Variance, Standard Deviation and Mixed Practice, Worked Solutions

Ques­tion 1: Mean and vari­ance of raw data#

The prob­lem#

Find the mean and vari­ance: $2, 4, 6, 8, 10, 12$.

Under­stand­ing the prob­lem#

There are $n = 6$ ungrouped obser­va­tions. You need the mean $\bar x$ and the vari­ance $\sigma^2$.

The idea#

Find $\bar x$, then the squared devi­a­tions $(x_i - \bar x)^2$, then aver­age them: $\displaystyle \sigma^2 = \frac1n\sum(x_i - \bar x)^2$.

Step-by-step solu­tion#

Step 1. Mean.

$$\displaystyle \bar x = \frac{2 + 4 + 6 + 8 + 10 + 12}{6} = \frac{42}{6} = 7.$$

Step 2. Devi­a­tions from the mean and their squares.

Step 3. Add the squares and divide by $6$.

$$\displaystyle \sigma^2 = \frac{25 + 9 + 1 + 1 + 9 + 25}{6} = \frac{70}{6} = \frac{35}{3} \approx 11.67.$$

Check­ing the answer#

Short-cut for­mula: $\displaystyle \sum x^2 = 4 + 16 + 36 + 64 + 100 + 144 = 364$, and $\displaystyle \frac{364}{6} - 49 = \frac{364 - 294}{6} = \frac{70}{6}$. Same result.

Answer#

Mean $= 7$, vari­ance $\displaystyle = \frac{35}{3}$.

Ques­tion 2: Vari­ance of the first $n$ even num­bers#

The prob­lem#

Find the vari­ance of the first $n$ even nat­ural num­bers.

Under­stand­ing the prob­lem#

The data are $2, 4, 6, \ldots, 2n$. You need a for­mula for their vari­ance in terms of $n$.

The idea#

These num­bers are exactly $2 \times (1, 2, \ldots, n)$. The les­son showed that the vari­ance of $1, 2, \ldots, n$ is $\displaystyle \frac{n^2 - 1}{12}$, and that mul­ti­ply­ing every obser­va­tion by $k$ mul­ti­plies the vari­ance by $k^2$.

Step-by-step solu­tion#

Step 1. Vari­ance of $1, 2, \ldots, n$ (from the les­son):

$$\displaystyle \sigma^2 = \frac{n^2 - 1}{12}.$$

Step 2. Mul­ti­ply­ing by $2$ mul­ti­plies $\sigma$ by $2$, so it mul­ti­plies $\sigma^2$ by $2^2 = 4$.

$$\displaystyle \sigma^2_{\text{even}} = 4 \cdot \frac{n^2 - 1}{12} = \frac{n^2 - 1}{3}.$$

Check­ing the answer#

$n = 3$: data $2, 4, 6$, mean $4$, squared devi­a­tions $4, 0, 4$, vari­ance $\displaystyle \frac83$. For­mula: $\displaystyle \frac{9 - 1}{3} = \frac83$. Cor­rect.

Answer#

The vari­ance is $\displaystyle \frac{n^2 - 1}{3}$.

Com­mon mis­take to avoid#

Dou­bling the data dou­bles the stan­dard devi­a­tion but mul­ti­plies the vari­ance by $4$, not $2$.

Ques­tion 3: Mean and stan­dard devi­a­tion of a fre­quency table#

The prob­lem#

Find mean and stan­dard devi­a­tion:

Under­stand­ing the prob­lem#

Each value $x_i$ occurs $f_i$ times. You need $\bar x$ and $\sigma$.

The idea#

Use $\displaystyle \bar x = \frac{\sum f_ix_i}{N}$ and $\displaystyle \sigma^2 = \frac{1}{N}\sum f_ix_i^2 - \bar x^2$, build­ing the columns $f_ix_i$ and $f_ix_i^2$.

Step-by-step solu­tion#

Step 1. Build the work­ing table.

Step 2. Mean.

$$\displaystyle \bar x = \frac{420}{30} = 14.$$

Step 3. Vari­ance.

$$\displaystyle \sigma^2 = \frac{7254}{30} - 14^2 = 241.8 - 196 = 45.8.$$

Step 4. Stan­dard devi­a­tion.

$$\sigma = \sqrt{45.8} \approx 6.77.$$

Check­ing the answer#

Most of the data lie between $8$ and $20$, within about one stan­dard devi­a­tion of $14$ ($14 \pm 6.77$ is $7.23$ to $20.77$). That is rea­son­able.

