Chapter 13: Statistics

The given data is $$6, 7, 10, 12, 13, 4, 8, 12$$, with $$n = 8$$ observations.

Step 1 — Find the mean.

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

Step 2 — Find the absolute deviations from the mean.

$$\sum_{i=1}^{8} |x_i - \bar{x}| = 3+2+1+3+4+5+1+3 = 22$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1}{n}\sum_{i=1}^{n} |x_i - \bar{x}| = \frac{22}{8} = 2.75$$

The given data has $$n = 20$$ observations.

Step 1 — Find the mean.

$$\sum x_i = 12+3+18+17+4+9+17+19+20+15+8+17+2+3+16+11+3+1+0+5 = 200$$

$$\bar{x} = \frac{\sum x_i}{n} = \frac{200}{20} = 10$$

Step 2 — Find the absolute deviations from the mean.

The deviations $$|x_i - 10|$$ for the observations, taken in the given order, are:

$$2,\ 7,\ 8,\ 7,\ 6,\ 1,\ 7,\ 9,\ 10,\ 5,\ 2,\ 7,\ 8,\ 7,\ 6,\ 1,\ 7,\ 9,\ 10,\ 5$$

$$\sum |x_i - \bar{x}| = 2+7+8+7+6+1+7+9+10+5+2+7+8+7+6+1+7+9+10+5 = 124$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1}{n}\sum |x_i - \bar{x}| = \frac{124}{20} = 6.2$$

The given data is $$3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21$$, with $$n = 11$$ observations.

Step 1 — Arrange the data in ascending order and find the median.

$$3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21$$

Since $$n = 11$$ is odd, the median is the $$\left(\frac{n+1}{2}\right)^{\text{th}} = 6^{\text{th}}$$ observation.

$$M = 9$$

Step 2 — Find the absolute deviations from the median.

$$\sum |x_i - M| = 6+6+5+4+2+0+1+3+9+10+12 = 58$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{n}\sum |x_i - M| = \frac{58}{11} \approx 5.27$$

Step 1 — Find the mean.

$$N = \sum f_i = 2+8+10+7+8+5 = 40$$

$$\sum f_i x_i = 2(2)+5(8)+6(10)+8(7)+10(8)+12(5) = 4+40+60+56+80+60 = 300$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{300}{40} = 7.5$$

Step 2 — Compute $$f_i|x_i - \bar{x}|$$.

$$\sum f_i|x_i - \bar{x}| = 11+20+15+3.5+20+22.5 = 92$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1}{N}\sum f_i|x_i - \bar{x}| = \frac{92}{40} = 2.3$$

Step 1 — Find the median.

$$N = \sum f_i = 3+4+5+2+4+5+4+3 = 30$$

The cumulative frequencies (c.f.) are:

Since $$N = 30$$ is even, the median is the mean of the $$15^{\text{th}}$$ and $$16^{\text{th}}$$ observations. From the c.f. column both of these observations equal $$13$$.

$$M = \frac{13 + 13}{2} = 13$$

Step 2 — Compute $$f_i|x_i - M|$$.

$$\sum f_i|x_i - M| = 30+28+20+2+0+10+32+27 = 149$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{N}\sum f_i|x_i - M| = \frac{149}{30} \approx 4.97$$

Step 1 — Find the class mid-points and the mean.

The mid-point of each class is $$x_i = \dfrac{\text{lower limit} + \text{upper limit}}{2}$$.

$$N = \sum f_i = 40, \qquad \sum f_i x_i = 1800$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{1800}{40} = 45$$

Step 2 — Compute $$f_i|x_i - \bar{x}|$$.

$$\sum f_i|x_i - \bar{x}| = 60+60+80+0+80+60+60 = 400$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1}{N}\sum f_i|x_i - \bar{x}| = \frac{400}{40} = 10$$

Step 1 — Find the median.

$$N = 50, \quad \frac{N}{2} = 25$$

The cumulative frequency just greater than $$25$$ is $$28$$, so the median class is $$20$$-$$30$$.

Here $$l = 20$$, $$C = 13$$ (c.f. before the median class), $$f = 15$$, $$h = 10$$.

$$M = l + \frac{\frac{N}{2} - C}{f}\times h = 20 + \frac{25 - 13}{15}\times 10 = 20 + 8 = 28$$

Step 2 — Compute $$f_i|x_i - M|$$ using class mid-points.

$$\sum f_i|x_i - M| = 138+91+45+112+68+54 = 508$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{N}\sum f_i|x_i - M| = \frac{508}{50} = 10.16$$

The given data is $$6, 8, 10, 12, 14, 16, 18, 20, 22, 24$$, with $$n = 10$$.

