Dispersion — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for dispersion.

UT-1: Variance vs Standard Deviation Units

Question:

The heights of 100 students have mean $165$ cm and variance $49$ cm$^2$. A new student of height $181$ cm joins the group. Find the new mean and new variance.

Solution:

Original: $n = 100$, $\bar{x} = 165$, $\sigma^2 = 49$.

Sum of original data: $\sum x_i = 100 \times 165 = 16500$.

Sum of squares of original data: $\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 100(49 + 27225) = 100 \times 27274 = 2727400$.

After adding 181:

New sum: $16500 + 181 = 16681$.

New $n = 101$.

New mean: $\bar{x}_{\text{new}} = \dfrac{16681}{101} \approx 165.16$.

New sum of squares: $2727400 + 181^2 = 2727400 + 32761 = 2760161$.

New variance: $\sigma^2_{\text{new}} = \dfrac{2760161}{101} - (165.16)^2 \approx 27328.33 - 27277.83 \approx 50.50$ cm$^2$.

Note: variance has units cm$^2$, while standard deviation has units cm.

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UT-2: Coding Effect on Mean and Standard Deviation

Question:

A set of data has mean $m$ and standard deviation $s$. If each value is transformed by $y = 3x - 5$, find the new mean and new standard deviation in terms of $m$ and $s$.

Solution:

For the transformation $y = ax + b$:

• New mean $= a \times \text{old mean} + b = 3m - 5$.
• New standard deviation $= |a| \times \text{old standard deviation} = 3s$.
• New variance $= a^2 \times \text{old variance} = 9s^2$.

The additive constant $-5$ affects the mean but NOT the standard deviation.

A common mistake is thinking the $-5$ affects the standard deviation.

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UT-3: Grouped Data Midpoints

Question:

Find the estimated mean and estimated standard deviation from the following grouped data:

Class	Frequency
$0 \leq x \lt 10$	5
$10 \leq x \lt 20$	8
$20 \leq x \lt 30$	12
$30 \leq x \lt 40$	5

Solution:

Midpoints: $5, 15, 25, 35$.

$x$	$f$	$fx$	$fx^2$
5	5	25	125
15	8	120	1800
25	12	300	7500
35	5	175	6125
Total	30	620	15550

Estimated mean: $\bar{x} = \dfrac{620}{30} = \dfrac{62}{3} \approx 20.67$.

Estimated variance: $\dfrac{15550}{30} - \left(\dfrac{62}{3}\right)^2 = \dfrac{1555}{3} - \dfrac{3844}{9} = \dfrac{4665 - 3844}{9} = \dfrac{821}{9} \approx 91.22$.

Estimated SD $= \sqrt{\dfrac{821}{9}} = \dfrac{\sqrt{821}}{3} \approx 9.56$.

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UT-4: Box Plot Interpretation

Question:

A box plot has minimum $= 10$, $Q_1 = 25$, median $= 35$, $Q_3 = 50$, maximum $= 80$. Find the interquartile range and identify any outliers.

Solution:

$\text{IQR} = Q_3 - Q_1 = 50 - 25 = 25$.

Lower fence: $Q_1 - 1.5 \times \text{IQR} = 25 - 37.5 = -12.5$.

Upper fence: $Q_3 + 1.5 \times \text{IQR} = 50 + 37.5 = 87.5$.

Since all values ($10$ to $80$) fall within $[-12.5,\; 87.5]$, there are no outliers.

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UT-5: Combined Data Sets

Question:

Set $A$ has $n_A = 8$, $\bar{x}_A = 10$, $s_A^2 = 4$. Set $B$ has $n_B = 12$, $\bar{x}_B = 15$, $s_B^2 = 9$. Find the mean and variance of the combined set $A \cup B$.

Solution:

Combined $n = 8 + 12 = 20$.

Combined mean: $\bar{x} = \dfrac{8 \times 10 + 12 \times 15}{20} = \dfrac{80 + 180}{20} = \dfrac{260}{20} = 13$.

