Dispersion

Statistics is the branch of mathematics concerned with the collection, organisation, analysis, and Interpretation of data. In the DSE compulsory syllabus, we focus on descriptive ./4-statistics-and-probability/2_statistics — Summarising a dataset through measures of central tendency and measures of dispersion. This page Also covers grouped data techniques and graphical representations such as box-and-whisker plots. These tools are frequently combined with probability concepts in exam questions.

Measures of Central Tendency

A measure of central tendency identifies a single value that is representative of an entire dataset.

Mean

The mean (arithmetic average) of a dataset $\{x_1, x_2, \ldots, x_n\}$ is defined as:

\begin{aligned} \bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \end{aligned}

The mean uses every data value, making it sensitive to outliers. It is the only measure of central Tendency that lends itself to algebraic manipulation (e.g., combining datasets).

Examples

- The scores of $5$ students are $72, 85, 90, 68, 80$. The mean is $\bar{x} = \frac{72+85+90+68+80}{5} = \frac{395}{5} = 79$. - If every score is increased by $5$ bonus marks, the new mean is $79 + 5 = 84$.

Median

The median is the middle value of an ordered dataset. For $n$ data values sorted in ascending Order:

If $n$ is odd, the median is the value at position $\dfrac{n+1}{2}$ .
If $n$ is even, the median is the average of the values at positions $\dfrac{n}{2}$ and $\dfrac{n}{2}+1$ .

The median is robust to outliers since only on the position of data points, not their Magnitude.

Examples

- Dataset: $\{3, 7, 1, 9, 5\}$. Sorted: $\{1, 3, 5, 7, 9\}$. Median = $5$ (position $3$ of $5$). - Dataset: $\{2, 4, 6, 8, 10, 12\}$. Median = $\frac{6+8}{2} = 7$ (average of positions $3$ and $4$). - Salaries: $\{18000, 20000, 22000, 25000, 150000\}$. Median = $22000$Which is far more representative than the mean of $47000$.

Mode

The mode is the value that occurs most frequently in a dataset. A dataset may be unimodal (one Mode), bimodal (two modes), multimodal, or have no mode at all.

The mode is the only measure of central tendency applicable to nominal (categorical) data.

Examples

- $\{4, 2, 7, 4, 3, 4, 8\}$: mode = $4$ (appears $3$ times). - $\{5, 5, 8, 8, 10\}$: bimodal, modes are $5$ and $8$. - $\{1, 2, 3, 4, 5\}$: no mode.

Comparison of the Three Measures

Measure	Uses all values	Affected by outliers	Applicable to categorical data	Unique value
Mean	Yes	Yes	No	Yes
Median	No	No	No	Yes
Mode	No	No	Yes	No

Measures of Dispersion

Measures of dispersion (spread) quantify how far individual data values deviate from the centre. Two Datasets can share the same mean yet have very different spreads.

Range

\begin{aligned} \mathrm{Range} = \mathrm{Maximum value} - \mathrm{Minimum value} \end{aligned}

The range is simple to compute but uses only two data points, making it highly sensitive to Outliers.

Examples

- $\{12, 15, 18, 22, 25\}$: range $= 25 - 12 = 13$. - $\{5, 10, 10, 10, 10, 100\}$: range $= 95$Heavily distorted by the single outlier.

Interquartile Range (IQR)

The quartiles divide an ordered dataset into four equal parts:

$Q_1$ (lower quartile): the median of the lower half.
$Q_2$ (median): the middle value.
$Q_3$ (upper quartile): the median of the upper half.

\begin{aligned} \mathrm{IQR} = Q_3 - Q_1 \end{aligned}

The IQR is resistant to outliers since it ignores the most extreme $50\%$ of data.

Examples

- Dataset: $\{3, 5, 7, 8, 12, 14, 18, 20, 25\}$ ($n=9$Odd). - Lower half: $\{3, 5, 7, 8\}$, $Q_1 = \frac{5+7}{2} = 6$. - $Q_2 = 12$. - Upper half: $\{14, 18, 20, 25\}$, $Q_3 = \frac{18+20}{2} = 19$. - IQR $= 19 - 6 = 13$.

Variance

Variance measures the average squared deviation from the mean. There are two versions depending on Whether the data represents the entire population or a sample drawn from a larger Population.