Step 1 — Find the mean.

$$\bar{x} = \frac{\sum x_i}{n} = \frac{6+8+10+12+14+16+18+20+22+24}{10} = \frac{150}{10} = 15$$

Step 2 — Compute the squared deviations $$(x_i - \bar{x})^2$$.

$$\sum (x_i - \bar{x})^2 = 81+49+25+9+1+1+9+25+49+81 = 330$$

Step 3 — Apply the variance formula.

$$\sigma^2 = \frac{1}{n}\sum (x_i - \bar{x})^2 = \frac{330}{10} = 33$$

Step 1 — Find the mean.

$$N = \sum f_i = 3+5+9+5+4+3+1 = 30$$

$$\sum f_i x_i = 4(3)+8(5)+11(9)+17(5)+20(4)+24(3)+32(1) = 12+40+99+85+80+72+32 = 420$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{420}{30} = 14$$

Step 2 — Compute $$f_i(x_i - \bar{x})^2$$.

$$\sum f_i(x_i - \bar{x})^2 = 300+180+81+45+144+300+324 = 1374$$

Step 3 — Find the variance and standard deviation.

$$\sigma^2 = \frac{1}{N}\sum f_i(x_i - \bar{x})^2 = \frac{1374}{30} = 45.8$$

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

Step 1 — Find the class mid-points and the mean.

$$N = 50, \qquad \sum f_i x_i = 3100$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{3100}{50} = 62$$

Step 2 — Compute $$f_i(x_i - \bar{x})^2$$.

$$\sum f_i(x_i - \bar{x})^2 = 2187+2023+588+135+1352+1587+2178 = 10050$$

Step 3 — Find the variance and standard deviation.

$$\sigma^2 = \frac{1}{N}\sum f_i(x_i - \bar{x})^2 = \frac{10050}{50} = 201$$

$$\sigma = \sqrt{201} \approx 14.18$$

We use the step-deviation method. Take the assumed mean $$A = 13$$ and common width $$h = 5$$, and define $$d_i = \dfrac{x_i - A}{h}$$.

$$N = \sum f_i = 7+10+15+10+6 = 48$$

$$\sum f_i d_i = -14-10+0+10+12 = -2, \qquad \sum f_i d_i^2 = 28+10+0+10+24 = 72$$

Apply the step-deviation formula for standard deviation.

$$\sigma = h\sqrt{\frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2}$$

$$\sigma = 5\sqrt{\frac{72}{48} - \left(\frac{-2}{48}\right)^2} = 5\sqrt{1.5 - 0.0017}$$

$$\sigma = 5\sqrt{1.4983} = 5(1.2241) \approx 6.12$$

We use the step-deviation method with assumed mean $$A = 65$$ and class width $$h = 10$$; define $$y_i = \dfrac{x_i - A}{h}$$, where $$x_i$$ is the class mid-point.

$$N = \sum f_i = 50$$

$$\sum f_i y_i = -15, \qquad \sum f_i y_i^2 = 105$$

Step 1 — Mean.

$$\bar{x} = A + h\left(\frac{\sum f_i y_i}{N}\right) = 65 + 10\left(\frac{-15}{50}\right) = 65 - 3 = 62$$

Step 2 — Variance.

$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$

$$\sigma^2 = \frac{100}{2500}\left[50(105) - (-15)^2\right] = \frac{1}{25}\left[5250 - 225\right] = \frac{5025}{25} = 201$$

Step 3 — Standard deviation.

$$\sigma = \sqrt{201} \approx 14.18$$

The data is $$4, 7, 8, 9, 10, 12, 13, 17$$, with $$n = 8$$.

Step 1 — Find the mean.

$$\bar{x} = \frac{\sum x_i}{n} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$$

Step 2 — Find the absolute deviations from the mean.

$$\sum |x_i - \bar{x}| = 6+3+2+1+0+2+3+7 = 24$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1}{n}\sum |x_i - \bar{x}| = \frac{24}{8} = 3$$

The data has $$n = 10$$ observations.

Step 1 — Find the mean.

$$\sum x_i = 38+70+48+40+42+55+63+46+54+44 = 500$$

$$\bar{x} = \frac{500}{10} = 50$$

Step 2 — Find the absolute deviations from the mean.

$$\sum |x_i - \bar{x}| = 12+20+2+10+8+5+13+4+4+6 = 84$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{84}{10} = 8.4$$

The data has $$n = 12$$ observations.

