Why an aver­age is not enough

Two bats­men both aver­age 4040 runs, yet one is steady match after match while the other swings between ducks and cen­turies. Which one would you pick for a tight chase? The aver­age can­not tell you; it hides exactly this dif­fer­ence. A mea­sure of dis­per­sion brings it out. We will start with the range, then meet the mean devi­a­tion, both about the mean and about the median, and com­pute it for raw data, dis­crete fre­quency tables and grouped data.

Range

Range == largest value −- small­est value. It takes a sec­ond to find, but it looks only at the two extremes and ignores every­thing in between.

Mean devi­a­tion

The mean devi­a­tion about a cen­tral value aa is sim­ply the aver­age dis­tance of the obser­va­tions from aa:

MD(a)=1n∑∣xi−a∣or, with frequencies,1N∑fi∣xi−a∣.\displaystyle \text{MD}(a) = \frac{1}{n}\sum \lvert x_i - a \rvert \quad \text{or, with frequencies,} \quad \frac{1}{N}\sum f_i \lvert x_i - a \rvert.

For aa we usu­ally take the mean xˉ\bar x or the median MM. (Of all choices, the median gives the small­est pos­si­ble mean devi­a­tion.)

Exam­ple 1. Data 4,7,8,9,10,12,13,174, 7, 8, 9, 10, 12, 13, 17: xˉ=808=10\displaystyle \bar x = \tfrac{80}{8} = 10. The devi­a­tions are 6,3,2,1,0,2,3,76, 3, 2, 1, 0, 2, 3, 7, which add up to 2424, so MD(xˉ)=3(\bar x) = 3.

Dot plot of the data 4, 7, 8, 9, 10, 12, 13, 17 above a number line, each joined to a dashed line at the mean 10 and labelled with its distance: 6, 3, 2, 1, 0, 2, 3, 7; MD = 24/8 = 3.
Exam­ple 1: the aver­age dis­tance from the mean 10 is 3

Exam­ple 2. Data 13,17,16,14,11,13,10,16,11,18,12,1713, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17. Sort them first; the two mid­dle val­ues are 1313 and 1414, so M=13.5M = 13.5. The devi­a­tions add up to 2828, so MD(M)=2812=2.33\displaystyle (M) = \tfrac{28}{12} = 2.33.

Dis­crete fre­quency dis­tri­b­u­tions

Exam­ple 3.

xix_i 22 55 66 88 1010 1212
fif_i 22 88 1010 77 88 55

N=40N = 40, ∑fixi=300\displaystyle \sum f_i x_i = 300, xˉ=7.5\bar x = 7.5. ∑fi∣xi−7.5∣=11+20+15+3.5+20+22.5=92\displaystyle \sum f_i \lvert x_i - 7.5 \rvert = 11 + 20 + 15 + 3.5 + 20 + 22.5 = 92: MD =2.3= 2.3.

Bar chart of the frequency table: x = 2, 5, 6, 8, 10, 12 with frequencies 2, 8, 10, 7, 8, 5, and a dashed line at the mean 7.5.
Exam­ple 3: the fre­quency dis­tri­b­u­tion with its mean 7.5

To find the median of a fre­quency table, write down the cumu­la­tive fre­quen­cies and see where N2\displaystyle \tfrac{N}{2} falls.

Grouped data

With grouped data, use the class mid­points as the xix_i. For the median, M=l+N2−Cf×h\displaystyle M = l + \frac{\frac{N}{2} - C}{f} \times h, where ll is the lower limit of the median class, CC the cumu­la­tive fre­quency before it, ff its fre­quency and hh the class width.

Exam­ple 4.

Marks 00–1010 1010–2020 2020–3030 3030–4040 4040–5050 5050–6060
Stu­dents 66 88 1414 1616 44 22

Mid­points 5,15,25,35,45,555, 15, 25, 35, 45, 55; N=50N = 50; ∑fixi=30+120+350+560+180+110=1350\displaystyle \sum f_i x_i = 30 + 120 + 350 + 560 + 180 + 110 = 1350; xˉ=27\bar x = 27. ∑fi∣xi−27∣=132+96+28+128+72+56=512\displaystyle \sum f_i\lvert x_i - 27 \rvert = 132 + 96 + 28 + 128 + 72 + 56 = 512; MD(xˉ)=10.24(\bar x) = 10.24.

Histogram of marks in classes 0-10 to 50-60 with 6, 8, 14, 16, 4 and 2 students, and a dashed line at the mean mark 27.
Exam­ple 4: marks of 50 stu­dents, mean 27

Where it falls short. Absolute val­ues are clumsy to han­dle in alge­bra, and the mean devi­a­tion treats a large devi­a­tion no more seri­ously than its size alone. That is why, in the next les­son, we will usu­ally pre­fer the vari­ance.

Try these your­self

  1. Find the range and the mean devi­a­tion about the mean: 6,7,10,12,13,4,8,126, 7, 10, 12, 13, 4, 8, 12.
  2. Find the mean devi­a­tion about the median: 36,72,46,42,60,45,53,46,51,4936, 72, 46, 42, 60, 45, 53, 46, 51, 49.
  3. Find the mean devi­a­tion about the mean:
xix_i 1010 3030 5050 7070 9090
fif_i 44 2424 2828 1616 88
  1. Find the mean devi­a­tion about the median:
xix_i 55 77 99 1010 1212 1515
fif_i 88 66 22 22 22 66
  1. Find the mean devi­a­tion about the mean for:
Class 1010–2020 2020–3030 3030–4040 4040–5050 5050–6060
ff 55 1010 2020 1010 55
  1. Why is the mean devi­a­tion about the median never larger than that about the mean?

Answers to check against

Show answers
  1. Range 99; xˉ=9\bar x = 9; devi­a­tions 3,2,1,3,4,5,1,33, 2, 1, 3, 4, 5, 1, 3: MD =2.75= 2.75.
  2. Sorted 36,42,45,46,46,49,51,53,60,7236, 42, 45, 46, 46, 49, 51, 53, 60, 72; M=47.5M = 47.5; devi­a­tions sum 7070: MD =7= 7.
  3. N=80N = 80, xˉ=400080=50\displaystyle \bar x = \tfrac{4000}{80} = 50; ∑f∣d∣=160+480+0+320+320=1280\displaystyle \sum f\lvert d \rvert = 160 + 480 + 0 + 320 + 320 = 1280: MD =16= 16.
  4. N=26N = 26, median is the aver­age of the 1313th and 1414th val­ues: both 77; ∑f∣x−7∣=16+0+4+6+10+48=84\displaystyle \sum f\lvert x - 7 \rvert = 16 + 0 + 4 + 6 + 10 + 48 = 84: MD =8426≈3.23\displaystyle = \tfrac{84}{26} \approx 3.23.
  5. The table is sym­met­ric, so xˉ=35\bar x = 35; ∑f∣d∣=100+100+0+100+100=400\displaystyle \sum f\lvert d \rvert = 100 + 100 + 0 + 100 + 100 = 400; N=50N = 50: MD =8= 8.
  6. ∑∣xi−a∣\displaystyle \sum \lvert x_i - a \rvert is small­est when aa is the median.