Population variance (divides by $n$ ):

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

Sample variance (divides by $n-1$ ):

\begin{aligned} s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2 \end{aligned}

An equivalent computational formula is:

\begin{aligned} \sigma^2 = \frac{1}{n}\left[\sum_{i=1}^{n}x_i^2 - \frac{1}{n}\left(\sum_{i=1}^{n}x_i\right)^2\right] \end{aligned}

Why $n$ vs $n-1$ ? Dividing by $n-1$ (Bessel”s correction) provides an unbiased estimator of The population variance when working with a sample. Using only $n$ data points, the sample mean $\bar{x}$ is closer to the data points than the true population mean $\mu$ So the squared Deviations tend to underestimate the true spread. Dividing by $n-1$ compensates for this. In the DSE Syllabus, unless the problem explicitly identifies the data as a sample, the population formula (dividing by $n$ ) is expected.

Examples

- Dataset: $\{2, 4, 4, 4, 5, 5, 7, 9\}$ ($n=8$). - $\bar{x} = \frac{40}{8} = 5$. - $\sum(x_i - \bar{x})^2 = 9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 = 32$. - Population variance: $\sigma^2 = \frac{32}{8} = 4$. - Sample variance: $s^2 = \frac{32}{7} \approx 4.57$.

Standard Deviation

The standard deviation is the positive square root of the variance, restoring the units to match The original data:

\begin{aligned} \sigma = \sqrt{\sigma^2}, \qquad s = \sqrt{s^2} \end{aligned}

Since the standard deviation is in the same units as the data, it is more interpretable than the Variance for comparing spread.

Examples

- Following the previous example: $\sigma = \sqrt{4} = 2$, $s = \sqrt{\frac{32}{7}} \approx 2.14$. - Two machines produce rods of length $10$ cm. Machine A has $\sigma = 0.1$ cm, Machine B has $\sigma = 0.5$ cm. Machine A is more precise.

Grouped Data

When data is presented in a grouped frequency distribution, individual values are not available. We work with class intervals instead.

Key Definitions

Class boundaries: The endpoints of each class interval, with no gaps between consecutive classes. For example, if raw intervals are $10$ — $19$ and $20$ — $29$ The class boundaries are $9.5$ — $19.5$ and $19.5$ — $29.5$ .
Class width: The difference between the upper and lower class boundaries.
Class mark (midpoint): $x_i = \dfrac{\mathrm{lower boundary} + \mathrm{upper boundary}}{2}$ used as the representative value for all data in the class.

Mean of Grouped Data

\begin{aligned} \bar{x} = \frac{\sum_{i=1}^{k} f_i x_i}{\sum_{i=1}^{k} f_i} \end{aligned}

Where $k$ is the number of classes, $f_i$ is the frequency of class $i$ And $x_i$ is the class Mark.

Assumed Mean Method (Coding Method)

When class marks are equally spaced, let $h$ be the common class width and $A$ be the class mark of A convenient class (the assumed mean). Define $d_i = \dfrac{x_i - A}{h}$ . Then:

\begin{aligned} \bar{x} = A + \frac{\sum_{i=1}^{k} f_i d_i}{\sum_{i=1}^{k} f_i} \times h \end{aligned}

This method simplifies calculation by working with small integer values of $d_i$ .

Examples

- The following frequency distribution records the marks of $40$ students:

Class interval	$f_i$	Class mark $x_i$	$d_i$	$f_i d_i$
30 — 39	4	34.5	$-2$	$-8$
40 — 49	8	44.5	$-1$	$-8$
50 — 59	14	54.5	$0$	$0$
60 — 69	10	64.5	$1$	$10$
70 — 79	4	74.5	$2$	$8$

Here $A = 54.5$ , $h = 10$ .

\begin{aligned} \bar{x} &= 54.5 + \frac{-8-8+0+10+8}{40} \times 10 \\ &= 54.5 + \frac{2}{40} \times 10 \\ &= 54.5 + 0.5 = 55 \end{aligned}

Variance of Grouped Data

For grouped data, the population variance is:

\begin{aligned} \sigma^2 = \frac{\sum_{i=1}^{k} f_i(x_i - \bar{x})^2}{\sum_{i=1}^{k} f_i} \end{aligned}

Or equivalently:

\begin{aligned} \sigma^2 = \frac{1}{n}\left[\sum f_i x_i^2 - \frac{1}{n}\left(\sum f_i x_i\right)^2\right], \quad n = \sum f_i \end{aligned}

Histogram Estimation

In a histogram, the area of each bar represents the frequency of the corresponding class. If class Widths are unequal, the height of each bar is the frequency density:

\begin{aligned} \mathrm{Frequency density} = \frac{\mathrm{Frequency}}{\mathrm{Class width}} \end{aligned}

The median, quartiles, and other percentiles can be estimated from a cumulative frequency curve (ogive) by linear interpolation within the relevant class.

