Measures of dispersion

Unit 12 Measures of dispersion

Key unit competence: Evaluate measures of dispersion for real-life data to formulate logical conclusions.

Key ideas

• Measures of central tendency (mean, median, mode) show the centre; measures of dispersion show the spread.
• Variance of $x_1,\dots ,x_n$: ${\sigma }^2=\frac{1}{n}\sum (x_i-\bar{x}{)}^2=\frac{1}{n}\sum x_i^2-{\bar{x}}^2$. With frequencies: ${\sigma }^2=\frac{\sum f(x-\bar{x}{)}^2}{\sum f}=\frac{\sum f{x}^2}{\sum f}-{\bar{x}}^2$.
• Standard deviation $\sigma =\sqrt{\text{variance}}$ (same units as the data).
• Combined sets: total mean $\bar{x}=\frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}$; total variance from $\sum x^2=n({\sigma }^2+{\bar{x}}^2)$ for each set.
• Coefficient of variation $CV=\frac{\sigma }{\bar{x}}\times 100\%$: compares the relative spread of sets with different means or units. The smaller CV is more consistent.
• Effects: adding $c$ to every value does not change $\sigma$; multiplying every value by $k$ multiplies $\sigma$ by $|k|$ and the variance by $k^2$.

Formulas to know

Common misconceptions

• "Standard deviation and variance have the same units." No: the variance is in squared units.
• "Adding 5 to every mark increases the standard deviation by 5." No: the spread is unchanged.
• "A larger standard deviation always means less consistent." Compare CVs when the means differ.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find the mean and standard deviation of 3, 5, 7, 9, 11.

   Show answer

   $\bar{x}=7$; deviations $-4,-2,0,2,4$; ${\sigma }^2=\frac{40}{5}=8$; $\sigma =2\sqrt{2}\approx 2.83$.

   Common mistake: Dividing by $n-1$ instead of $n$ (the book uses $n$).

2. Which data set is more spread out: A = {10, 20, 30} or B = {19, 20, 21}? Justify with the standard deviation.

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   Both have mean 20. ${\sigma }_A=\sqrt{\frac{200}{3}}\approx 8.16$; ${\sigma }_B=\sqrt{\frac{2}{3}}\approx 0.82$. A is far more spread out.

   Common mistake: Concluding they are equal because the means are equal.

3. Calculate the variance from the frequency table: $x$: 1, 2, 3, 4; $f$: 3, 5, 8, 4.

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   $\sum f=20$, $\sum fx=3+10+24+16=53$, $\bar{x}=2.65$; $\sum f{x}^2=3+20+72+64=159$; ${\sigma }^2=7.95-7.0225=0.9275$; $\sigma \approx 0.96$.

   Common mistake: Using $\sum x^2$ without the frequencies.

4. Show that $\frac{1}{n}\sum (x-\bar{x}{)}^2=\frac{1}{n}\sum x^2-{\bar{x}}^2$.

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   Expand: $\frac{1}{n}\sum (x^2-2x\bar{x}+{\bar{x}}^2)=\frac{1}{n}\sum x^2-2\bar{x}\cdot \bar{x}+{\bar{x}}^2=\frac{1}{n}\sum x^2-{\bar{x}}^2$.

   Common mistake: Forgetting that $\frac{1}{n}\sum x=\bar{x}$.

5. Two machines fill 500 g bags. Machine A: mean 498 g, σ 4 g. Machine B: mean 503 g, σ 2 g. Comment.

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   B is more consistent ($CV$ about 0.4% vs 0.8%); A is closer to 500 g on average but more variable.

   Common mistake: Looking only at the means.

6. Compute the coefficient of variation for monthly incomes with mean 250 000 FRW and σ 50 000 FRW, and interpret it.

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   $CV=20\%$: incomes typically vary by about a fifth of the mean.

   Common mistake: Multiplying instead of dividing.

7. Explain the effect on the mean and standard deviation of adding 5 to every mark and then doubling.

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   Mean: $2(\bar{x}+5)=2\bar{x}+10$. Standard deviation: $2\sigma$ (the shift has no effect).

   Common mistake: Adding 10 to the standard deviation.

8. Group A (15 students): mean 12, σ 2. Group B (10 students): mean 17, σ 3. Find the combined mean and standard deviation.

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   Mean $=\frac{180+170}{25}=14$. $\sum x^2$: A $15(4+144)=2220$, B $10(9+289)=2980$; total 5200; ${\sigma }^2=208-196=12$; $\sigma \approx 3.46$.