Step 1 — Arrange in ascending order and find the median.

$$10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$$

Since $$n = 12$$ is even, the median is the mean of the $$6^{\text{th}}$$ and $$7^{\text{th}}$$ observations.

$$M = \frac{13 + 14}{2} = 13.5$$

Step 2 — Find the absolute deviations from the median.

$$\sum |x_i - M| = 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5 = 28$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{n}\sum |x_i - M| = \frac{28}{12} \approx 2.33$$

The data has $$n = 10$$ observations.

Step 1 — Arrange in ascending order and find the median.

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

Since $$n = 10$$ is even, the median is the mean of the $$5^{\text{th}}$$ and $$6^{\text{th}}$$ observations.

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

Step 2 — Find the absolute deviations from the median.

$$\sum |x_i - M| = 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5 = 70$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{70}{10} = 7$$

Step 1 — Find the mean.

$$N = \sum f_i = 7+4+6+3+5 = 25$$

$$\sum f_i x_i = 5(7)+10(4)+15(6)+20(3)+25(5) = 35+40+90+60+125 = 350$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$$

Step 2 — Compute $$f_i|x_i - \bar{x}|$$.

$$\sum f_i|x_i - \bar{x}| = 63+16+6+18+55 = 158$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1}{N}\sum f_i|x_i - \bar{x}| = \frac{158}{25} = 6.32$$

Step 1 — Find the mean.

$$N = \sum f_i = 4+24+28+16+8 = 80$$

$$\sum f_i x_i = 10(4)+30(24)+50(28)+70(16)+90(8) = 40+720+1400+1120+720 = 4000$$

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

Step 2 — Compute $$f_i|x_i - \bar{x}|$$.

$$\sum f_i|x_i - \bar{x}| = 160+480+0+320+320 = 1280$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1280}{80} = 16$$

Step 1 — Find the median.

$$N = \sum f_i = 8+6+2+2+2+6 = 26$$

Since $$N = 26$$ is even, the median is the mean of the $$13^{\text{th}}$$ and $$14^{\text{th}}$$ observations. From the c.f. column, both observations equal $$7$$.

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

Step 2 — Compute $$f_i|x_i - M|$$.

$$\sum f_i|x_i - M| = 16+0+4+6+10+48 = 84$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{N}\sum f_i|x_i - M| = \frac{84}{26} \approx 3.23$$

Step 1 — Find the median.

$$N = \sum f_i = 3+5+6+7+8 = 29$$

Since $$N = 29$$ is odd, the median is the $$\left(\frac{N+1}{2}\right)^{\text{th}} = 15^{\text{th}}$$ observation. The c.f. first reaches $$15$$ at the c.f. value $$21$$, so the $$15^{\text{th}}$$ observation is $$30$$.

$$M = 30$$

Step 2 — Compute $$f_i|x_i - M|$$.

$$\sum f_i|x_i - M| = 45+45+18+0+40 = 148$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{N}\sum f_i|x_i - M| = \frac{148}{29} \approx 5.10$$

Step 1 — Find the class mid-points and the mean.

$$N = 50, \qquad \sum f_i x_i = 17900$$

$$\bar{x} = \frac{17900}{50} = 358$$

Step 2 — Compute $$f_i|x_i - \bar{x}|$$.

$$\sum f_i|x_i - \bar{x}| = 1232+1664+972+80+644+960+1168+1176 = 7896$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{7896}{50} = 157.92$$

Step 1 — Find the class mid-points and the mean.

$$N = 100, \qquad \sum f_i x_i = 12530$$

$$\bar{x} = \frac{12530}{100} = 125.3$$

Step 2 — Compute $$f_i|x_i - \bar{x}|$$.

$$\sum f_i|x_i - \bar{x}| = 227.7+198.9+137.8+141.0+176.4+247.0 = 1128.8$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(\bar{x}) = \frac{1128.8}{100} = 11.288$$

Step 1 — Find the median.

$$N = 50, \quad \frac{N}{2} = 25$$

The c.f. just greater than $$25$$ is $$28$$, so the median class is $$20$$-$$30$$, with $$l = 20$$, $$C = 14$$, $$f = 14$$, $$h = 10$$.

$$M = l + \frac{\frac{N}{2} - C}{f}\times h = 20 + \frac{25 - 14}{14}\times 10 = 20 + \frac{110}{14} \approx 27.86$$

Step 2 — Compute $$f_i|x_i - M|$$ using class mid-points.

$$\sum f_i|x_i - M| = 137.14+102.86+40.00+114.29+68.57+54.29 \approx 517.14$$

Step 3 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{N}\sum f_i|x_i - M| = \frac{517.14}{50} \approx 10.34$$

Step 1 — Convert to a continuous frequency distribution.