Combined variance using the formula:

$s^2 = \frac{n_A(s_A^2 + d_A^2) + n_B(s_B^2 + d_B^2)}{n_A + n_B}$

where $d_A = \bar{x}_A - \bar{x} = 10 - 13 = -3$ and $d_B = \bar{x}_B - \bar{x} = 15 - 13 = 2$.

$s^2 = \frac{8(4 + 9) + 12(9 + 4)}{20} = \frac{8 \times 13 + 12 \times 13}{20} = \frac{104 + 156}{20} = \frac{260}{20} = 13$

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Integration Tests

Tests synthesis of dispersion with other topics.

IT-1: Dispersion and Probability (with Probability)

Question:

A random variable $X$ takes values $1, 2, 3, 4, 5$ with probabilities $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{3}{15}$, $\dfrac{4}{15}$, $\dfrac{5}{15}$ respectively. Find $E(X)$ and $\text{Var}(X)$.

Solution:

$E(X) = 1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 3 \times \frac{3}{15} + 4 \times \frac{4}{15} + 5 \times \frac{5}{15}$

$= \frac{1 + 4 + 9 + 16 + 25}{15} = \frac{55}{15} = \frac{11}{3}$

$E(X^2) = \frac{1 + 8 + 27 + 64 + 125}{15} = \frac{225}{15} = 15$

$\text{Var}(X) = E(X^2) - [E(X)]^2 = 15 - \frac{121}{9} = \frac{135 - 121}{9} = \frac{14}{9}$

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IT-2: Dispersion and Inequalities (with Inequalities)

Question:

A set of 50 numbers has mean $20$ and standard deviation $4$. Using Chebyshev's inequality (or empirical reasoning), at least what percentage of the data lies within 2 standard deviations of the mean?

Solution:

Within 2 standard deviations: $20 \pm 2(4) = [12,\; 28]$.

By Chebyshev's inequality, at least $1 - \dfrac{1}{k^2} = 1 - \dfrac{1}{4} = \dfrac{3}{4} = 75\%$ of data lies within $k = 2$ standard deviations.

If the data is approximately normally distributed, the empirical rule gives approximately $95\%$, but Chebyshev gives the guaranteed minimum of $75\%$.

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IT-3: Dispersion and Combinatorics (with Combinatorics)

Question:

All possible samples of size 2 are drawn with replacement from the population $\\{2, 4, 6\\}$. Find the mean and variance of the sampling distribution of the sample mean.

Solution:

Population: $\\{2, 4, 6\\}$, $\mu = 4$, $\sigma^2 = \dfrac{(4+0+4)}{3} = \dfrac{8}{3}$.

All samples of size 2 with replacement (9 samples):

$(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)$.

Sample means: $2, 3, 4, 3, 4, 5, 4, 5, 6$.

Mean of sample means: $\dfrac{2+3+4+3+4+5+4+5+6}{9} = \dfrac{36}{9} = 4 = \mu$.

Variance of sample means: $\dfrac{(2-4)^2 + 2(3-4)^2 + 3(4-4)^2 + 2(5-4)^2 + (6-4)^2}{9}$

$= \dfrac{4 + 2 + 0 + 2 + 4}{9} = \dfrac{12}{9} = \dfrac{4}{3} = \dfrac{\sigma^2}{n}$.

This confirms: $E(\bar{X}) = \mu$ and $\text{Var}(\bar{X}) = \dfrac{\sigma^2}{n}$.

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Worked Examples

WE-1: Effect of Adding a Constant to All Data

Question:

A set of 5 numbers has mean $10$ and standard deviation $3$. If $7$ is added to each number, find the new mean, new standard deviation, and new variance.

Solution:

Transformation: $y = x + 7$ (i.e. $a = 1$, $b = 7$).

New mean $= 1 \times 10 + 7 = 17$.

New standard deviation $= |1| \times 3 = 3$.

New variance $= 1^2 \times 9 = 9$.

The additive constant $7$ shifts the data but does not affect the spread.