Properties of Variance

Linear Transformation

For a dataset $X$ and constants $a, b$ :

\begin{aligned} \mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X) \end{aligned}

Adding a constant $b$ shifts all values equally and does not affect spread. Multiplying by $a$ Scales the spread by $|a|$ .

For the mean: $\overline{aX+b} = a\bar{x} + b$ .

Examples

- If $\bar{x} = 50$ and $\sigma^2 = 16$Then for $Y = 3X - 4$: $\bar{y} = 3(50)-4 = 146$ and $\mathrm{Var}(Y) = 9 \times 16 = 144$. - Temperatures recorded in Celsius have mean $25$ and standard deviation $3$. In Fahrenheit ($F = 1.8C + 32$): mean $= 1.8(25)+32 = 77$Standard deviation $= 1.8 \times 3 = 5.4$.

Combined Variance

Given two datasets $X$ and $Y$ with sizes $n_1$ and $n_2$ Means $\bar{x}_1$ and $\bar{x}_2$ And Variances $\sigma_1^2$ and $\sigma_2^2$ The combined variance of the pooled dataset is:

\begin{aligned} \sigma_c^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2 + n_1(\bar{x}_1 - \bar{x}_c)^2 + n_2(\bar{x}_2 - \bar{x}_c)^2}{n_1 + n_2} \end{aligned}

Where the combined mean is:

\begin{aligned} \bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2} \end{aligned}

The additional terms $n_1(\bar{x}_1 - \bar{x}_c)^2$ and $n_2(\bar{x}_2 - \bar{x}_c)^2$ account for The between-group variation caused by the difference in means.

Examples

- Group A: $n_1 = 6$$\bar{x}_1 = 10$$\sigma_1^2 = 4$. - Group B: $n_2 = 4$$\bar{x}_2 = 20$$\sigma_2^2 = 9$.

Combined mean: $\bar{x}_c = \frac{6(10)+4(20)}{10} = 14$ .

\begin{aligned} \sigma_c^2 &= \frac{6(4) + 4(9) + 6(10-14)^2 + 4(20-14)^2}{10} \\ &= \frac{24 + 36 + 96 + 144}{10} = \frac{300}{10} = 30 \end{aligned}

Applications

Coefficient of Variation

The coefficient of variation (CV) allows comparison of variability between datasets measured in Different units or with vastly different means:

\begin{aligned} \mathrm{CV} = \frac{\sigma}{\bar{x}} \times 100\% \end{aligned}

A larger CV indicates greater relative dispersion.

Examples

- Investment A: mean return $= 8\%$Standard deviation $= 2\%$. CV $= \frac{2}{8} \times 100\% = 25\%$. - Investment B: mean return $= 15\%$Standard deviation $= 5\%$. CV $= \frac{5}{15} \times 100\% \approx 33.3\%$. - Investment A has lower relative risk.

Box-and-Whisker Plots

A box-and-whisker plot is a standardised graphical display of the five-number summary: minimum, $Q_1$$Q_2$ (median), $Q_3$ And maximum.

Construction:

Draw a rectangular box from $Q_1$ to $Q_3$ .
Draw a line inside the box at $Q_2$ .
Extend “whiskers” to the minimum and maximum values.

Identifying outliers: A data point is considered a potential outlier if it falls below $Q_1 - 1.5 \times \mathrm{IQR}$ or above $Q_3 + 1.5 \times \mathrm{IQR}$ .

Examples

- Dataset: $\{5, 8, 12, 15, 18, 20, 24, 28, 35, 42, 58\}$ ($n=11$). - $Q_2 = 18$. - Lower half: $\{5, 8, 12, 15, 18\}$$Q_1 = 12$. - Upper half: $\{18, 20, 24, 28, 35, 42, 58\}$$Q_3 = 28$. - IQR $= 28 - 12 = 16$. - Lower fence: $12 - 1.5(16) = -12$. - Upper fence: $28 + 1.5(16) = 52$. - Since $58 > 52$The value $58$ is an outlier. The upper whisker extends to $42$ instead.

Skewness (DSE awareness)

While not computed algebraically in the compulsory syllabus, students should recognise:

Positively skewed: mean $>$ median, the right tail is longer.
Negatively skewed: mean $<$ median, the left tail is longer.
Symmetrical: mean $=$ median $=$ mode (for unimodal distributions).

