Measures of Central Tendency and Dispersion | IB Math AA SL

SL 4.3 · Core Foundations

1. Measures of Central Tendency and Skewness

Measures of central tendency indicate the center or typical value of a dataset. The choice of measure depends on data distribution symmetry and presence of outliers.

Mean ($\bar{x}$): The arithmetic average $\bar{x} = \frac{\sum x_i}{n} = \frac{\sum f_i x_i}{\sum f_i}$. Incorporates every data point, but is sensitive to extreme values (outliers).
Median: The middle value of an ordered dataset ($50^{\text{th}}$ percentile). Highly resistant to outliers and skewed tails.
Mode / Modal Class: The most frequently occurring value or interval. Useful for categorical data or bimodal distributions.
Skewness and the Mean-Median Relationship:
- Symmetric: $\text{Mean} \approx \text{Median} \approx \text{Mode}$
- Positive Skew (Right-skewed): The long right tail pulls the mean upward: $\text{Mode} < \text{Median} < \text{Mean}$
- Negative Skew (Left-skewed): The long left tail pulls the mean downward: $\text{Mean} < \text{Median} < \text{Mode}$

Worked Example 1.1 · Grouped Mean Calculation

The scores of 50 students are grouped into classes as shown: $0 \le x < 10$ (freq 6), $10 \le x < 20$ (freq 14), $20 \le x < 30$ (freq 20), $30 \le x < 40$ (freq 10). Estimate the mean score.

Find class midpoints $x_i$: $5, 15, 25, 35$.
Calculate products $f_i x_i$: $$\sum f_i x_i = 6(5) + 14(15) + 20(25) + 10(35) = 30 + 210 + 500 + 350 = 1090$$
Divide by total frequency $n = 50$: $$\bar{x} = \frac{1090}{50} = 21.8$$

Answer: Estimated mean score $= 21.8$.

SL 4.3 · Dispersion

2. Measures of Dispersion: Spread and Variability

Measures of dispersion quantify the degree of variation or spread of the data around the central value.

Range: $\text{Range} = x_{\max} - x_{\min}$. Simple to find, but completely determined by the two most extreme values.
Interquartile Range ($\text{IQR}$): $\text{IQR} = Q_3 - Q_1$. Measures the spread of the middle $50\%$ of the distribution; completely resistant to outliers.
Population Variance ($\sigma^2$) and Standard Deviation ($\sigma$): $$\sigma^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n} = \frac{\sum_{i=1}^n x_i^2}{n} - \bar{x}^2$$ $$\sigma = \sqrt{\frac{\sum_{i=1}^n x_i^2}{n} - \bar{x}^2}$$ For frequency tables: $\sigma^2 = \frac{\sum f_i x_i^2}{n} - \bar{x}^2$.
IB GDC Note: In IB examinations, use the population standard deviation ($\sigma_x$ on your GDC) when the question specifies standard deviation of a given dataset, or sample standard deviation ($s_x$) if specifically requested as an unbiased sample estimate.

Worked Example 1.2 · Variance from Summary Statistics

A sample of $n = 25$ measurements yields $\sum x = 450$ and $\sum x^2 = 8520$. Calculate the mean and the standard deviation.

Calculate the mean: $$\bar{x} = \frac{\sum x}{n} = \frac{450}{25} = 18$$
Calculate the variance using $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$: $$\sigma^2 = \frac{8520}{25} - 18^2 = 340.8 - 324 = 16.8$$
Calculate standard deviation: $$\sigma = \sqrt{16.8} \approx 4.0988 \approx 4.10\text{ (3 s.f.)}$$

Answer: $\bar{x} = 18$; $\sigma \approx 4.10$.

SL 4.3 · Transformations

3. Effect of Linear Transformations: $y = ax + b$

A fundamental IB concept is understanding how scaling ($a$) and translating ($b$) affect central tendency versus dispersion.