Subtract $$0.5$$ from each lower limit and add $$0.5$$ to each upper limit.

Step 2 — Find the median.

$$N = 100, \quad \frac{N}{2} = 50$$

The c.f. just greater than $$50$$ is $$63$$, so the median class is $$35.5$$-$$40.5$$, with $$l = 35.5$$, $$C = 37$$, $$f = 26$$, $$h = 5$$.

$$M = l + \frac{\frac{N}{2} - C}{f}\times h = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$$

Step 3 — Compute $$f_i|x_i - M|$$.

$$\sum f_i|x_i - M| = 100+90+120+70+0+60+160+135 = 735$$

Step 4 — Apply the mean deviation formula.

$$\text{M.D.}(M) = \frac{1}{N}\sum f_i|x_i - M| = \frac{735}{100} = 7.35$$

The data is $$6, 7, 10, 12, 13, 4, 8, 12$$, with $$n = 8$$.

Step 1 — Find the mean.

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

Step 2 — Compute the squared deviations from the mean.

For each observation we find the deviation $$x_i - \bar{x}$$ and its square $$(x_i - \bar{x})^2$$.

$$\sum (x_i - \bar{x})^2 = 9+4+1+9+16+25+1+9 = 74$$

Step 3 — Apply the variance formula.

$$\sigma^2 = \frac{1}{n}\sum (x_i - \bar{x})^2 = \frac{74}{8} = 9.25$$

The data is $$1, 2, 3, \ldots, n$$.

Step 1 — Find the mean.

Using $$1+2+\cdots+n = \dfrac{n(n+1)}{2}$$,

$$\bar{x} = \frac{1}{n}\sum x_i = \frac{1}{n}\cdot\frac{n(n+1)}{2} = \frac{n+1}{2}$$

Step 2 — Find the variance.

We use $$\sigma^2 = \dfrac{1}{n}\sum x_i^2 - \bar{x}^2$$, with $$1^2+2^2+\cdots+n^2 = \dfrac{n(n+1)(2n+1)}{6}$$.

$$\frac{1}{n}\sum x_i^2 = \frac{1}{n}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$$

$$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$$

Taking $$(n+1)$$ common,

$$\sigma^2 = (n+1)\left[\frac{2n+1}{6} - \frac{n+1}{4}\right] = (n+1)\cdot\frac{2(2n+1) - 3(n+1)}{12}$$

$$\sigma^2 = (n+1)\cdot\frac{4n+2 - 3n-3}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$$

The first $$10$$ multiples of $$3$$ are $$3, 6, 9, 12, 15, 18, 21, 24, 27, 30$$, with $$n = 10$$.

Step 1 — Find the mean.

$$\sum x_i = 3+6+9+12+15+18+21+24+27+30 = 165$$

$$\bar{x} = \frac{\sum x_i}{n} = \frac{165}{10} = 16.5$$

Step 2 — Compute the squared deviations $$(x_i - \bar{x})^2$$.

$$\sum (x_i - \bar{x})^2 = 182.25+110.25+56.25+20.25+2.25+2.25+20.25+56.25+110.25+182.25 = 742.5$$

Step 3 — Apply the variance formula.

$$\sigma^2 = \frac{1}{n}\sum (x_i - \bar{x})^2 = \frac{742.5}{10} = 74.25$$

Step 1 — Find the mean.

$$N = \sum f_i = 2+4+7+12+8+4+3 = 40$$

$$\sum f_i x_i = 6(2)+10(4)+14(7)+18(12)+24(8)+28(4)+30(3)$$

$$\sum f_i x_i = 12+40+98+216+192+112+90 = 760$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{760}{40} = 19$$

Step 2 — Compute $$f_i(x_i - \bar{x})^2$$.

$$\sum f_i(x_i - \bar{x})^2 = 338+324+175+12+200+324+363 = 1736$$

Step 3 — Apply the variance formula.

$$\sigma^2 = \frac{1}{N}\sum f_i(x_i - \bar{x})^2 = \frac{1736}{40} = 43.4$$

Step 1 — Find the mean.