Wrap-up Questions

1. **Question:** The marks of $7$ students are $56, 62, 45, 78, 83, 71, 65$. Find the mean, median, and mode. ### Details

Answer

- Sorted: $\{45, 56, 62, 65, 71, 78, 83\}$. - Mean: $\bar{x} = \frac{460}{7} \approx 65.7$. - Median (position $4$ of $7$): $65$. - Mode: none (all values are distinct).

Question: A dataset has mean $20$ and variance $36$ . Find the mean and variance of the Transformed dataset $Y = \dfrac{X - 20}{6}$ .

Answer

$\bar{y} = \frac{1}{6}(20) - \frac{20}{6} = \frac{20-20}{6} = 0$ .
$\mathrm{Var}(Y) = \left(\frac{1}{6}\right)^2 \times 36 = \frac{1}{36} \times 36 = 1$ .

Question: For the grouped frequency distribution below, find the mean and standard deviation using the coding method.

Class	Frequency
10 — 19	5
20 — 29	12
30 — 39	18
40 — 49	10
50 — 59	5

Answer

- Class marks: $14.5, 24.5, 34.5, 44.5, 54.5$. Let $A = 34.5$$h = 10$. - $d_i$: $-2, -1, 0, 1, 2$. - $\sum f_i = 50$$\sum f_i d_i = 5(-2) + 12(-1) + 18(0) + 10(1) + 5(2) = -10 + (-12) + 0 + 10 + 10 = -2$. - $\bar{x} = 34.5 + \frac{-2}{50} \times 10 = 34.5 - 0.4 = 34.1$. - $\sum f_i d_i^2 = 5(4) + 12(1) + 18(0) + 10(1) + 5(4) = 20 + 12 + 0 + 10 + 20 = 62$. - $\sigma_d^2 = \frac{62}{50} - \left(\frac{-2}{50}\right)^2 = 1.24 - 0.0016 = 1.2384$. - $\sigma^2 = 1.2384 \times 10^2 = 123.84$. So $\sigma = \sqrt{123.84} \approx 11.13$.

Question: Two classes sat the same test. Class A ( $n_1 = 30$$\bar{x}_1 = 72$ $\sigma_1 = 8$ ). Class B ( $n_2 = 20$$\bar{x}_2 = 80$$\sigma_2 = 6$ ). Find the combined mean and Combined standard deviation.

Answer

Combined mean: $\bar{x}_c = \frac{30(72)+20(80)}{50} = \frac{2160+1600}{50} = \frac{3760}{50} = 75.2$ .
Combined variance: $,$ \begin{aligned} \sigma_c^2 &= \frac{30(64) + 20(36) + 30(72-75.2)^2 + 20(80-75.2)^2}{50} \ &= \frac{1920 + 720 + 30(10.24) + 20(23.04)}{50} \ &= \frac{1920 + 720 + 307.2 + 460.8}{50} \ &= \frac{3408}{50} = 68.16 \end{aligned} $,$
Combined standard deviation: $\sigma_c = \sqrt{68.16} \approx 8.26$ .

Question: The following are the lifetimes (in hours) of $10$ light bulbs: $820, 790, 810, 780, 830, 800, 795, 815, 805, 855$ . Determine the range, IQR, and identify any Outliers.

Answer

Sorted: $\{780, 790, 795, 800, 805, 810, 815, 820, 830, 855\}$ .
Range $= 855 - 780 = 75$ .
$Q_2 = \frac{805+810}{2} = 807.5$ .
Lower half: $\{780, 790, 795, 800, 805\}$$Q_1 = 795$ .
Upper half: $\{810, 815, 820, 830, 855\}$$Q_3 = 820$ .
IQR $= 820 - 795 = 25$ .
Lower fence: $795 - 1.5(25) = 757.5$ . Upper fence: $820 + 1.5(25) = 857.5$ .
No outliers (all values lie within $[757.5, 857.5]$ ).

Question: A farmer records the yields (in kg) of two varieties of wheat over several seasons. Variety A: mean $= 45$ Standard deviation $= 5$ . Variety B: mean $= 60$ Standard deviation $= 9$ . Which variety has more consistent yield?

Answer

CV $_A = \frac{5}{45} \times 100\% \approx 11.1\%$ .
CV $_B = \frac{9}{60} \times 100\% = 15.0\%$ .
Since CV $_A <$ CV $_B$ Variety A has more consistent (less variable) yield relative to its mean.

Question: Given the dataset $\{a, b, c\}$ with mean $10$ and variance $8$ Find the value of $a^2 + b^2 + c^2$ .

Answer

$\bar{x} = \frac{a+b+c}{3} = 10 \implies a+b+c = 30$ .
$\sigma^2 = \frac{a^2+b^2+c^2}{3} - \bar{x}^2 = 8$ .
$\frac{a^2+b^2+c^2}{3} - 100 = 8 \implies a^2+b^2+c^2 = 324$ .