   Common mistake: Averaging the two standard deviations.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the variance of $2,4,6,8$.	a. $\sigma =12$, variance $144$
   2. Midpoints 10, 20, 30 have frequencies 1, 2, 1. Find the standard deviation.	b. $5$
   3. A data set has $\sigma =4$. Every value is multiplied by 3. Find the new standard deviation and variance.	c. $\approx 7.07$
   4. The values $5,7,x$ have mean 7. Find $x$ and the variance.	d. $x=9$, variance $\frac{8}{3}$

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   1 → b ($5$) · 2 → c ($\approx 7.07$) · 3 → a ($\sigma =12$, variance $144$) · 4 → d ($x=9$, variance $\frac{8}{3}$)

   • Step 1. Mean $=\frac{20}{4}=5$.
     Step 2. Squared deviations: $9,1,1,9$, sum 20.
     Step 3. ${\sigma }^2=\frac{20}{4}=5$.
   • Step 1. Mean $=\frac{10+40+30}{4}=20$.
     Step 2. ${\sigma }^2=\frac{1(100)+2(0)+1(100)}{4}=50$.
     Step 3. $\sigma =\sqrt{50}\approx 7.07$.
   • Step 1. $\sigma$ is multiplied by 3: $12$.
     Step 2. The variance is multiplied by $3^2$: $16\times 9=144$.
   • Step 1. $5+7+x=21$, so $x=9$.
     Step 2. Squared deviations $4,0,4$: ${\sigma }^2=\frac{8}{3}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $5$  ·  $2$  ·  $2\sqrt{2}\approx 2.83$  ·  $4$
   1. Four values have variance 5. A fifth value equal to their mean is added. The new variance is .
   2. A mark of 70 is in a class with mean 60 and $\sigma =5$. Its standard score is .
   3. The marks 4, 8, 6, 2, 10 have standard deviation .
   Show answer

   1: $4$ · 2: $2$ · 3: $2\sqrt{2}\approx 2.83$

   • Step 1. Sum of squared deviations $=4\times 5=20$; the new value adds 0.
     Step 2. New variance $=\frac{20}{5}=4$.
   • Step 1. $z=\frac{x-\bar{x}}{\sigma }$.
     Step 2. $\frac{70-60}{5}=2$.
   • Step 1. Mean $=\frac{30}{5}=6$.
     Step 2. Squared deviations $4,4,0,16,16$, sum 40; ${\sigma }^2=8$.
     Step 3. $\sigma =\sqrt{8}=2\sqrt{2}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Student A scored 70 (class mean 60, $\sigma =5$); student B scored 80 (class mean 75, $\sigma =10$). Who did relatively better?	a. $0.61$
   2. Group 1: 10 values, mean 6. Group 2: 15 values, mean 8. Find the combined mean.	b. $7.2$
   3. Values 1, 2, 3 have frequencies 2, 3, 5. Find the variance.	c. A ($z=2$ against $z=0.5$)
   4. $x$ has mean 5 and $\sigma =2$. Find the mean and standard deviation of $y=2x-3$.	d. Mean 7, $\sigma =4$

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   1 → c (A ($z=2$ against $z=0.5$)) · 2 → b ($7.2$) · 3 → a ($0.61$) · 4 → d (Mean 7, $\sigma =4$)

   • Step 1. $z_A=\frac{10}{5}=2$; $z_B=\frac{5}{10}=0.5$.
     Step 2. A is further above the class mean in standard deviations.
   • Step 1. Totals: $10\times 6=60$ and $15\times 8=120$.
     Step 2. $\frac{60+120}{25}=7.2$.
   • Step 1. $\sum f=10$, $\sum fx=2+6+15=23$, mean $2.3$.
     Step 2. $\sum f{x}^2=2+12+45=59$.
     Step 3. ${\sigma }^2=\frac{59}{10}-2.3^2=5.9-5.29=0.61$.
   • Step 1. $\bar{y}=2(5)-3=7$.
     Step 2. ${\sigma }_y=|2|\times 2=4$ (the $-3$ does not change spread).

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $8$  ·  $4$  ·  $2$  ·  $3$
   1. A data set has $\sigma =4$. Every value is increased by 10. The new standard deviation is .
   2. Find the interquartile range of $3,5,7,9,11,13,15,17$. →
   3. Find the variance of $1,2,3,4,5$. →
   Show answer

   1: $4$ · 2: $8$ · 3: $2$

   • Step 1. Adding a constant shifts all values together.
     Step 2. The distances from the mean do not change: $\sigma$ stays 4.
   • Step 1. Lower half $3,5,7,9$: $Q_1=\frac{5+7}{2}=6$.
     Step 2. Upper half $11,13,15,17$: $Q_3=\frac{13+15}{2}=14$.
     Step 3. $IQR=14-6=8$.
   • Step 1. Mean 3; squared deviations $4,1,0,1,4$, sum 10.
     Step 2. ${\sigma }^2=\frac{10}{5}=2$.

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