Range and Mean Deviation, Worked Solutions

Ques­tion 1: Range and mean devi­a­tion about the mean#

The prob­lem#

Find the range and the mean devi­a­tion about the mean: $6, 7, 10, 12, 13, 4, 8, 12$.

Under­stand­ing the prob­lem#

You have $n = 8$ raw obser­va­tions (no fre­quency table). You need the range, which depends only on the largest and small­est val­ues, and the mean devi­a­tion about the mean, which is the aver­age dis­tance of the obser­va­tions from $\bar x$.

The idea#

Range $=$ largest $-$ small­est. For the mean devi­a­tion use

$$\displaystyle \text{MD}(\bar x) = \frac{1}{n}\sum \lvert x_i - \bar x \rvert.$$

Step-by-step solu­tion#

Step 1. Find the range. The largest value is $13$ and the small­est is $4$.

$$\text{Range} = 13 - 4 = 9$$

Step 2. Find the mean.

$$\displaystyle \bar x = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9$$

Step 3. Find each dis­tance from the mean. The absolute value makes every dis­tance pos­i­tive.

Step 4. Add the dis­tances.

$$\displaystyle \sum \lvert x_i - 9 \rvert = 3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 = 22$$

Step 5. Divide by $n = 8$.

$$\displaystyle \text{MD}(\bar x) = \frac{22}{8} = 2.75$$

Check­ing the answer#

With­out absolute val­ues, the devi­a­tions from the mean always add to $0$: $-3 - 2 + 1 + 3 + 4 - 5 - 1 + 3 = 0$. This con­firms the mean $9$ is right. A mean devi­a­tion of $2.75$ is less than half the range, which is rea­son­able.

Answer#

Range $= 9$; mean devi­a­tion about the mean $= 2.75$.

Com­mon mis­take to avoid#

For­get­ting the absolute value. With­out it the sum is always $0$ and tells you noth­ing.

Ques­tion 2: Mean devi­a­tion about the median#

The prob­lem#

Find the mean devi­a­tion about the median: $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.

Under­stand­ing the prob­lem#

There are $n = 10$ obser­va­tions, given in no par­tic­u­lar order. The cen­tral value to use is the median, so you must first sort the data.

The idea#

For an even num­ber of obser­va­tions, the median is the aver­age of the two mid­dle val­ues (the $\displaystyle \tfrac{n}{2}$th and $\displaystyle \left(\tfrac{n}{2} + 1\right)$th). Then

$$\displaystyle \text{MD}(M) = \frac{1}{n}\sum \lvert x_i - M \rvert.$$

Step-by-step solu­tion#

Step 1. Arrange the data in increas­ing order.

$$36, \; 42, \; 45, \; 46, \; 46, \; 49, \; 51, \; 53, \; 60, \; 72$$

Step 2. With $n = 10$, the mid­dle val­ues are the $5$th and $6$th: $46$ and $49$.

$$\displaystyle M = \frac{46 + 49}{2} = 47.5$$

Step 3. Find each dis­tance from $47.5$.

Step 4. Add them. Adding the first five and last five sep­a­rately helps: $11.5 + 5.5 + 2.5 + 1.5 + 1.5 = 22.5$ and $1.5 + 3.5 + 5.5 + 12.5 + 24.5 = 47.5$.

$$\displaystyle \sum \lvert x_i - 47.5 \rvert = 22.5 + 47.5 = 70$$

Step 5. Divide by $n = 10$.

$$\displaystyle \text{MD}(M) = \frac{70}{10} = 7$$

Check­ing the answer#

Because the median has equal num­bers of val­ues on each side, the sum of dis­tances equals (sum of top five) $-$ (sum of bot­tom five): $(49 + 51 + 53 + 60 + 72) - (36 + 42 + 45 + 46 + 46) = 285 - 215 = 70$. It agrees.

Answer#

Mean devi­a­tion about the median $= 7$.

Ques­tion 3: Mean devi­a­tion about the mean for a fre­quency table#

The prob­lem#

Find the mean devi­a­tion about the mean:

Under­stand­ing the prob­lem#

Each value $x_i$ occurs $f_i$ times. For exam­ple, $30$ occurs $24$ times. The total num­ber of obser­va­tions is $\displaystyle N = \sum f_i$.