Question: A set of $20$ numbers has mean $15$ and standard deviation $3$ . If each number is Multiplied by $2$ and then $5$ is added, find the new mean and new standard deviation.

Answer

New mean: $2(15) + 5 = 35$ .
New variance: $2^2 \times 3^2 = 36$ .
New standard deviation: $\sqrt{36} = 6$ .

Question: The histogram below (described verbally) shows the distribution of weights of $50$ apples. The class intervals and frequencies are:

Weight (g)	Frequency
100 — 119	6
120 — 139	14
140 — 159	20
160 — 179	8
180 — 199	2

Estimate the median weight from the cumulative frequency distribution.

Answer

Cumulative frequencies: $6, 20, 40, 48, 50$ .
The median is the $\frac{50}{2} = 25$ Th value, which lies in the class $140$ — $159$ (cumulative $20$ to $40$ ).
Using linear interpolation within the class: $,$ \begin{aligned} \mathrm{Median} &= 139.5 + \frac{25-20}{40-20} \times (159.5 - 139.5) \ &= 139.5 + \frac{5}{20} \times 20 \ &= 139.5 + 5 = 144.5 \mathrm{ g} \end{aligned} $,$

Question: For the dataset $\{3, 7, 7, 2, 9, 5, 1, 8, 6, 4\}$ Find $\sum x_i$$\sum x_i^2$ The mean, and the population variance. Verify your variance using both the definition formula and The computational formula.

Answer

$\sum x_i = 3+7+7+2+9+5+1+8+6+4 = 52$ .
$\sum x_i^2 = 9+49+49+4+81+25+1+64+36+16 = 334$ .
$\bar{x} = \frac{52}{10} = 5.2$ .
Definition formula: $\begin{aligned} \sigma^2 &= \frac{(3-5.2)^2 + (7-5.2)^2 + (7-5.2)^2 + (2-5.2)^2 + (9-5.2)^2 + (5-5.2)^2 + (1-5.2)^2 + (8-5.2)^2 + (6-5.2)^2 + (4-5.2)^2}{10} \\ &= \frac{4.84+3.24+3.24+10.24+14.44+0.04+17.64+7.84+0.64+1.44}{10} \\ &= \frac{63.6}{10} = 6.36 \end{aligned}$
Computational formula: $\begin{aligned} \sigma^2 &= \frac{334}{10} - \left(\frac{52}{10}\right)^2 = 33.4 - 27.04 = 6.36 \quad \checkmark \end{aligned}$

Question: The weekly wages (in dollars) of $8$ workers in a small factory are $3200, 3500, 3800, 4200, 4500, 4800, 5200, 12000$ . The factory owner claims the average wage is USD 5150. Is this claim misleading? Explain using an appropriate measure of central tendency and Dispersion.

Answer

Mean: $\bar{x} = \frac{41200}{8} = 5150$ . The owner’s figure is arithmetically correct.
Sorted: $\{3200, 3500, 3800, 4200, 4500, 4800, 5200, 12000\}$ .
Median: $\frac{4200+4500}{2} = 4350$ .
The median ( $4350$ ) is a far more representative measure here. The single extreme value of USD 12000 (likely the owner’s own salary or a manager’s) inflates the mean by USD 800. The median is resistant to outliers and better reflects what a typical worker earns.
The range ( $12000 - 3200 = 8800$ ) and the large gap between the mean and median both indicate Significant skewness, confirming the mean is a poor choice of summary statistic.

Question: A set of data has variance $25$ and mean $0$ . A new set is formed by removing the Value $10$ from the original set. If the original set had $n = 6$ values, find the new mean and new Variance.

Answer

Original: $\bar{x} = 0$$\sigma^2 = 25$$n = 6$ .
$\sum x_i = 0$ So $\sum x_i^2 = n\sigma^2 + \frac{(\sum x_i)^2}{n} = 6(25) + 0 = 150$ .
After removing $10$ : new sum $= 0 - 10 = -10$ New $n' = 5$ .
New mean: $\bar{x}' = \frac{-10}{5} = -2$ .
New sum of squares: $150 - 100 = 50$ .
New variance: $\sigma'^2 = \frac{50}{5} - (-2)^2 = 10 - 4 = 6$ .

For the A-Level treatment of this topic, see Data Representation.