$$N = \sum f_i = 3+2+3+2+6+3+3 = 22$$

$$\sum f_i x_i = 92(3)+93(2)+97(3)+98(2)+102(6)+104(3)+109(3)$$

$$\sum f_i x_i = 276+186+291+196+612+312+327 = 2200$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2200}{22} = 100$$

Step 2 — Compute $$f_i(x_i - \bar{x})^2$$.

$$\sum f_i(x_i - \bar{x})^2 = 192+98+27+8+24+48+243 = 640$$

Step 3 — Apply the variance formula.

$$\sigma^2 = \frac{1}{N}\sum f_i(x_i - \bar{x})^2 = \frac{640}{22} \approx 29.09$$

We use the short-cut (assumed mean) method. Take the assumed mean $$A = 64$$ and define the deviations $$d_i = x_i - A$$.

$$N = \sum f_i = 2+1+12+29+25+12+10+4+5 = 100$$

$$\sum f_i d_i = -8-3-24-29+0+12+20+12+20 = 0$$

$$\sum f_i d_i^2 = 32+9+48+29+0+12+40+36+80 = 286$$

Step 1 — Mean.

$$\bar{x} = A + \frac{\sum f_i d_i}{N} = 64 + \frac{0}{100} = 64$$

Step 2 — Standard deviation.

$$\sigma = \sqrt{\frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2}$$

$$\sigma = \sqrt{\frac{286}{100} - \left(\frac{0}{100}\right)^2} = \sqrt{2.86} \approx 1.69$$

We use the step-deviation method with assumed mean $$A = 105$$ and class width $$h = 30$$; define $$y_i = \dfrac{x_i - A}{h}$$, where $$x_i$$ is the class mid-point.

$$N = \sum f_i = 2+3+5+10+3+5+2 = 30$$

$$\sum f_i y_i = -6-6-5+0+3+10+6 = 2, \qquad \sum f_i y_i^2 = 18+12+5+0+3+20+18 = 76$$

Step 1 — Mean.

$$\bar{x} = A + h\left(\frac{\sum f_i y_i}{N}\right) = 105 + 30\left(\frac{2}{30}\right) = 105 + 2 = 107$$

Step 2 — Variance.

$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$

$$\sigma^2 = \frac{(30)^2}{(30)^2}\left[30(76) - (2)^2\right] = 1\cdot\left[2280 - 4\right] = 2276$$

Step 1 — Find the class mid-points and the mean.

$$N = \sum f_i = 50, \qquad \sum f_i x_i = 25+120+375+560+270 = 1350$$

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{1350}{50} = 27$$

Step 2 — Compute $$f_i(x_i - \bar{x})^2$$.

$$\sum f_i(x_i - \bar{x})^2 = 2420+1152+60+1024+1944 = 6600$$

Step 3 — Apply the variance formula.

$$\sigma^2 = \frac{1}{N}\sum f_i(x_i - \bar{x})^2 = \frac{6600}{50} = 132$$

We use the step-deviation (short-cut) method with assumed mean $$A = 92.5$$ and class width $$h = 5$$; define $$y_i = \dfrac{x_i - A}{h}$$, where $$x_i$$ is the class mid-point.

$$N = \sum f_i = 3+4+7+7+15+9+6+6+3 = 60$$

$$\sum f_i y_i = -12-12-14-7+0+9+12+18+12 = 6$$

$$\sum f_i y_i^2 = 48+36+28+7+0+9+24+54+48 = 254$$

Step 1 — Mean.

$$\bar{x} = A + h\left(\frac{\sum f_i y_i}{N}\right) = 92.5 + 5\left(\frac{6}{60}\right) = 92.5 + 0.5 = 93$$

Step 2 — Variance.

$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$

$$\sigma^2 = \frac{(5)^2}{(60)^2}\left[60(254) - (6)^2\right] = \frac{25}{3600}\left[15240 - 36\right] = \frac{25 \times 15204}{3600} \approx 105.58$$

Step 3 — Standard deviation.

$$\sigma = \sqrt{105.58} \approx 10.27$$

Step 1 — Make the data continuous.

As suggested in the hint, subtract $$0.5$$ from each lower limit and add $$0.5$$ to each upper limit, giving the continuous classes $$32.5$$-$$36.5$$, $$36.5$$-$$40.5$$, $$40.5$$-$$44.5$$, $$44.5$$-$$48.5$$, $$48.5$$-$$52.5$$.