Adding a Constant $b$ (Translation / Shift):
- Measures of location shift by $b$: $\bar{y} = \bar{x} + b$, $\text{Median}_y = \text{Median}_x + b$, $\text{Mode}_y = \text{Mode}_x + b$.
- Measures of spread are COMPLETELY UNCHANGED: $\sigma_y = \sigma_x$, $\text{Var}_y = \text{Var}_x$, $\text{IQR}_y = \text{IQR}_x$, $\text{Range}_y = \text{Range}_x$.
Multiplying by a Scale Factor $a$ (Dilation / Scaling):
- Measures of location scale by $a$: $\bar{y} = a\bar{x}$, $\text{Median}_y = a\text{Median}_x$.
- Measures of spread scale by $|a|$: $\sigma_y = |a|\sigma_x$, $\text{IQR}_y = |a|\text{IQR}_x$, $\text{Range}_y = |a|\text{Range}_x$.
- Variance scales by $a^2$: $\sigma_y^2 = a^2 \sigma_x^2$.
Combined Transformation $y = ax + b$: $$\bar{y} = a\bar{x} + b, \quad \sigma_y = |a|\sigma_x, \quad \sigma_y^2 = a^2\sigma_x^2, \quad \text{IQR}_y = |a|\text{IQR}_x$$

Figure 4.2: Skewness ordering of central tendency and the differential impact of linear transformations on center versus spread.

Worked Example 1.3 · Linear Data Coding Transformations

A chemistry student measures temperatures $T$ in $^\circ\text{C}$ with mean $\bar{T} = 24.5^\circ\text{C}$ and standard deviation $\sigma_T = 3.2^\circ\text{C}$. The measurements are converted to degrees Fahrenheit using $F = 1.8T + 32$.
(a) Find the mean temperature in $^\circ\text{F}$. [2 marks]
(b) Find the standard deviation of the temperatures in $^\circ\text{F}$. [2 marks]
(c) Find the variance of the temperatures in $^\circ\text{F}$. [2 marks]

Part (a): Transformed mean $\bar{F} = 1.8\bar{T} + 32$: $$\bar{F} = 1.8(24.5) + 32 = 44.1 + 32 = 76.1^\circ\text{F} \quad [M1][A1]$$
Part (b): Transformed standard deviation $\sigma_F = |a|\sigma_T$: $$\sigma_F = 1.8 \times 3.2 = 5.76^\circ\text{F} \quad [M1][A1]$$ Notice: Adding $+32$ translates the data but does NOT change the spread!
Part (c): Transformed variance $\sigma_F^2$: $$\sigma_F^2 = (5.76)^2 = 33.1776 \approx 33.2 \quad [M1][A1]$$

Answers: (a) $76.1^\circ\text{F}$; (b) $5.76^\circ\text{F}$; (c) $33.2$.

Worked Example 1.4 · Combined Weighted Mean

Class A contains $18$ students with a mean test score of $74\%$. Class B contains $22$ students with a mean test score of $82\%$.
(a) Calculate the combined mean score for all $40$ students. [3 marks]
(b) A student from Class A who scored $92\%$ transfers to Class B. Find the new mean score of Class A. [3 marks]

Part (a): Combined weighted mean: $$\text{Sum for Class A: } \sum x_A = 18 \times 74 = 1332 \quad [M1]$$ $$\text{Sum for Class B: } \sum x_B = 22 \times 82 = 1804 \quad [A1]$$ $$\text{Total Sum} = 1332 + 1804 = 3136$$ $$\bar{x}_{\text{comb}} = \frac{3136}{18 + 22} = \frac{3136}{40} = 78.4\% \quad [A1]$$
Part (b): New mean of Class A after transfer: $$\text{New sum for Class A} = 1332 - 92 = 1240 \quad [M1]$$ $$\text{New count} = 18 - 1 = 17 \quad [A1]$$ $$\bar{x}_{\text{new}} = \frac{1240}{17} \approx 72.94\% \implies 72.9\% \quad [A1]$$

Answers: (a) $78.4\%$; (b) $72.9\%$.

Worked Example 1.5 · Variance and Sum of Squares Formulation

A sample of $12$ measurements satisfies $\sum x = 144$ and $\sum x^2 = 1884$.
(a) Calculate the sample mean $\bar{x}$. [1 mark]
(b) Calculate the population standard deviation $\sigma$. [3 marks]
(c) Each measurement is multiplied by $3$ and then $5$ is subtracted. Find the new standard deviation. [2 marks]

Part (a): Mean $\bar{x} = \frac{\sum x}{n}$: $$\bar{x} = \frac{144}{12} = 12 \quad [A1]$$
Part (b): Variance formula $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$: $$\sigma^2 = \frac{1884}{12} - 12^2 = 157 - 144 = 13 \quad [M1][A1]$$ $$\sigma = \sqrt{13} \approx 3.61 \quad [A1]$$
Part (c): Transformed standard deviation for $y = 3x - 5$: $$\sigma_y = |3|\sigma = 3\sqrt{13} \approx 10.8 \quad [M1][A1]$$

Answers: (a) $12$; (b) $\sqrt{13} \approx 3.61$; (c) $3\sqrt{13} \approx 10.8$.