:::tip Diagnostic Test Ready to test your understanding of Dispersion? The contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Dispersion with other DSE mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

DSE Exam Technique

Showing Working

For ./4-statistics-and-probability/2_statistics problems in DSE Paper 1:

When computing the mean, show the sum divided by $n$ before giving the decimal.
When computing variance, use the computational formula $\sigma^2 = \dfrac{\sum x_i^2}{n} - \bar{x}^2$ and show both terms.
For grouped data, show the class marks and the coding method in a table.
For the coding method, state the assumed mean $A$ and class width $h$ .
For box plots, label all five values (min, $Q_1$$Q_2$$Q_3$ Max).

Significant Figures

The DSE requires answers to be given to 3 significant figures unless the question specifies otherwise. Exact fractions are preferred when they arise .

Common DSE Question Types

Combined mean and variance of two groups.
Grouped data mean and standard deviation using the coding method.
Effect of linear transformations on mean and variance.
Box-and-whisker plots with outlier identification.
Coefficient of variation for comparing relative dispersion.

Additional Worked Examples

Worked Example 13: Effect of adding a data point

A dataset $\{2, 5, 8, 11, 14\}$ has mean $\bar{x} = 8$ and variance $\sigma^2 = 20$ . Find the new mean and variance if the value $20$ is added.

Solution

New $n' = 6$ . New sum $= 40 + 20 = 60$ . New mean $\bar{x}' = \dfrac{60}{6} = 10$ .

$\sum x_i^2 = 4 + 25 + 64 + 121 + 196 = 410$ . New $\sum x_i^2 = 410 + 400 = 810$ .

New variance: $\sigma'^2 = \dfrac{810}{6} - 10^2 = 135 - 100 = 35$ .

Worked Example 14: Standardised scores

In an exam, the mean is 60 and the standard deviation is 10. Student A scores 75 and Student B scores 55. Express each score as a standardised score (z-score).

Solution

$z_A = \frac{75 - 60}{10} = 1.5$

$z_B = \frac{55 - 60}{10} = -0.5$

Student A scored 1.5 standard deviations above the mean; Student B scored 0.5 standard deviations below.

Worked Example 15: Finding data from summary ./4-statistics-and-probability/2_statistics

A dataset of 5 positive integers has mean 6 and variance 4. Find all possible datasets.

Solution

$\sum x_i = 30$ and $\dfrac{\sum x_i^2}{5} - 36 = 4 \implies \sum x_i^2 = 200$ .

We need five positive integers summing to 30 with squares summing to 200.

If the data is symmetric around 6: try $\{4, 5, 6, 7, 8\}$ .

Sum $= 30$ . $\sum x_i^2 = 16 + 25 + 36 + 49 + 64 = 190 \neq 200$ .

Try $\{2, 6, 6, 6, 10\}$ : sum $= 30$$\sum x_i^2 = 4 + 36 + 36 + 36 + 100 = 212 \neq 200$ .

Try $\{4, 4, 6, 8, 8\}$ : sum $= 30$$\sum x_i^2 = 16 + 16 + 36 + 64 + 64 = 196 \neq 200$ .

Try $\{3, 5, 7, 7, 8\}$ : sum $= 30$$\sum x_i^2 = 9 + 25 + 49 + 49 + 64 = 196 \neq 200$ .

Try $\{4, 4, 8, 6, 8\}$ : sum $= 30$$\sum x_i^2 = 16 + 16 + 64 + 36 + 64 = 196$ .

Try $\{2, 6, 6, 8, 8\}$ : sum $= 30$$\sum x_i^2 = 4 + 36 + 36 + 64 + 64 = 204$ .

Try $\{4, 6, 6, 6, 8\}$ : sum $= 30$$\sum x_i^2 = 16 + 36 + 36 + 36 + 64 = 188$ .

There may be no solution with 5 positive integers. Let me try $\{2, 5, 7, 7, 9\}$ : sum $= 30$$\sum x_i^2 = 4 + 25 + 49 + 49 + 81 = 208$ .

$\{3, 5, 6, 8, 8\}$ : sum $= 30$$\sum x_i^2 = 9 + 25 + 36 + 64 + 64 = 198$ .

$\{4, 5, 6, 7, 8\}$ : $\sum x_i^2 = 190$ . Need 200. The deficit is 10. If we change 5 to 6 and 6 to 5: $\{4, 6, 5, 7, 8\}$ : same sum of squares.

If we change 4 to 5 and 8 to 7: $\{5, 5, 6, 7, 7\}$ : $\sum x_i^2 = 25 + 25 + 36 + 49 + 49 = 184$ .

The minimum $\sum x_i^2$ for sum 30 with 5 positive integers is achieved by values closest to 6.

