,358. (See detailed calculations below.) b) The ordered list is:36427, 36919, 41661, 42578, 44205. Median = $41,661, the 3th value in the ordered list.
No mode. c) range = 44205 – 36427 = $7,778; MAD = $2,948
2 = 9,741,020; = √9741020 = $3,121
55. In the chapter, the mean was described as a “balance point” for a set of data, equalizing distances (deviations) of values to the left of the mean with distances of values to the right. Show that this is the case for the data in
a) Exercise 50: The mean, is 100, so summing the x – distances gives
∑ ( x−μ )= 0+(−10)+10+(−20)+20+40+0+(−40)= 0
b) Exercise 53: The mean, x̄ , here is 106, so summing the x –x̄ distances gives
56. The “central tendency” of team performance is the same for both teams: the mean and the median in each case are 75. The teams obviously differ in terms of “consistency:” The range for Team A is 10, vs. 50 for Team B. The standard deviation for Team A is 3.16, while the standard deviation for Team B is 18.4.
Most would choose the more consistent team, Team A. However, an argument could be made that even though assigning the job to Team B is more risky, Team B also offers the chance for superior performance, as indicated by the high score of 90.
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Team A appears the safer bet, but has a lower ceiling. Team B has a much higher “upside,” but a more severe “downside.”
57. a) The median must be halfway between 13 and 25: 19b) If the MAD is 9, then the distance of each value from the mean must be 9 units. The sum of the squared distances, then, must be 92+ 92= 162. The
variance, therefore, is 162/2 = 81, making the standard deviation √81 = 9.c) The standard deviation is 4. For a data set of two values, the standard deviation is precisely equal to the average distance of the two values from the mean. (See part b).)The values must be 20 – 4 = 16 and 20 + 4 = 24.
58. a) The MAD must be 4. b) The range must be 10.c) If the mode is 200, there must be two values of 200, making the third value 245 in order to have the mean of the three numbers be 215. The sum of the squared distances from the mean, therefore, must be (−15)2 + (−15)2 + 302 =
1350. The standard deviation, then, is √13503 = 21.2.
59. a)
b) Mean = 21.4 shifts; Variance = 1.42; standard deviation = √1 .42= 1.19 shifts. (See the details below.)c) Median = 21 shifts, the value halfway between the 65th and 66th values ((130+1)/2 = 65.5) in the list.
Nurse's Shifts for the Past Month
05
101520253035404550
20 21 22 23 24
no. of shifts
no. o
f nur
ses
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Mode= 21.
d) This is a unimodal, positively skewed distribution
60. a)
b) Mean = 42.5 officers; Variance = 3.93; standard deviation = √3 . 93= 1.98 officers. (See the details below.)
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c) Median = 42.5, the value halfway between the 25 th and 26th values ((50+1)/2 = 25.5) in the list. Modes are 40 and 45.d) This is a symmetric, bi−modal distribution.
61. a)
b) Mean = 3.88 teeth; Variance = 2.0; standard deviation =√2 . 0= 1.41 teeth. (See the details below.).c) Median = 4, the 38th value ((75+1)/2 = 38) in the list. Mode = 4.
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62. a) The chart shows two distinct groups of applicants, one that performed rather poorly on the test and another that performed fairly well. It might be useful for the company totry to identify what particular factors caused this sort of result: Was it a difference in education levels for the two groups? A difference in prior experience? A difference in the conditions under which the tests were administered? The particular administrators who conducted the tests? Etc. b) The mean, as the balance point for the data, appears to be around 6. The median—the 50-50 marker for the data—appears to be 8. At least half the values look to be at or below 8; at least half look to be at or above 8.c) The MAD for the distribution appears to be approximately 3. It’s the average distance of the values in the data set from the mean. Among the other possible answers here, 1 appears to be too small, and 7 and 10 appear too large to fit this definition.d) The standard deviation for the distribution appears to be approximately 4. It should be roughly equal to the MAD, but will almost always be larger.
63. a) Clearly the company isn’t meeting the 3-day standard for a significant number of its deliveries. In fact, in a number of instances, delivery time was at least double the 3-day standard.b) The mean, as the balance point for the data, appears to be about halfway between 3 and 4 days. The median—the 50-50 marker for the data—appears to be 3 days. At least half the timeslook to be at or below 3; at least half look to be at or above 3.c) The MAD for the distribution looks to be approximately 1.5 days. It’s the average distance of the values in the data set from the mean. The other possible answers here, 5.5, 8.5 and 15, all appear too large to fit this definition.d) The standard deviation for the distribution looks to be approximately 2 days.It should be roughly equal to the MAD, but will almost always be larger.