Topic 4.3 Practice Exercises

10 tiered problems covering mean, median, standard deviation, grouped variance, skewness, and transformations.

Question 1

Consider the dataset: $5, 8, 9, 12, 14, 18, 20$. Calculate the mean, median, and interquartile range ($\text{IQR}$).

Show Step-by-Step Solution

Step 1: Mean:

$$\sum x = 5 + 8 + 9 + 12 + 14 + 18 + 20 = 86$$ $$\bar{x} = \frac{86}{7} \approx 12.2857 \approx 12.3\text{ (3 s.f.)}$$

Step 2: Median: With $n = 7$, median is the $4^{\text{th}}$ ordered value: $\text{Median} = 12$.

Step 3: Quartiles and IQR:

Lower half: $5, 8, 9 \implies Q_1 = 8$.
Upper half: $14, 18, 20 \implies Q_3 = 18$.
$$\text{IQR} = Q_3 - Q_1 = 18 - 8 = 10$$

Final Answer: Mean $\approx 12.3$; Median $= 12$; $\text{IQR} = 10$.

Question 2

A dataset has mean $\bar{x} = 40$ and standard deviation $\sigma_x = 6$. A constant $5$ is added to every score. State the new mean and the new standard deviation.

Show Step-by-Step Solution

Step 1: New Mean:
Adding a constant shifts the mean: $\bar{y} = \bar{x} + 5 = 40 + 5 = 45$.

Step 2: New Standard Deviation:
Adding a constant does not alter the spread of data around its mean:
$$\sigma_y = \sigma_x = 6$$

Final Answer: New Mean $= 45$; New Standard Deviation $= 6$.

Question 3

The temperature $C$ in degrees Celsius has mean $15^\circ\text{C}$ and standard deviation $4^\circ\text{C}$. The temperature is converted to Fahrenheit via $F = 1.8C + 32$. Find the mean and standard deviation of $F$.

Show Step-by-Step Solution

Step 1: Mean in Fahrenheit:

$$\bar{F} = 1.8\bar{C} + 32 = 1.8(15) + 32 = 27 + 32 = 59^\circ\text{F}$$

Step 2: Standard Deviation in Fahrenheit:
The additive constant $+32$ does not affect dispersion:

$$\sigma_F = 1.8 \sigma_C = 1.8(4) = 7.2^\circ\text{F}$$

Final Answer: Mean $= 59^\circ\text{F}$; Standard Deviation $= 7.2^\circ\text{F}$.

Question 4

For 8 numbers, $\sum x = 96$ and $\sum x^2 = 1248$. Calculate the exact variance and standard deviation.

Show Step-by-Step Solution

Step 1: Mean: $\bar{x} = \frac{96}{8} = 12$.

Step 2: Variance:

$$\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 = \frac{1248}{8} - 12^2 = 156 - 144 = 12$$

Step 3: Standard Deviation:

$$\sigma = \sqrt{12} = 2\sqrt{3} \approx 3.46\text{ (3 s.f.)}$$

Final Answer: Variance $= 12$; Standard Deviation $= 2\sqrt{3} \approx 3.46$.

Question 5

A group of 20 runners has a mean 10 km time of 48.0 minutes. When a 21st runner joins the group, the new mean increases to 48.5 minutes. Determine the 10 km time of the 21st runner.

Show Step-by-Step Solution

Step 1: Initial sum of times:

$$\sum_{i=1}^{20} x_i = 20 \times 48.0 = 960\text{ minutes}$$

Step 2: New total sum of times:

$$\sum_{i=1}^{21} x_i = 21 \times 48.5 = 1018.5\text{ minutes}$$

Step 3: Time of 21st runner:

$$x_{21} = 1018.5 - 960 = 58.5\text{ minutes}$$

Final Answer: 58.5 minutes

Question 6

The heights of 10 basketball players have mean $195\text{ cm}$ and standard deviation $6\text{ cm}$. The height of player $A$, which was $190\text{ cm}$, was incorrectly recorded as $180\text{ cm}$.
(a) Find the corrected mean height.
(b) Find the corrected standard deviation.