The constraints may not be satisfiable with integers. In an exam, this would be solved numerically.

Worked Example 16: Grouped data variance with coding

For the frequency distribution below, find the standard deviation using the assumed mean method.

Class	Frequency
5 — 9	3
10 — 14	7
15 — 19	12
20 — 24	5
25 — 29	3

Solution

Class marks: $7$$12$$17$$22$$27$ . $A = 17$$h = 5$ .

$d_i$ : $-2$$-1$$0$$1$$2$ .

$f_i$	$d_i$	$f_id_i$	$f_id_i^2$
3	$-2$	$-6$	12
7	$-1$	$-7$	7
12	0	0	0
5	1	5	5
3	2	6	12

$n = 30$$\sum f_id_i = -2$$\sum f_id_i^2 = 36$ .

$\bar{d} = \dfrac{-2}{30} = -\dfrac{1}{15}$ .

$\sigma_d^2 = \dfrac{36}{30} - \left(-\dfrac{1}{15}\right)^2 = 1.2 - \dfrac{1}{225} = \dfrac{269}{225}$ .

$\sigma^2 = \dfrac{269}{225} \times 25 = \dfrac{269}{9} \approx 29.89$ .

$\sigma = \sqrt{\dfrac{269}{9}} = \dfrac{\sqrt{269}}{3} \approx 5.47$ .

DSE Exam-Style Questions

DSE Practice 1. Two groups of students took the same test. Group A: $n_1 = 40$$\bar{x}_1 = 65$$\sigma_1 = 8$ . Group B: $n_2 = 60$$\bar{x}_2 = 72$$\sigma_2 = 10$ . Find the overall mean and standard deviation.

Solution

Combined mean: $\bar{x}_c = \dfrac{40(65) + 60(72)}{100} = \dfrac{2600 + 4320}{100} = 69.2$ .

Combined variance:

$\sigma_c^2 = \frac{40(64) + 60(100) + 40(65 - 69.2)^2 + 60(72 - 69.2)^2}{100}$

$= \frac{2560 + 6000 + 40(17.64) + 60(7.84)}{100}$

$= \frac{8560 + 705.6 + 470.4}{100} = \frac{9736}{100} = 97.36$

$\sigma_c = \sqrt{97.36} \approx 9.87$

DSE Practice 2. The heights (in cm) of 8 students are: 158, 162, 165, 168, 170, 172, 175, 180. After converting to feet (1 cm = 0.03281 ft), find the mean and standard deviation in feet.

Solution

Let $Y = 0.03281X$ . Then $\bar{y} = 0.03281\bar{x}$ and $\sigma_Y = 0.03281\sigma_X$ .

$\bar{x} = \dfrac{158 + 162 + 165 + 168 + 170 + 172 + 175 + 180}{8} = \dfrac{1350}{8} = 168.75$ cm.

$\bar{y} = 0.03281 \times 168.75 \approx 5.537$ ft.

$\sigma_X = \sqrt{\dfrac{\sum x_i^2}{8} - 168.75^2}$ .

$\sum x_i^2 = 24964 + 26244 + 27225 + 28224 + 28900 + 29584 + 30625 + 32400 = 228166$ .

$\sigma_X^2 = \dfrac{228166}{8} - 28476.5625 = 28520.75 - 28476.5625 = 44.1875$ .

$\sigma_X = \sqrt{44.1875} \approx 6.647$ cm.

$\sigma_Y = 0.03281 \times 6.647 \approx 0.2181$ ft.

DSE Practice 3. For the dataset $\{1, 3, 5, 7, 9, 11, 13\}$ Find the mean deviation (mean absolute deviation) and compare it with the standard deviation.

Solution

$\bar{x} = \dfrac{49}{7} = 7$ .

Mean deviation $= \dfrac{|1-7| + |3-7| + |5-7| + |7-7| + |9-7| + |11-7| + |13-7|}{7} = \dfrac{6 + 4 + 2 + 0 + 2 + 4 + 6}{7} = \dfrac{24}{7} \approx 3.43$ .

$\sigma^2 = \dfrac{1 + 9 + 25 + 49 + 81 + 121 + 169}{7} - 49 = \dfrac{455}{7} - 49 = 65 - 49 = 16$ .

$\sigma = 4$ .

The standard deviation (4) is greater than the mean deviation (3.43), which is always the case for datasets that are not constant.

DSE Practice 4. A set of data has $\bar{x} = 50$ and $\sigma = 4$ . If every value is increased by $k$ The new standard deviation becomes 10. Find $k$ and explain your answer.