Answer#

Mean $= 14$, stan­dard devi­a­tion $= \sqrt{45.8} \approx 6.77$.

Ques­tion 4: Grouped data by the short-cut method#

The prob­lem#

Find mean and stan­dard devi­a­tion by the short-cut method:

Under­stand­ing the prob­lem#

For grouped data each class is rep­re­sented by its mid­point. The short-cut (step-devi­a­tion) method uses small num­bers $y_i$ instead of the mid­points, to keep the arith­metic easy.

The idea#

Take $A = 65$ (the mid­dle mid­point) and $h = 10$ (the class width), set $\displaystyle y_i = \frac{x_i - A}{h}$, and use

$$\displaystyle \bar x = A + h\bar y, \qquad \sigma_x = h\,\sigma_y, \qquad \sigma_y^2 = \frac{\sum fy^2}{N} - \bar y^2.$$

Step-by-step solu­tion#

Step 1. Mid­points, $y$ val­ues and prod­ucts.

Step 2. Mean of $y$ and of $x$.

$$\displaystyle \bar y = \frac{-15}{50} = -0.3, \qquad \bar x = 65 + 10(-0.3) = 65 - 3 = 62.$$

Step 3. Vari­ance of $y$.

$$\displaystyle \sigma_y^2 = \frac{105}{50} - (-0.3)^2 = 2.1 - 0.09 = 2.01.$$

Step 4. Stan­dard devi­a­tion of $x$.

$$\sigma_x = 10\sqrt{2.01} \approx 10 \times 1.4177 \approx 14.18.$$

Check­ing the answer#

The mean $62$ lies in the class $60$–$70$, which has the high­est fre­quency, and slightly below $65$ because more obser­va­tions lie below than above. Sen­si­ble.

Answer#

Mean $= 62$, stan­dard devi­a­tion $\approx 14.18$.

Com­mon mis­take to avoid#

Remem­ber to mul­ti­ply $\sigma_y$ by $h = 10$. The shift $A$ does not affect $\sigma$, but the scale $h$ does.

Ques­tion 5: Which plant is more con­sis­tent?#

The prob­lem#

Two plants' daily out­put: plant P mean $63$, $\sigma = 9$; plant Q mean $54$, $\sigma = 7.2$. Which is more con­sis­tent?

Under­stand­ing the prob­lem#

"More con­sis­tent" means less vari­able rel­a­tive to its size. Because the means dif­fer, com­par­ing $\sigma$ alone is not fair.

The idea#

Com­pare coef­fi­cients of vari­a­tion, $\displaystyle \text{CV} = \frac{\sigma}{\bar x} \times 100$. The smaller CV is the more con­sis­tent.

Step-by-step solu­tion#

Step 1. CV of plant P.

$$\displaystyle \text{CV}_P = \frac{9}{63} \times 100 = \frac{100}{7} \approx 14.29\%.$$

Step 2. CV of plant Q.

$$\displaystyle \text{CV}_Q = \frac{7.2}{54} \times 100 = \frac{40}{3} \approx 13.33\%.$$

Step 3. Com­pare: $13.33\% < 14.29\%$.

Check­ing the answer#

$\displaystyle \frac{9}{63} = \frac17 \approx 0.143$ and $\displaystyle \frac{7.2}{54} = \frac{2}{15} \approx 0.133$. The order is con­firmed.

Answer#

Plant Q is more con­sis­tent (CV $13.33\%$ against $14.29\%$).

Ques­tion 6: Cor­rect­ing a wrongly recorded obser­va­tion#

The prob­lem#

The mean and stan­dard devi­a­tion of $25$ obser­va­tions are $12$ and $3$. One obser­va­tion was recorded as $9$ instead of $14$. Find the cor­rect mean and stan­dard devi­a­tion.

Under­stand­ing the prob­lem#

The mean and $\sigma$ were com­puted with the wrong value $9$. You must replace $9$ by $14$ and recom­pute. Here $n = 25$.