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WE-2: Standard Deviation from a Frequency Distribution

Question:

Find the mean and standard deviation of the following data:

$x$	2	4	6	8	10
$f$	3	5	8	4	2

Solution:

$x$	$f$	$fx$	$fx^2$
2	3	6	12
4	5	20	80
6	8	48	288
8	4	32	256
10	2	20	200
Total	22	126	836

$\bar{x} = \frac{126}{22} = \frac{63}{11} \approx 5.727$

$\sigma^2 = \frac{836}{22} - \left(\frac{63}{11}\right)^2 = \frac{418}{11} - \frac{3969}{121} = \frac{4598 - 3969}{121} = \frac{629}{121}$

$\sigma = \sqrt{\frac{629}{121}} = \frac{\sqrt{629}}{11} \approx 2.28$

DSE Exam Technique: Always set up the table with columns for $fx$ and $fx^2$. This earns method marks even if there is a minor calculation error. Leave standard deviation in exact form unless asked for a decimal.

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WE-3: Coding with Grouped Data

Question:

Using the coding $y = \dfrac{x - 25}{5}$, the coded data has mean $\bar{y} = 1.2$ and variance $s_y^2 = 4.8$. Find the mean and standard deviation of the original data.

Solution:

The coding is $y = \dfrac{x - 25}{5} = \dfrac{1}{5}x - 5$, so $a = \dfrac{1}{5}$ and $b = -5$.

Original mean: $\bar{x} = \dfrac{\bar{y} - b}{a} = \dfrac{1.2 + 5}{1/5} = 6.2 \times 5 = 31$.

Original standard deviation: $\sigma_x = \dfrac{\sigma_y}{|a|} = \dfrac{\sqrt{4.8}}{1/5} = 5\sqrt{4.8} = 5 \times 2\sqrt{1.2} = 10\sqrt{1.2} = 2\sqrt{30} \approx 10.95$.

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WE-4: Interquartile Range from Raw Data

Question:

Find the median, quartiles, and interquartile range of the data set:

$12, 15, 18, 20, 22, 25, 28, 30, 35, 40, 45$

Solution:

There are $n = 11$ data values (odd).

Median position: $\dfrac{11 + 1}{2} = 6$. Median $= 25$.

Lower half: $12, 15, 18, 20, 22$ (5 values).

$Q_1$ position: $\dfrac{5 + 1}{2} = 3$. $Q_1 = 18$.

Upper half: $28, 30, 35, 40, 45$ (5 values).

$Q_3$ position: $\dfrac{5 + 1}{2} = 3$. $Q_3 = 35$.

$\text{IQR} = Q_3 - Q_1 = 35 - 18 = 17$

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WE-5: Comparing Distributions Using Standard Deviation

Question:

Class A has test scores with mean $65$ and standard deviation $8$. Class B has test scores with mean $68$ and standard deviation $15$. Which class has more consistent performance? Justify your answer.

Solution:

The coefficient of variation (CV) measures relative dispersion:

$\text{CV}_A = \frac{8}{65} \times 100\% \approx 12.3\%$

$\text{CV}_B = \frac{15}{68} \times 100\% \approx 22.1\%$

Since $\text{CV}_A < \text{CV}_B$, Class A has more consistent performance.

Alternatively, comparing standard deviations directly: Class A has $\sigma = 8$ and Class B has $\sigma = 15$. The smaller standard deviation of Class A indicates less variability, i.e. more consistency.

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WE-6: Removing an Outlier

Question:

A data set is $\{5, 8, 10, 12, 15, 18, 45\}$. The value $45$ is suspected to be an outlier.

(a) Calculate the mean and standard deviation of all 7 values. (b) Calculate the mean and standard deviation after removing $45$. (c) Comment on the effect of the outlier.

Solution:

(a) Sum $= 113$, $\bar{x} = \dfrac{113}{7} \approx 16.14$.

$\sum x^2 = 25 + 64 + 100 + 144 + 225 + 324 + 2025 = 2907$.

$\sigma^2 = \dfrac{2907}{7} - \left(\dfrac{113}{7}\right)^2 = 415.29 - 260.31 = 154.98$.

$\sigma \approx 12.45$.

(b) After removing $45$: sum $= 68$, $n = 6$, $\bar{x} = \dfrac{68}{6} = \dfrac{34}{3} \approx 11.33$.

$\sum x^2 = 882$, $\sigma^2 = \dfrac{882}{6} - \left(\dfrac{34}{3}\right)^2 = 147 - 128.44 = 18.56$.