Show Step-by-Step Solution

Part (a): Corrected Mean:
Original sum: $\sum x = 10 \times 195 = 1950\text{ cm}$.
Corrected sum: $\sum x_{\text{new}} = 1950 - 180 + 190 = 1960\text{ cm}$.
$$\bar{x}_{\text{new}} = \frac{1960}{10} = 196\text{ cm}$$

Part (b): Corrected Standard Deviation:
From $\sigma_x^2 = \frac{\sum x^2}{n} - \bar{x}^2$ with $\sigma = 6$ and $\bar{x} = 195$:
$$36 = \frac{\sum x^2}{10} - 195^2 = \frac{\sum x^2}{10} - 38025 \implies \frac{\sum x^2}{10} = 38061 \implies \sum x^2 = 380610$$ Adjust $\sum x^2$: replace $180^2 = 32400$ with $190^2 = 36100$:
$$\sum x_{\text{new}}^2 = 380610 - 32400 + 36100 = 384310$$ New variance: $$\sigma_{\text{new}}^2 = \frac{384310}{10} - 196^2 = 38431 - 38416 = 15$$ New standard deviation: $$\sigma_{\text{new}} = \sqrt{15} \approx 3.87\text{ cm}$$

Final Answer: (a) Corrected Mean $= 196\text{ cm}$; (b) Corrected Standard Deviation $= \sqrt{15} \approx 3.87\text{ cm}$.

Question 7

A discrete distribution of marks $x \in \{1, 2, 3, 4, 5\}$ has frequencies $f \in \{3, k, 8, 5, 2\}$.
(a) Given that the mean mark is $2.7$, find the value of $k$.
(b) Calculate the variance of the marks.

Show Step-by-Step Solution

Part (a): Solve for $k$:
Total frequency: $n = 3 + k + 8 + 5 + 2 = 18 + k$.
Sum of marks: $\sum f_i x_i = 3(1) + 2k + 8(3) + 5(4) + 2(5) = 57 + 2k$ Given $\bar{x} = 2.7$: $\frac{57 + 2k}{18 + k} = 2.7 \implies 57 + 2k = 2.7(18 + k) = 48.6 + 2.7k$ $57 - 48.6 = 2.7k - 2k \implies 8.4 = 0.7k \implies k = \frac{8.4}{0.7} = 12 \quad [M1][A1]$

Part (b): Variance:
With $k = 12$, total frequency $n = 30$.
$\sum f_i x_i^2 = 3(1^2) + 12(2^2) + 8(3^2) + 5(4^2) + 2(5^2) = 3 + 48 + 72 + 80 + 50 = 253$ $\sigma^2 = \frac{\sum f_i x_i^2}{n} - \bar{x}^2 = \frac{253}{30} - 2.7^2 \approx 8.433 - 7.29 = 1.14 \quad [M1][A1]$

Final Answer: (a) $k = 12$; (b) $\sigma^2 \approx 1.14$.

Question 8

An employee receives an annual bonus $B = 0.15S - 2000$, where $S$ is their sales volume. Over a 5-year period, their sales volume had a mean of $\$120{,}000$ and a standard deviation of $\$18{,}000$.
(a) Find the mean bonus.
(b) Find the standard deviation of the bonus.

Show Step-by-Step Solution

Part (a): Mean Bonus:

$$\bar{B} = 0.15\bar{S} - 2000 = 0.15(120{,}000) - 2000 = 18{,}000 - 2000 = \$16{,}000$$

Part (b): Standard Deviation of Bonus:
The constant $-\$2000$ does not affect spread:

$$\sigma_B = 0.15 \sigma_S = 0.15(18{,}000) = \$2{,}700$$

Final Answer: (a) Mean Bonus $= \$16{,}000$; (b) Standard Deviation $= \$2{,}700$.

Question 9

A school combines the mathematics exam results of two classes:
• Class A: $n_A = 20$, mean $\bar{x}_A = 70$, standard deviation $\sigma_A = 8$
• Class B: $n_B = 30$, mean $\bar{x}_B = 80$, standard deviation $\sigma_B = 6$
(a) Calculate the combined mean $\bar{x}_C$ of the entire cohort of 50 students.
(b) Calculate the combined standard deviation $\sigma_C$ of the entire cohort.

