Descriptive Statistics Practice Questions

Q 01

Build a Frequency Table topic: frequency distribution · level: easy

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A teacher recorded the number of books read last month by 15 students:
3, 1, 4, 2, 3, 5, 2, 3, 1, 4, 3, 2, 5, 1, 3

Construct a frequency table including absolute frequency, relative frequency (as a proportion), and cumulative frequency.

1

Identify unique values and count occurrences.
Values: 1, 2, 3, 4, 5. Count each: 1 appears 3×, 2 appears 3×, 3 appears 5×, 4 appears 2×, 5 appears 2×.

2

Total \(n = 15\). Relative frequency \(= f_i / n\). Cumulative frequency adds from top down.

3

Completed table:

Value (x)	Freq (f)	Rel. Freq (f/n)	Cumul. Freq
1	3	0.20	3
2	3	0.20	6
3	5	0.33	11
4	2	0.13	13
5	2	0.13	15
Total	15	1.00	—

✓ The mode is 3 (highest frequency = 5). Relative frequencies sum to 1.00 ✓

Q 02

Calculate the Mean topic: central tendency · level: easy

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The daily temperatures (°C) recorded over one week were:
18, 21, 19, 25, 23, 20, 22

Calculate the arithmetic mean temperature.

1

Formula:

\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]

2

Sum all values:

\[18 + 21 + 19 + 25 + 23 + 20 + 22 = 148\]

3

Divide by \(n = 7\):

\[\bar{x} = \frac{148}{7} \approx 21.14\,°C\]

✓ Mean temperature = 21.14 °C

Q 03

Find the Median and Mode topic: central tendency · level: easy

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A shop recorded the number of customers per hour over 9 hours:
12, 7, 15, 9, 12, 18, 7, 12, 10

Find (a) the median and (b) the mode.

1

Sort the data in ascending order:

\[7,\ 7,\ 9,\ 10,\ \mathbf{12},\ 12,\ 12,\ 15,\ 18\]

2

(a) Median: With \(n = 9\) (odd), the median is the \(\frac{n+1}{2} = 5^{th}\) value.

\[\text{Median} = x_{(5)} = 12\]

3

(b) Mode: The value that appears most often.

7 appears 2×, 12 appears 3×, all others appear once.

✓ Median = 12 customers | Mode = 12 customers

Q 04

Range and Grouped Frequency Table topic: frequency tables · dispersion · level: intermediate

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Exam scores for 20 students (out of 100):
45, 67, 72, 55, 88, 91, 60, 74, 83, 50, 76, 69, 95, 58, 80, 42, 77, 65, 87, 73

(a) Find the range. (b) Group into class intervals of width 10 starting at 40 and build a grouped frequency table.

1

(a) Range

\[\text{Range} = x_{\max} - x_{\min} = 95 - 42 = 53\]

2

(b) Grouped frequency table: Count scores falling in each interval [a, b).

Class Interval	Tally	Frequency	Rel. Freq.
[40 – 50)	\|\|\|	3	0.15
[50 – 60)	\|\|\|	3	0.15
[60 – 70)	\|\|\|\|	4	0.20
[70 – 80)	\|\|\|\| \|	6	0.30
[80 – 90)	\|\|\|\|	4	0.20
[90 – 100]	\|\|	2	0.10
Total		20	1.00

✓ Range = 53 | Modal class: [70 – 80) with 6 students

Q 05

Mean from a Frequency Table topic: central tendency · level: intermediate

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The following frequency table shows the number of siblings reported by 30 students:

Siblings (x)	Frequency (f)
0	5
1	12
2	8
3	4
4	1

Calculate the mean number of siblings.

1

Formula for mean from frequency table:

\[\bar{x} = \frac{\sum f_i \cdot x_i}{\sum f_i}\]

2

Compute \(f \cdot x\) for each row:

x	f	f · x
0	5	0
1	12	12
2	8	16
3	4	12
4	1	4
Σ	30	44

3

Compute the mean:

\[\bar{x} = \frac{44}{30} \approx 1.47 \text{ siblings}\]

✓ Mean = 1.47 siblings

Q 06

Variance and Standard Deviation topic: dispersion · level: intermediate

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The ages of 6 employees in a startup are:
24, 28, 31, 25, 30, 26

Calculate (a) the population variance and (b) the population standard deviation.