The idea#

Recover the wrong $\displaystyle \sum x$ and $\displaystyle \sum x^2$ from the given mean and $\sigma$, cor­rect both sums, then recom­pute. Use

$$\displaystyle \sum x = n\bar x, \qquad \sum x^2 = n(\sigma^2 + \bar x^2).$$

Step-by-step solu­tion#

Step 1. Wrong sums.

$$\displaystyle \sum x = 25 \times 12 = 300, \qquad \sum x^2 = 25(3^2 + 12^2) = 25 \times 153 = 3825.$$

Step 2. Cor­rect the sums: remove $9$, add $14$.

$$\displaystyle \begin{aligned} \sum x &= 300 - 9 + 14 = 305, \\ \sum x^2 &= 3825 - 81 + 196 = 3940. \end{aligned}$$

Step 3. Cor­rect mean.

$$\displaystyle \bar x = \frac{305}{25} = 12.2.$$

Step 4. Cor­rect vari­ance and stan­dard devi­a­tion.

$$\displaystyle \sigma^2 = \frac{3940}{25} - 12.2^2 = 157.6 - 148.84 = 8.76, \qquad \sigma = \sqrt{8.76} \approx 2.96.$$

Check­ing the answer#

Rais­ing one value by $5$ should raise the mean by $\displaystyle \frac{5}{25} = 0.2$: $12 \to 12.2$. Cor­rect. The new value $14$ is closer to the mean than $9$ was, so the spread should fall slightly: $3 \to 2.96$. Sen­si­ble.

Answer#

Cor­rect mean $= 12.2$, cor­rect stan­dard devi­a­tion $= \sqrt{8.76} \approx 2.96$.

Ques­tion 7: Effect of $3x + 5$ on mean and SD#

The prob­lem#

If each obser­va­tion is mul­ti­plied by $3$ and then increased by $5$, how do the mean and stan­dard devi­a­tion change?

Under­stand­ing the prob­lem#

Each new obser­va­tion is $y_i = 3x_i + 5$. You must express $\bar y$ and $\sigma_y$ in terms of $\bar x$ and $\sigma_x$.

The idea#

From the les­son: adding a con­stant shifts the mean but not the spread; mul­ti­ply­ing by $k$ mul­ti­plies both the mean and $\sigma$ by $k$ (the SD by $\lvert k \rvert$).

Step-by-step solu­tion#

Step 1. Mean.

$$\displaystyle \bar y = \frac{1}{n}\sum(3x_i + 5) = 3 \cdot \frac1n\sum x_i + 5 = 3\bar x + 5.$$

Step 2. Devi­a­tions.

$$y_i - \bar y = (3x_i + 5) - (3\bar x + 5) = 3(x_i - \bar x).$$

Step 3. Vari­ance and SD.

$$\displaystyle \sigma_y^2 = \frac1n\sum 9(x_i - \bar x)^2 = 9\sigma_x^2 \;\Rightarrow\; \sigma_y = 3\sigma_x.$$

Check­ing the answer#

Data $1, 3$: mean $2$, $\sigma = 1$. New data $8, 14$: mean $11 = 3(2) + 5$, $\sigma = 3$. Cor­rect.

Answer#

The mean becomes $3\bar x + 5$; the stan­dard devi­a­tion becomes $3\sigma$ (the $+5$ has no effect on it).

Ques­tion 8: Find­ing two miss­ing obser­va­tions#

The prob­lem#

The mean and vari­ance of $7$ obser­va­tions are $8$ and $16$. Five of them are $2, 4, 10, 12, 14$. Find the other two.

Under­stand­ing the prob­lem#

Call the miss­ing val­ues $a$ and $b$. The mean gives their sum; the vari­ance gives the sum of their squares. Then solve.

The idea#

Use $\displaystyle \sum x = n\bar x$ and $\displaystyle \sum x^2 = n(\sigma^2 + \bar x^2)$, then the iden­tity $(a + b)^2 = a^2 + b^2 + 2ab$ to find $ab$.

Step-by-step solu­tion#

Step 1. Sum of all seven.

$$\displaystyle \sum x = 7 \times 8 = 56.$$

The five known val­ues add to $2 + 4 + 10 + 12 + 14 = 42$, so

$$a + b = 56 - 42 = 14.$$

Step 2. Sum of squares of all seven.

$$\displaystyle \sum x^2 = 7(16 + 64) = 560.$$

The known squares add to $4 + 16 + 100 + 144 + 196 = 460$, so

$$a^2 + b^2 = 560 - 460 = 100.$$

Step 3. Find $ab$.

$$2ab = (a + b)^2 - (a^2 + b^2) = 196 - 100 = 96 \;\Rightarrow\; ab = 48.$$

Step 4. $a$ and $b$ are roots of $t^2 - 14t + 48 = 0$.

$$(t - 6)(t - 8) = 0 \;\Rightarrow\; t = 6 \text{ or } 8.$$

Check­ing the answer#

Data $2, 4, 6, 8, 10, 12, 14$: mean $8$; squared devi­a­tions $36, 16, 4, 0, 4, 16, 36$ add to $112$, and $\displaystyle \frac{112}{7} = 16$. Cor­rect.