Solution

Adding a constant $k$ does not change the standard deviation. Therefore, the new standard deviation should still be $\sigma = 4$ Not $10$ .

There is no value of $k$ that changes the standard deviation from 4 to 10 by addition alone. To change the standard deviation, we would need to multiply by a constant. If $Y = aX + b$ Then $\sigma_Y = |a|\sigma_X$ . For $\sigma_Y = 10$ : $|a| = \dfrac{10}{4} = 2.5$ .

The question likely intends a multiplication, not just addition. If $Y = 2.5X + k$ Then $\sigma_Y = 10$ for any $k$ .

DSE Practice 5. The table shows the distribution of marks in a test.

Marks	Frequency
0 — 19	4
20 — 39	10
40 — 59	22
60 — 79	14
80 — 100	5

Estimate the mean and standard deviation.

Solution

Class marks: $9.5$ , $29.5$ , $49.5$ , $69.5$ , $89.5$ . Class widths: 20, 20, 20, 20, 21.

For the coding method with equal class widths (using width 20): $A = 49.5$ , $h = 20$ .

$d_i$ : $-2$ , $-1$ , $0$ , $1$ , $2$ (approximately; the last class has width 21).

Using approximate equal widths:

$f_i$	$x_i$	$d_i$	$f_id_i$	$f_id_i^2$
4	9.5	$-2$	$-8$	16
10	29.5	$-1$	$-10$	10
22	49.5	0	0	0
14	69.5	1	14	14
5	89.5	2	10	20

$n = 55$$\sum f_id_i = 6$$\sum f_id_i^2 = 60$ .

$\bar{x} = 49.5 + \dfrac{6}{55} \times 20 = 49.5 + 2.18 = 51.68 \approx 51.7$ .

$\sigma_d^2 = \dfrac{60}{55} - \left(\dfrac{6}{55}\right)^2 = 1.0909 - 0.0119 = 1.079$ .

$\sigma^2 = 1.079 \times 400 = 431.6$ . $\sigma = \sqrt{431.6} \approx 20.8$ .

Common Pitfalls

Confusing population and sample standard deviation. The population SD divides by $N$ ; the sample SD divides by $n - 1$ (Bessel’s correction). Fix: Check the question wording: “population” or “estimate from a sample.”
Forgetting that standard deviation is always non-negative. $\sigma \geq 0$ , with $\sigma = 0$ only when all values are identical. Fix: If your calculation gives a negative value, check for arithmetic errors.
Wrong formula for grouped data variance. Use $\sigma^2 = \frac{\sum f x^2}{\sum f} - \bar{x}^2$ for grouped data with class midpoints. Fix: Use midpoints $x$ and frequencies $f$ ; do not use class boundaries directly.

Summary

Mean $\bar{x} = \frac{\sum x_i}{n}$ ; variance $\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n}$ ; standard deviation $\sigma = \sqrt{\sigma^2}$ .
Computational formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ .
For grouped data: use class midpoints and frequencies.
Sample variance uses $n - 1$ in the denominator (Bessel’s correction).

Worked Examples

Example 1: Calculating mean and standard deviation

Problem. Find the mean and standard deviation of: $4, 7, 8, 10, 11$ .

Solution. $\bar{x} = \frac{4+7+8+10+11}{5} = \frac{40}{5} = 8$ .

$\sigma^2 = \frac{(4-8)^2 + (7-8)^2 + (8-8)^2 + (10-8)^2 + (11-8)^2}{5} = \frac{16 + 1 + 0 + 4 + 9}{5} = \frac{30}{5} = 6$

$\sigma = \sqrt{6} \approx 2.45$

$\blacksquare$

Example 2: Grouped data

Problem. Find the mean and variance of the following grouped data:

Class	Frequency
0–10	3
10–20	7
20–30	5

Solution. Midpoints: $x = 5, 15, 25$ . Frequencies: $f = 3, 7, 5$ . Total: $n = 15$ .

$\bar{x} = \frac{3(5) + 7(15) + 5(25)}{15} = \frac{15 + 105 + 125}{15} = \frac{245}{15} = \frac{49}{3}$ .

$\sigma^2 = \frac{\sum fx^2}{n} - \bar{x}^2 = \frac{3(25) + 7(225) + 5(625)}{15} - \left(\frac{49}{3}\right)^2 = \frac{4655}{15} - \frac{2401}{9} = \frac{931}{3} - \frac{2401}{9} = \frac{2793 - 2401}{9} = \frac{392}{9} pprox 43.6$ .

$\blacksquare$

Cross-References

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