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b) Estimated Mean = 25.7 employees.Estimated Variance = 252.89
Estimated Standard Deviation = √Variance= 15.9 employees.
m (midpoint
) f(x) mf(x) m− (m−)2 (m−2fx)
5 20 100.00 −20.7 428.49 8569.80
15 50 750.00 −10.7 114.49 5724.50
25 30 750.00 −0.7 0.49 14.70
35 20 700.00 9.3 86.49 1729.80
45 15 675.00 19.3 372.49 5587.35
55 10 550.00 29.3 858.49 8584.90
65 5 325.00 39.31544.4
9 7722.45totals 150 3850.00 37933.50
= 3850/150
2 = 37933.5/150
= 25.7 = 252.89
75. a)
25 35 45 55 65 80
age
Age of Tabloid Buyers
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* Note: The area of each bar in a histogram should be proportional to the fraction of the observations that fall in the bar’s class. Above we’reshowing a graph in which heightsare used to represent relative frequency, even though we have classes of unequal width. Although technically incorrect, this is fairly common practice. The graph below shows the more technically correct picture, with areas representing frequency. To give the proper proportional area, the first bar height has been raised in response to the bar’s relatively narrowwidth; the height of the last bar has been lowered because of itsrelatively greater width. These adjustments give a picture in which areas and not heights are proportional to the frequency of class membership.. Proper bar heights can be set by dividing class frequencies by class widths. (The vertical scale on the histogram would be a somewhat cumbersome “frequency per unit width.”)
b) Estimated Mean = 43.42 years of age.Estimated Variance = 256.16
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* As noted in the solution to Exercise 57, the area of each bar in a histogram should be proportional to the fraction of the observations that fall in its class. Above we’re showing a graph in which heights are used to represent relative frequency, even though we have classes of unequal width. Although technically incorrect, this is fairly common practice. The graph below shows the more technically correct picture, with areas representing frequency. Notice that the bar heights have been loweredas the classes get wider so that area is kept proportional to frequency.Proper bar heights can be set by dividing class frequencies by class widths. (The vertical scale on the histogram would be a somewhat cumbersome “relative frequency per unit width.”)
10 20 30 40 50 60 75 100 125 150 200
$1000s
Household Income
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b) Estimated Mean = 54.77 (times $1000). Estimated Variance = 2311.4 (times $10002).
Estimated Standard Deviation = √Variance= 48.08 (times $1000).
77. a) There is a significant cluster of customers in the 12 to 18 age group and another significant cluster of customers in their late 20s and early 30s. There is a substantial gap in the late teens/early 20s age group.b) There is a clear indication here that the store attracts high income shoppers. The number of customers with family incomes of $60,000 or less is relatively small.c) The store appears to attract customers who are relatively infrequent visitors to the mall. More than 50% of the shoppers made no more than 10 visits—a rate of less than once a month.(Of course it’s possible that there just aren’t that many people in general who visit the mall more than 10 time or so in a year.)
Based on what you see in these charts, what recommendations might you make to the owner of Kari H Junior Fashions?
Kari H might find a line that appeals to shoppers in their late teens and early 20s. Clearly this is an age group that is not well represented among Kari H shoppers.Furthermore, the store can either continue to cultivate its appeal to higher end shoppers ortry to find ways to reach low to moderate income shoppers as well. Finally, the store doesn’t appear to be attracting the more frequent mall visitors,
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which may mean more frequent changes in displays and/or merchandise may be appropriate.
78. a) 23 looks to be the approximate balance point for the data.b) 27 appears to be the best answer among the possibilities. More so than any of the other choices, it looks like the approximate50-50 marker. At least half the values are at or above; at least half are at or below.c) The standard deviation—roughly the average distance of the values from the mean—looks to be around 6. Among the other answers, 3 looks much too small to be the average distance; 12 and 18 look too large.
79. a) 12 looks to be the approximate “balance point” for the data.b) 7 appears to be the best answer among the possibilities. More so than any of the other choices, it looks like the approximate 50-50 marker:at least half the values appear to be at or above that point; at least half at or below.c) The standard deviation—roughly the average distance of the values from the mean—looks to be closest to 13. The other answers all appear to be much too small to be measuring, even roughly, this sort of average distance.