Answer#

The other two obser­va­tions are $6$ and $8$.

Ques­tion 9: Vari­ance zero means all val­ues equal#

The prob­lem#

Show that the vari­ance of a set of equal num­bers is $0$, and con­versely.

Under­stand­ing the prob­lem#

Two direc­tions must be proved. (i) If every $x_i$ is the same, then $\sigma^2 = 0$. (ii) If $\sigma^2 = 0$, then every $x_i$ is the same.

The idea#

Use the def­i­n­i­tion $\displaystyle \sigma^2 = \frac1n\sum(x_i - \bar x)^2$ and the fact that a sum of squares of real num­bers is zero only when each square is zero.

Step-by-step solu­tion#

Step 1 (equal num­bers give zero vari­ance). Sup­pose $x_1 = x_2 = \cdots = x_n = c$. Then

$$\displaystyle \bar x = \frac{nc}{n} = c, \qquad x_i - \bar x = 0 \text{ for every } i,$$

so $\displaystyle \sigma^2 = \frac1n(0 + 0 + \cdots + 0) = 0$.

Step 2 (zero vari­ance gives equal num­bers). Sup­pose $\sigma^2 = 0$. Then

$$\displaystyle \sum_{i=1}^{n}(x_i - \bar x)^2 = 0.$$

Step 3. Each term $(x_i - \bar x)^2 \ge 0$. A sum of non-neg­a­tive terms is $0$ only if every term is $0$. So $x_i - \bar x = 0$, that is $x_i = \bar x$ for every $i$. All the num­bers are equal. $\blacksquare$

Check­ing the answer#

Data $5, 5, 5$: mean $5$, devi­a­tions all $0$, vari­ance $0$. Data $5, 5, 6$ has vari­ance $\displaystyle \frac29 \ne 0$, so even one dif­fer­ent value makes the vari­ance pos­i­tive.

Answer#

$\displaystyle \sigma^2 = \frac1n\sum(x_i - \bar x)^2 = 0$ exactly when every $x_i = \bar x$, that is, when all the num­bers are equal.

Ques­tion 10: Mean devi­a­tion against stan­dard devi­a­tion#

The prob­lem#

Com­pute the mean devi­a­tion about the mean and the stan­dard devi­a­tion of $3, 5, 7, 9, 11$. Which is larger?

Under­stand­ing the prob­lem#

Mean devi­a­tion uses absolute devi­a­tions $\lvert x_i - \bar x \rvert$; stan­dard devi­a­tion uses squared devi­a­tions and then a square root. Com­pute both and com­pare.

The idea#

$$\displaystyle \text{MD} = \frac1n\sum\lvert x_i - \bar x \rvert, \qquad \sigma = \sqrt{\frac1n\sum(x_i - \bar x)^2}.$$

Step-by-step solu­tion#

Step 1. Mean.

$$\displaystyle \bar x = \frac{3 + 5 + 7 + 9 + 11}{5} = \frac{35}{5} = 7.$$

Step 2. Devi­a­tions.

Step 3. Mean devi­a­tion.

$$\displaystyle \text{MD} = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4.$$

Step 4. Stan­dard devi­a­tion.

$$\displaystyle \sigma^2 = \frac{16 + 4 + 0 + 4 + 16}{5} = \frac{40}{5} = 8, \qquad \sigma = \sqrt{8} = 2\sqrt2 \approx 2.83.$$

Step 5. Com­pare: $2.83 > 2.4$.

Check­ing the answer#

Squar­ing gives extra weight to the larger devi­a­tions ($4$ becomes $16$), so $\sigma$ is nor­mally at least as large as the mean devi­a­tion. The result fits.

Answer#

MD $= 2.4$, $\sigma = 2\sqrt2 \approx 2.83$; the stan­dard devi­a­tion is larger.