We use the step-deviation method with assumed mean $$A = 42.5$$ and class width $$h = 4$$; define $$y_i = \dfrac{x_i - A}{h}$$, where $$x_i$$ is the class mid-point.

$$N = \sum f_i = 15+17+21+22+25 = 100$$

$$\sum f_i y_i = -30-17+0+22+50 = 25, \qquad \sum f_i y_i^2 = 60+17+0+22+100 = 199$$

Step 2 — Mean diameter.

$$\bar{x} = A + h\left(\frac{\sum f_i y_i}{N}\right) = 42.5 + 4\left(\frac{25}{100}\right) = 42.5 + 1 = 43.5$$

Step 3 — Variance and standard deviation.

$$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$$

$$\sigma^2 = \frac{(4)^2}{(100)^2}\left[100(199) - (25)^2\right] = \frac{16}{10000}\left[19900 - 625\right] = \frac{16 \times 19275}{10000} = 30.84$$

$$\sigma = \sqrt{30.84} \approx 5.55$$

Let the $$20$$ observations be $$x_1, x_2, \ldots, x_{20}$$ with mean $$\bar{x}$$ and variance $$\sigma^2 = 5$$.

Step 1 — New mean.

When each observation is multiplied by $$2$$, the new observations are $$2x_1, 2x_2, \ldots, 2x_{20}$$. Their mean is

$$\frac{1}{n}\sum (2x_i) = 2\cdot\frac{1}{n}\sum x_i = 2\bar{x}$$

Step 2 — New variance.

The variance of the new observations is the mean of the squared deviations from the new mean $$2\bar{x}$$:

$$\sigma_{\text{new}}^2 = \frac{1}{n}\sum (2x_i - 2\bar{x})^2 = \frac{1}{n}\sum 4(x_i - \bar{x})^2 = 4\cdot\frac{1}{n}\sum (x_i - \bar{x})^2 = 4\sigma^2$$

Therefore

$$\sigma_{\text{new}}^2 = 4\times 5 = 20$$

In general, if each observation is multiplied by a constant $$k$$, the variance becomes $$k^2$$ times the original variance.

Let the two unknown observations be $$x$$ and $$y$$. The five observations are $$1, 2, 6, x, y$$.

Step 1 — Use the mean.

$$\bar{x} = \frac{1+2+6+x+y}{5} = 4.4$$

$$9 + x + y = 22 \quad\Rightarrow\quad x + y = 13 \qquad \text{...(1)}$$

Step 2 — Use the variance.

$$\sigma^2 = \frac{1}{5}\sum (x_i - \bar{x})^2 = 8.24 \quad\Rightarrow\quad \sum (x_i - 4.4)^2 = 41.2$$

For the three known observations:

$$(1-4.4)^2 + (2-4.4)^2 + (6-4.4)^2 = 11.56 + 5.76 + 2.56 = 19.88$$

Hence

$$(x-4.4)^2 + (y-4.4)^2 = 41.2 - 19.88 = 21.32 \qquad \text{...(2)}$$

Step 3 — Solve the equations.

Expand (2):

$$x^2 + y^2 - 8.8(x+y) + 2(4.4)^2 = 21.32$$

Using $$x+y = 13$$ and $$(4.4)^2 = 19.36$$:

$$x^2 + y^2 - 8.8(13) + 38.72 = 21.32$$

$$x^2 + y^2 = 21.32 + 114.4 - 38.72 = 97$$

Now $$(x+y)^2 = x^2 + y^2 + 2xy$$, so

$$13^2 = 97 + 2xy \quad\Rightarrow\quad 2xy = 169 - 97 = 72 \quad\Rightarrow\quad xy = 36$$

With $$x + y = 13$$ and $$xy = 36$$, the numbers are the roots of

$$t^2 - 13t + 36 = 0 \quad\Rightarrow\quad (t-4)(t-9) = 0$$

$$t = 4 \quad\text{or}\quad t = 9$$

Hence the other two observations are $$4$$ and $$9$$.

Let the original observations be $$x_1, x_2, \ldots, x_n$$ with mean $$\bar{x}$$ and variance

$$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar{x})^2$$

Step 1 — Find the new mean.

When each observation is increased by $$a$$, the new observations are $$x_i + a$$. Their mean is

$$\bar{y} = \frac{1}{n}\sum_{i=1}^{n} (x_i + a) = \frac{1}{n}\sum_{i=1}^{n} x_i + \frac{1}{n}(na) = \bar{x} + a$$

Step 2 — Find the new variance.