Show Step-by-Step Solution

Part (a): Combined Mean:

$$\sum x_A = 20 \times 70 = 1400, \quad \sum x_B = 30 \times 80 = 2400$$ $$\sum x_C = 1400 + 2400 = 3800$$ $$\bar{x}_C = \frac{3800}{50} = 76$$

Part (b): Combined Standard Deviation:
We find $\sum x^2$ for each class using $\sum x^2 = n(\sigma^2 + \bar{x}^2)$:
For Class A: $$\sum x_A^2 = 20(8^2 + 70^2) = 20(64 + 4900) = 20(4964) = 99{,}280$$ For Class B: $$\sum x_B^2 = 30(6^2 + 80^2) = 30(36 + 6400) = 30(6436) = 193{,}080$$ Combined sum of squares: $$\sum x_C^2 = 99{,}280 + 193{,}080 = 292{,}360$$ Combined variance: $$\sigma_C^2 = \frac{292{,}360}{50} - 76^2 = 5847.2 - 5776 = 71.2$$ Combined standard deviation: $$\sigma_C = \sqrt{71.2} \approx 8.438 \approx 8.44$$ Insight: Notice that $\sigma_C = 8.44$ is larger than both individual standard deviations ($8$ and $6$) because the difference between the two class means ($70$ vs $80$) introduces substantial between-group variation!

Final Answer: (a) Combined Mean $= 76$; (b) Combined Standard Deviation $\approx 8.44$.

Question 10

Let $x_1, x_2, \dots, x_n$ have mean $\bar{x}$ and variance $\sigma_x^2$.
(a) Prove algebraically from first principles that the variance of $y_i = x_i + c$ is identical to $\sigma_x^2$.
(b) Prove that for $z_i = ax_i$, the variance is $\sigma_z^2 = a^2 \sigma_x^2$.

Show Step-by-Step Solution

Part (a) Proof for translation:
By definition of mean, $\bar{y} = \frac{1}{n}\sum (x_i + c) = \frac{1}{n}\sum x_i + \frac{nc}{n} = \bar{x} + c$.
Now consider the variance definition $\sigma_y^2 = \frac{1}{n}\sum (y_i - \bar{y})^2$:
Substitute $y_i = x_i + c$ and $\bar{y} = \bar{x} + c$: $$y_i - \bar{y} = (x_i + c) - (\bar{x} + c) = x_i - \bar{x}$$ Therefore: $$\sigma_y^2 = \frac{1}{n}\sum (x_i - \bar{x})^2 = \sigma_x^2 \quad \blacksquare$$

Part (b) Proof for scaling:
$\bar{z} = \frac{1}{n}\sum (ax_i) = a\bar{x}$.
Deviation from mean: $z_i - \bar{z} = ax_i - a\bar{x} = a(x_i - \bar{x})$.
Squared deviation: $(z_i - \bar{z})^2 = a^2(x_i - \bar{x})^2$.
$$\sigma_z^2 = \frac{1}{n}\sum a^2(x_i - \bar{x})^2 = a^2 \cdot \frac{1}{n}\sum (x_i - \bar{x})^2 = a^2\sigma_x^2 \quad \blacksquare$$

Final Answer: Both proofs complete from first principles of deviation sums.

IB Formula Booklet References (Topic 4)

Mean of Grouped Data:

$$\bar{x} = \frac{\sum_{i=1}^k f_i x_i}{n}, \quad \text{where } n = \sum_{i=1}^k f_i$$

Interquartile Range:

$$\text{IQR} = Q_3 - Q_1$$

Must-Memorize Formulas

Variance (Computational Formulas):

$$\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 = \frac{\sum f_i x_i^2}{n} - \bar{x}^2$$

Standard Deviation:

$$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}$$

Linear Transformation Effects ($y = ax + b$):

$$\bar{y} = a\bar{x} + b, \quad \sigma_y = |a|\sigma_x, \quad \sigma_y^2 = a^2\sigma_x^2, \quad \text{IQR}_y = |a|\text{IQR}_x$$

Critical IB Examination Pitfalls

Confusing Variance with Standard Deviation: Always verify whether the question asks for $\sigma^2$ (variance) or $\sigma$ (standard deviation). Forgetting the square root is the #1 algebraic error on Paper 1.
Applying $+b$ to Standard Deviation: When computing $\sigma_{ax+b}$, students frequently add $b$. The shift $b$ translates all points equally, leaving distances between points and their mean unchanged ($\sigma_{ax+b} = |a|\sigma_x$).
Mixing GDC Output $\sigma_x$ and $s_x$: $\sigma_x$ divides by $n$ (population), while $s_x$ divides by $n-1$ (sample). In AA SL, standard deviation of a sample dataset refers to $\sigma_x$ unless specified otherwise.
Combining Cohorts Incorrectly: The combined standard deviation is NOT the average of the individual standard deviations! You must use $\sum x^2 = n(\sigma^2 + \bar{x}^2)$ to account for difference in cohort means.