$\sigma \approx 4.31$.

(c) The outlier $45$ significantly increases both the mean (from $11.33$ to $16.14$) and the standard deviation (from $4.31$ to $12.45$). The standard deviation is nearly tripled, showing that outliers have an exaggerated effect on measures of dispersion.

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WE-7: Using a Given Mean to Find Missing Frequency

Question:

The mean of the following data is $4.5$. Find the value of $k$.

$x$	1	2	3	4	5	6
$f$	2	4	$k$	6	3	1

Solution:

Total frequency: $n = 2 + 4 + k + 6 + 3 + 1 = 16 + k$.

Sum: $\sum fx = 2 + 8 + 3k + 24 + 15 + 6 = 55 + 3k$.

$\bar{x} = \frac{55 + 3k}{16 + k} = 4.5$

$55 + 3k = 4.5(16 + k) = 72 + 4.5k$

$1.5k = 17$

$k = \frac{17}{1.5} = \frac{34}{3}$

Since $k$ must be a non-negative integer, and $\dfrac{34}{3}$ is not an integer, there is no integer value of $k$ that gives a mean of exactly $4.5$. If the question allows non-integer frequencies, $k = \dfrac{34}{3}$.

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WE-8: Sheppard's Correction (Awareness)

Question:

For grouped data with class width $h$, state Sheppard's correction for the variance and explain when it is appropriate to use it.

Solution:

Sheppard's correction adjusts the grouped data variance:

$\sigma_{\text{corrected}}^2 = \sigma_{\text{grouped}}^2 - \frac{h^2}{12}$

This correction is appropriate when:

• The data is approximately normally distributed.
• The distribution is continuous.
• The class intervals are of equal width.

It accounts for the fact that using midpoints assumes data is concentrated at the centre of each class, which slightly overestimates the variance.

In DSE examinations, Sheppard's correction is generally not required unless explicitly asked.

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Common Pitfalls

1. Confusing population variance with sample variance. The population variance formula divides by $n$, while the sample variance divides by $n - 1$ (Bessel's correction). In DSE Maths, unless specified otherwise, use the population formula (divide by $n$).

2. Forgetting that variance has squared units. If data is in centimetres, the variance is in cm$^2$ and the standard deviation is in cm. Do not mix up units when writing conclusions.

3. Incorrectly applying coding formulas. For the transformation $y = ax + b$: new mean $= a\bar{x} + b$, new SD $= |a| \cdot s$. The additive constant $b$ does NOT affect the standard deviation. A common error is writing new SD $= as + b$.

4. Using the wrong formula for combined variance. When combining two data sets, do not simply average the variances. Use the correct formula involving the deviation of each set's mean from the combined mean.

5. Misidentifying quartile positions. Different textbooks use different conventions for finding $Q_1$ and $Q_3$. In DSE, the most common approach is: $Q_1$ is the median of the lower half and $Q_3$ is the median of the upper half.

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DSE Exam-Style Questions

DSE-1

The following table shows the distribution of marks obtained by 40 students in a test.

Marks	$1$--$10$	$11$--$20$	$21$--$30$	$31$--$40$	$41$--$50$
Frequency	3	8	14	10	5

(a) Estimate the mean mark. (3 marks) (b) Estimate the standard deviation of the marks. (3 marks) (c) If a student scored 35 marks, find the student's standardised score (z-score). (2 marks)

Solution:

(a) Midpoints: $5.5, 15.5, 25.5, 35.5, 45.5$.

$x$	$f$	$fx$	$fx^2$
5.5	3	16.5	90.75
15.5	8	124	1922
25.5	14	357	9103.5
35.5	10	355	12602.5
45.5	5	227.5	10351.25
Total	40	1080	34070

$\bar{x} = \frac{1080}{40} = 27$

(b) $\sigma^2 = \dfrac{34070}{40} - 27^2 = 851.75 - 729 = 122.75$.

$\sigma = \sqrt{122.75} \approx 11.08$.

(c) $z = \dfrac{35 - 27}{11.08} = \dfrac{8}{11.08} \approx 0.72$.