By definition, the variance of the new observations is

$$\sigma_{\text{new}}^2 = \frac{1}{n}\sum_{i=1}^{n} \big[(x_i + a) - \bar{y}\big]^2 = \frac{1}{n}\sum_{i=1}^{n} \big[(x_i + a) - (\bar{x} + a)\big]^2$$

The constant $$a$$ cancels inside the bracket:

$$\sigma_{\text{new}}^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar{x})^2 = \sigma^2$$

Since the value of $$a$$ (positive or negative) never appears in the final expression, the variance is unaffected by adding a constant to every observation. Hence the variance remains unchanged.

Here $$n = 100$$, the incorrect mean $$= 40$$ and the incorrect standard deviation $$= 5.1$$.

Step 1 — Correct the mean.

$$\text{Incorrect } \sum x_i = n \times \text{mean} = 100 \times 40 = 4000$$

The wrong value $$50$$ must be removed and the correct value $$40$$ added:

$$\text{Correct } \sum x_i = 4000 - 50 + 40 = 3990$$

$$\text{Correct mean} = \frac{3990}{100} = 39.9$$

Step 2 — Correct the sum of squares.

Using $$\sigma^2 = \dfrac{\sum x_i^2}{n} - (\bar{x})^2$$ with the incorrect data:

$$(5.1)^2 = \frac{\sum x_i^2}{100} - (40)^2$$

$$26.01 = \frac{\sum x_i^2}{100} - 1600 \quad\Rightarrow\quad \text{Incorrect } \sum x_i^2 = 100(1626.01) = 162601$$

Replace the wrong square by the correct one:

$$\text{Correct } \sum x_i^2 = 162601 - (50)^2 + (40)^2 = 162601 - 2500 + 1600 = 161701$$

Step 3 — Correct the standard deviation.

$$\text{Correct } \sigma^2 = \frac{161701}{100} - (39.9)^2 = 1617.01 - 1592.01 = 25$$

$$\text{Correct } \sigma = \sqrt{25} = 5$$

Let the two unknown observations be $$x$$ and $$y$$. The eight observations are $$6, 7, 10, 12, 12, 13, x, y$$.

Step 1 — Use the mean.

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

$$60 + x + y = 72 \quad\Rightarrow\quad x + y = 12 \qquad \text{...(1)}$$

Step 2 — Use the variance.

$$\sigma^2 = \frac{1}{8}\sum (x_i - \bar{x})^2 = 9.25 \quad\Rightarrow\quad \sum (x_i - 9)^2 = 74$$

For the six known observations:

$$(6-9)^2+(7-9)^2+(10-9)^2+(12-9)^2+(12-9)^2+(13-9)^2 = 9+4+1+9+9+16 = 48$$

Hence

$$(x-9)^2 + (y-9)^2 = 74 - 48 = 26 \qquad \text{...(2)}$$

Step 3 — Solve the equations.

Expand (2):

$$x^2 + y^2 - 18(x+y) + 2(81) = 26$$

Using $$x+y = 12$$:

$$x^2 + y^2 = 26 + 18(12) - 162 = 26 + 216 - 162 = 80$$

From $$(x+y)^2 = x^2 + y^2 + 2xy$$:

$$12^2 = 80 + 2xy \quad\Rightarrow\quad 2xy = 144 - 80 = 64 \quad\Rightarrow\quad xy = 32$$

With $$x + y = 12$$ and $$xy = 32$$, the observations are the roots of

$$t^2 - 12t + 32 = 0 \quad\Rightarrow\quad (t-4)(t-8) = 0$$

$$t = 4 \quad\text{or}\quad t = 8$$

Hence the remaining two observations are $$4$$ and $$8$$.

Let the two unknown observations be $$x$$ and $$y$$. The seven observations are $$2, 4, 10, 12, 14, x, y$$.

Step 1 — Use the mean.

$$\bar{x} = \frac{2+4+10+12+14+x+y}{7} = 8$$

$$42 + x + y = 56 \quad\Rightarrow\quad x + y = 14 \qquad \text{...(1)}$$

Step 2 — Use the variance.

$$\sigma^2 = \frac{1}{7}\sum (x_i - \bar{x})^2 = 16 \quad\Rightarrow\quad \sum (x_i - 8)^2 = 112$$

For the five known observations:

$$(2-8)^2+(4-8)^2+(10-8)^2+(12-8)^2+(14-8)^2 = 36+16+4+16+36 = 108$$

Hence

$$(x-8)^2 + (y-8)^2 = 112 - 108 = 4 \qquad \text{...(2)}$$

Step 3 — Solve the equations.