The idea#

Use the fre­quency ver­sions of the for­mu­las, as in Exam­ple 3:

$$\displaystyle \bar x = \frac{\sum f_i x_i}{N}, \qquad \text{MD}(\bar x) = \frac{1}{N}\sum f_i \lvert x_i - \bar x \rvert.$$

Step-by-step solu­tion#

Step 1. Find $N$.

$$N = 4 + 24 + 28 + 16 + 8 = 80$$

Step 2. Find $\displaystyle \sum f_i x_i$.

$$\displaystyle \sum f_i x_i = 40 + 720 + 1400 + 1120 + 720 = 4000$$

Step 3. Find the mean.

$$\displaystyle \bar x = \frac{4000}{80} = 50$$

Step 4. Build the table of dis­tances and weighted dis­tances.

Step 5. Add the last row and divide by $N$.

$$\displaystyle \sum f_i \lvert x_i - 50 \rvert = 160 + 480 + 0 + 320 + 320 = 1280, \qquad \text{MD}(\bar x) = \frac{1280}{80} = 16$$

Check­ing the answer#

The weighted devi­a­tions with sign should total $0$: below the mean $-160 - 480 = -640$; above $+320 + 320 = +640$. They bal­ance, so $\bar x = 50$ is right.

Answer#

Mean devi­a­tion about the mean $= 16$.

Com­mon mis­take to avoid#

Divid­ing by the num­ber of dis­tinct val­ues ($5$) instead of $N = 80$.

Ques­tion 4: Mean devi­a­tion about the median for a fre­quency table#

The prob­lem#

Find the mean devi­a­tion about the median:

Under­stand­ing the prob­lem#

The val­ues are already in increas­ing order, each with a fre­quency. To find the median you must locate the mid­dle obser­va­tion(s), which needs cumu­la­tive fre­quen­cies.

The idea#

Build the cumu­la­tive fre­quency (c.f.) col­umn. If $N$ is even, the median is the aver­age of the $\displaystyle \tfrac{N}{2}$th and $\displaystyle \left(\tfrac{N}{2} + 1\right)$th obser­va­tions; the c.f. col­umn tells you which value each falls on. Then use $\displaystyle \text{MD}(M) = \frac{1}{N}\sum f_i \lvert x_i - M \rvert$.

Step-by-step solu­tion#

Step 1. Build the cumu­la­tive fre­quen­cies.

So $N = 26$.

Step 2. Locate the mid­dle obser­va­tions. $N$ is even, so we need the $13$th and $14$th obser­va­tions. The c.f. shows obser­va­tions $1$ to $8$ are $5$, and obser­va­tions $9$ to $14$ are $7$. So both the $13$th and $14$th are $7$.

$$\displaystyle M = \frac{7 + 7}{2} = 7$$

Step 3. Find the dis­tances from $7$ and mul­ti­ply by fre­quen­cies.

Step 4. Add and divide by $N$.

$$\displaystyle \sum f_i \lvert x_i - 7 \rvert = 16 + 0 + 4 + 6 + 10 + 48 = 84$$

$$\displaystyle \text{MD}(M) = \frac{84}{26} = \frac{42}{13} \approx 3.23$$

Check­ing the answer#

$13$ obser­va­tions lie at or below $7$ and $13$ at or above it, so $7$ really splits the data in half. For com­par­i­son, the mean here is $9$ and the mean devi­a­tion about it works out to $\displaystyle \tfrac{88}{26} \approx 3.38$, which is larger, as Ques­tion 6 pre­dicts.

Answer#

Mean devi­a­tion about the median $\displaystyle = \tfrac{42}{13} \approx 3.23$.

Com­mon mis­take to avoid#

Tak­ing the median as the mid­dle of the list of dis­tinct val­ues ($9$ or $10$). The median depends on the fre­quen­cies.