Topic 4.3 Diagnostic Mini Test

Section A: 4 Diagnostic MCQs with immediate feedback. Section B: 2 Exam-style Free Response Questions with full markschemes.

Section A: Diagnostic Multiple Choice

Question 1: In a distribution that is heavily skewed to the right (positive skew), which ordering of central tendency is typically observed?

Question 2: A dataset $X$ has variance $\sigma_x^2 = 9$. If $Y = 3X - 5$, what is the variance of $Y$?

Question 3: Which measure of dispersion is unaffected when a constant $k = 10$ is subtracted from every data item?

Question 4: For 10 numbers, $\sum x = 80$ and $\sum x^2 = 730$. What is the standard deviation?

Section B: Examination-Style Free Response Questions

Question 5 (6 marks)

The daily sales $x$ (in thousands of dollars) of a shop over a period of $30$ days give:
$$\sum x = 180, \quad \sum x^2 = 1170$$ (a) Find the mean and standard deviation of daily sales. [3 marks]
(b) Daily profit $P$ is modeled by $P = 0.4x - 0.8$. Find the mean and standard deviation of daily profit. [3 marks]

Show Full Markscheme

(a) Mean and standard deviation:
$\bar{x} = \frac{180}{30} = 6$ ($6000$ dollars). [A1]
Variance $\sigma_x^2 = \frac{1170}{30} - 6^2 = 39 - 36 = 3$. [M1]
$\sigma_x = \sqrt{3} \approx 1.73$ thousand dollars. [A1]

(b) Profit transformation:
$\bar{P} = 0.4\bar{x} - 0.8 = 0.4(6) - 0.8 = 2.4 - 0.8 = 1.6$ thousand dollars. [M1 A1]
$\sigma_P = 0.4 \sigma_x = 0.4\sqrt{3} \approx 0.693$ thousand dollars. [A1]

Question 6 (7 marks)

A set of 5 positive integers has mean $6$, median $5$, unique mode $5$, and range $10$.
(a) Find the 5 integers. [4 marks]
(b) Calculate the standard deviation of these 5 integers. [3 marks]

Show Full Markscheme

(a) Determine the integers:
Let the 5 ordered integers be $x_1 \le x_2 \le x_3 \le x_4 \le x_5$.
• Median $x_3 = 5$. [A1]
• Mode is $5 \implies$ at least two $5$'s. Since $x_3 = 5$, either $x_2 = 5$ or $x_4 = 5$. If $x_2 = 5$, $x_1 < 5$.
• Range is $10 \implies x_5 - x_1 = 10 \implies x_5 = x_1 + 10$. [M1]
• Mean is $6 \implies \sum x = 5 \times 6 = 30$.
Substitute: $x_1 + 5 + 5 + x_4 + (x_1 + 10) = 30 \implies 2x_1 + x_4 + 20 = 30 \implies 2x_1 + x_4 = 10$.
Since integers are positive and $x_4 \ge x_3 = 5$:
If $x_4 = 5$, then mode would be three $5$'s ($2x_1 = 5 \implies x_1 = 2.5$ not an integer).
If $x_1 = 2 \implies x_4 = 10 - 2(2) = 6$.
Check order: $x_1 = 2, x_2 = 5, x_3 = 5, x_4 = 6, x_5 = 2 + 10 = 12$.
Sum: $2 + 5 + 5 + 6 + 12 = 30$ (Mean $6$). Median $5$. Mode $5$. Range $12 - 2 = 10$. All conditions satisfied! [A2]

(b) Standard deviation:
Deviations from mean $6$: $(2-6) = -4, (5-6) = -1, (5-6) = -1, (6-6) = 0, (12-6) = 6$. [M1]
Squared deviations: $16 + 1 + 1 + 0 + 36 = 54$.
Variance $\sigma^2 = \frac{54}{5} = 10.8$. [A1]
Standard deviation $\sigma = \sqrt{10.8} \approx 3.29$. [A1]