The student scored $0.72$ standard deviations above the mean.

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DSE-2

The weights (in kg) of 8 parcels are: $2.3, 3.1, 4.5, 5.2, 3.8, 4.1, 2.9, 3.6$.

(a) Find the mean and standard deviation. (4 marks) (b) If each parcel has a label weighing $0.1$ kg added, find the new mean and new standard deviation. (2 marks) (c) If the weight of each parcel is converted to grams, find the new mean and new variance. (2 marks)

Solution:

(a) Sum $= 29.5$, $\bar{x} = \dfrac{29.5}{8} = 3.6875$.

$\sum x^2 = 5.29 + 9.61 + 20.25 + 27.04 + 14.44 + 16.81 + 8.41 + 12.96 = 114.81$.

$\sigma^2 = \dfrac{114.81}{8} - (3.6875)^2 = 14.35125 - 13.59766 = 0.7536$.

$\sigma = \sqrt{0.7536} \approx 0.868$ kg.

(b) Transformation: $y = x + 0.1$.

New mean $= 3.6875 + 0.1 = 3.7875$ kg.

New SD $= 0.868$ kg (unchanged by addition).

(c) Transformation: $y = 1000x$.

New mean $= 1000 \times 3.6875 = 3687.5$ g.

New variance $= 1000^2 \times 0.7536 = 753600$ g$^2$.

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DSE-3

Two classes took the same examination. Class $A$ (30 students) had mean $72$ and variance $36$. Class $B$ (20 students) had mean $65$ and variance $64$.

(a) Find the overall mean of all 50 students. (2 marks) (b) Find the overall variance of all 50 students. (4 marks)

Solution:

(a) Combined mean $= \dfrac{30 \times 72 + 20 \times 65}{50} = \dfrac{2160 + 1300}{50} = \dfrac{3460}{50} = 69.2$.

(b) $d_A = 72 - 69.2 = 2.8$, $d_B = 65 - 69.2 = -4.2$.

$\sigma^2 = \frac{30(36 + 2.8^2) + 20(64 + 4.2^2)}{50} = \frac{30(36 + 7.84) + 20(64 + 17.64)}{50}$

$= \frac{30 \times 43.84 + 20 \times 81.64}{50} = \frac{1315.2 + 1632.8}{50} = \frac{2948}{50} = 58.96$

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DSE-4

The box-and-whisker diagram below summarises the daily temperatures (in $°$C) recorded in a city for 30 days:

Minimum $= 12$, $Q_1 = 18$, Median $= 22$, $Q_3 = 28$, Maximum $= 35$.

(a) Find the interquartile range. (1 mark) (b) Determine the lower and upper fences and identify any outliers. (3 marks) (c) What percentage of the data lies between $18$ and $28$? (1 mark)

Solution:

(a) $\text{IQR} = Q_3 - Q_1 = 28 - 18 = 10$.

(b) Lower fence $= Q_1 - 1.5 \times \text{IQR} = 18 - 15 = 3$.

Upper fence $= Q_3 + 1.5 \times \text{IQR} = 28 + 15 = 43$.

Since all values ($12$ to $35$) lie within $[3, 43]$, there are no outliers.

(c) By definition, $50\%$ of the data lies between $Q_1$ and $Q_3$.

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DSE-5

A set of data $x_1, x_2, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. A new data set is formed by removing the value $\mu$ from the original set.

(a) Find the mean of the new data set. (2 marks) (b) Find the variance of the new data set in terms of $\sigma$ and $n$. (4 marks)

Solution:

(a) Original sum $= n\mu$. After removing $\mu$: new sum $= (n-1)\mu$.

New mean $= \dfrac{(n-1)\mu}{n-1} = \mu$.

(b) Original $\sum x_i^2 = n(\sigma^2 + \mu^2)$.

After removing $\mu$: $\sum x_i^2 = n(\sigma^2 + \mu^2) - \mu^2 = n\sigma^2 + (n-1)\mu^2$.

New variance $= \dfrac{n\sigma^2 + (n-1)\mu^2}{n-1} - \mu^2 = \dfrac{n\sigma^2}{n-1}$.