Ques­tion 5: Mean devi­a­tion about the mean for grouped data#

The prob­lem#

Find the mean devi­a­tion about the mean for:

Under­stand­ing the prob­lem#

The data are grouped into classes of width $10$, so you do not know the exact val­ues. The stan­dard approach is to treat every obser­va­tion in a class as if it were at the class mid­point.

The idea#

Replace each class by its mid­point $\displaystyle x_i = \tfrac{\text{lower} + \text{upper}}{2}$ and then pro­ceed exactly as for a dis­crete fre­quency table (Exam­ple 4).

Step-by-step solu­tion#

Step 1. Find the mid­points: $\displaystyle \tfrac{10 + 20}{2} = 15$, and so on.

$$x_i = 15, \; 25, \; 35, \; 45, \; 55$$

Step 2. Find $N$ and $\displaystyle \sum f_i x_i$.

$$N = 5 + 10 + 20 + 10 + 5 = 50$$

$$\displaystyle \sum f_i x_i = 75 + 250 + 700 + 450 + 275 = 1750$$

Step 3. Find the mean.

$$\displaystyle \bar x = \frac{1750}{50} = 35$$

This makes sense: the fre­quen­cies $5, 10, 20, 10, 5$ are sym­met­ric about the mid­dle class, whose mid­point is $35$.

Step 4. Dis­tances from $35$ and weighted dis­tances.

Step 5. Add and divide by $N$.

$$\displaystyle \sum f_i\lvert x_i - 35 \rvert = 400, \qquad \text{MD}(\bar x) = \frac{400}{50} = 8$$

Check­ing the answer#

The answer is less than the class width of $10$ and most of the data sit in the mid­dle class, so an aver­age dis­tance of $8$ is sen­si­ble.

Answer#

Mean devi­a­tion about the mean $= 8$.

Ques­tion 6: Why the median gives the smaller mean devi­a­tion#

The prob­lem#

Why is the mean devi­a­tion about the median never larger than that about the mean?

Under­stand­ing the prob­lem#

You must explain a gen­eral fact: for any data set, $\text{MD}(M) \le \text{MD}(\bar x)$. This is a rea­son­ing ques­tion, not a cal­cu­la­tion.

The idea#

Think of $\displaystyle S(a) = \sum \lvert x_i - a \rvert$, the total dis­tance of all obser­va­tions from a point $a$. The les­son states that $S(a)$ is small­est when $a$ is the median. Since the mean is just one pos­si­ble choice of $a$, it can­not beat the median.

Step-by-step solu­tion#

Step 1. Define the total dis­tance from any point $a$:

$$\displaystyle S(a) = \sum \lvert x_i - a \rvert$$

Step 2. See what hap­pens when you move $a$. Sup­pose $a$ is moved a small amount $h$ to the right, and there are $p$ obser­va­tions to the left of $a$ and $q$ to its right. Each of the $p$ left obser­va­tions becomes $h$ far­ther away, and each of the $q$ right obser­va­tions becomes $h$ closer. So $S(a)$ changes by $(p - q)h$.

Step 3. Con­clude where $S(a)$ is least. If more obser­va­tions lie on the right ($q > p$), mov­ing right low­ers $S$; if more lie on the left, mov­ing left low­ers $S$. $S$ stops decreas­ing only when the obser­va­tions are bal­anced on each side, and that is exactly the median.

Step 4. Com­pare with the mean. The mean $\bar x$ is just one value of $a$, so

$$\displaystyle \sum \lvert x_i - M \rvert \le \sum \lvert x_i - \bar x \rvert$$

Step 5. Divide both sides by $n$ (or $N$):

$$\text{MD}(M) \le \text{MD}(\bar x)$$

Equal­ity holds when the mean and median coin­cide, as in Ques­tion 5.

Check­ing the answer#

In Ques­tion 4 the median is $7$ with MD $\approx 3.23$, while the mean is $9$ with MD $\approx 3.38$. The median gives the smaller value, as the argu­ment pre­dicts.

Answer#

Because the total dis­tance $\displaystyle \sum \lvert x_i - a \rvert$ is least when $a$ is the median, the mean devi­a­tion about the median is the small­est pos­si­ble, so it can never exceed the mean devi­a­tion about the mean.