Expand (2):

$$x^2 + y^2 - 16(x+y) + 2(64) = 4$$

Using $$x+y = 14$$:

$$x^2 + y^2 = 4 + 16(14) - 128 = 4 + 224 - 128 = 100$$

From $$(x+y)^2 = x^2 + y^2 + 2xy$$:

$$14^2 = 100 + 2xy \quad\Rightarrow\quad 2xy = 196 - 100 = 96 \quad\Rightarrow\quad xy = 48$$

With $$x + y = 14$$ and $$xy = 48$$, the observations are the roots of

$$t^2 - 14t + 48 = 0 \quad\Rightarrow\quad (t-6)(t-8) = 0$$

$$t = 6 \quad\text{or}\quad t = 8$$

Hence the remaining two observations are $$6$$ and $$8$$.

Let the six observations have mean $$\bar{x} = 8$$ and standard deviation $$\sigma = 4$$.

Step 1 — New mean.

When each observation is multiplied by $$3$$, the new observations are $$3x_1, 3x_2, \ldots, 3x_6$$. Their mean is

$$\bar{x}_{\text{new}} = \frac{1}{6}\sum (3x_i) = 3\cdot\frac{1}{6}\sum x_i = 3\bar{x} = 3\times 8 = 24$$

Step 2 — New standard deviation.

The new variance is

$$\sigma_{\text{new}}^2 = \frac{1}{6}\sum (3x_i - 3\bar{x})^2 = 9\cdot\frac{1}{6}\sum (x_i - \bar{x})^2 = 9\sigma^2$$

Hence

$$\sigma_{\text{new}} = \sqrt{9\sigma^2} = 3\sigma = 3\times 4 = 12$$

In general, multiplying every observation by a constant $$k$$ multiplies the mean by $$k$$ and the standard deviation by $$|k|$$.

Let the $$n$$ observations be $$x_1, x_2, \ldots, x_n$$ with mean $$\bar{x}$$ and variance $$\sigma^2$$, so that

$$\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \qquad\text{and}\qquad \sigma^2 = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar{x})^2$$

Step 1 — Mean of the new observations.

Consider the observations $$ax_1, ax_2, \ldots, ax_n$$, where $$a \neq 0$$. Their mean is

$$\overline{(ax)} = \frac{1}{n}\sum_{i=1}^{n} (ax_i) = a\cdot\frac{1}{n}\sum_{i=1}^{n} x_i = a\bar{x}$$

So the new mean is $$a\bar{x}$$.

Step 2 — Variance of the new observations.

By definition, the variance of $$ax_1, ax_2, \ldots, ax_n$$ is the mean of the squared deviations from their mean $$a\bar{x}$$:

$$\text{Variance} = \frac{1}{n}\sum_{i=1}^{n} (ax_i - a\bar{x})^2$$

Taking $$a$$ common inside the bracket and squaring:

$$= \frac{1}{n}\sum_{i=1}^{n} a^2(x_i - \bar{x})^2 = a^2\cdot\frac{1}{n}\sum_{i=1}^{n} (x_i - \bar{x})^2 = a^2\sigma^2$$

Hence the mean and variance of $$ax_1, ax_2, \ldots, ax_n$$ are $$a\bar{x}$$ and $$a^2\sigma^2$$ respectively.

For the original group: $$n = 100$$, mean $$= 20$$, standard deviation $$= 3$$.

Step 1 — Recover the original sums.

$$\sum x_i = 100 \times 20 = 2000$$

Using $$\sigma^2 = \dfrac{\sum x_i^2}{n} - (\text{mean})^2$$:

$$3^2 = \frac{\sum x_i^2}{100} - 20^2 \quad\Rightarrow\quad \sum x_i^2 = 100(9 + 400) = 40900$$

Step 2 — Omit the three incorrect observations $$21, 21, 18$$.

The number of observations becomes $$100 - 3 = 97$$.

$$\text{New } \sum x_i = 2000 - (21 + 21 + 18) = 2000 - 60 = 1940$$

$$\text{New } \sum x_i^2 = 40900 - (21^2 + 21^2 + 18^2) = 40900 - (441 + 441 + 324) = 40900 - 1206 = 39694$$

Step 3 — Correct mean and standard deviation.

$$\text{Correct mean} = \frac{1940}{97} = 20$$

$$\text{Correct variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2 = \frac{39694}{97} - (20)^2 = 409.22 - 400 = 9.22$$

$$\text{Correct standard deviation} = \sqrt{9.22} \approx 3.04$$