1

Calculate the mean first:

\[\bar{x} = \frac{24+28+31+25+30+26}{6} = \frac{164}{6} \approx 27.33\]

2

Compute squared deviations \((x_i - \bar{x})^2\):

\(x_i\)	\(x_i - \bar{x}\)	\((x_i - \bar{x})^2\)
24	−3.33	11.09
28	0.67	0.45
31	3.67	13.47
25	−2.33	5.43
30	2.67	7.13
26	−1.33	1.77
Σ	—	39.33

3

(a) Population variance:

\[\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{N} = \frac{39.33}{6} \approx 6.56\]

(b) Standard deviation:

\[\sigma = \sqrt{\sigma^2} = \sqrt{6.56} \approx 2.56 \text{ years}\]

Note: For a sample (not population), divide by \(n-1 = 5\) to get sample variance \(s^2 \approx 7.87\) and \(s \approx 2.81\).

✓ \(\sigma^2 \approx 6.56\) | \(\sigma \approx 2.56\) years

Q 07

Interquartile Range (IQR) topic: dispersion · level: intermediate

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The waiting times (in minutes) at a clinic for 12 patients were:
5, 8, 12, 6, 15, 9, 11, 7, 14, 10, 8, 13

Calculate the interquartile range (IQR).

1

Sort the data:

\[5,\ 6,\ 7,\ 8,\ 8,\ 9,\ 10,\ 11,\ 12,\ 13,\ 14,\ 15\]

\(n = 12\) (even number of values).

2

Find Q2 (Median): Average of 6th and 7th values.

\[Q_2 = \frac{9 + 10}{2} = 9.5\]

3

Find Q1: Median of the lower half \(\{5,6,7,8,8,9\}\). Average of 3rd and 4th values.

\[Q_1 = \frac{7 + 8}{2} = 7.5\]

Find Q3: Median of the upper half \(\{10,11,12,13,14,15\}\). Average of 3rd and 4th values.

\[Q_3 = \frac{12 + 13}{2} = 12.5\]

4

Compute IQR:

\[\text{IQR} = Q_3 - Q_1 = 12.5 - 7.5 = 5 \text{ min}\]

✓ \(Q_1 = 7.5\) | \(Q_2 = 9.5\) | \(Q_3 = 12.5\) | IQR = 5 min

Q 08

Outlier Detection with IQR Fence topic: outliers · dispersion · level: challenging

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A company's sales data (units) for 14 weeks:
120, 135, 128, 142, 119, 200, 131, 125, 138, 7, 145, 132, 140, 129

Use the IQR method (Tukey's fences) to identify any outliers.

1

Sort the data (\(n = 14\)):

\[7,\ 119,\ 120,\ 125,\ 128,\ 129,\ 131,\ 132,\ 135,\ 138,\ 140,\ 142,\ 145,\ 200\]

2

Lower half (positions 1–7): \(\{7, 119, 120, 125, 128, 129, 131\}\)
\(Q_1 =\) median of 7 values = 4th value \(= 125\)
Upper half (positions 8–14): \(\{132, 135, 138, 140, 142, 145, 200\}\)
\(Q_3 = 140\)

3

Compute IQR and Tukey's fences:

\[\text{IQR} = 140 - 125 = 15\]

\[\text{Lower fence} = Q_1 - 1.5 \times \text{IQR} = 125 - 22.5 = 102.5\]

\[\text{Upper fence} = Q_3 + 1.5 \times \text{IQR} = 140 + 22.5 = 162.5\]

4

Flag values outside \([102.5,\ 162.5]\):

7 < 102.5 → outlier (possible data entry error)
200 > 162.5 → outlier (exceptionally high sales week)

✓ Outliers detected: 7 (low) and 200 (high)

Q 09

Variance from a Grouped Frequency Table topic: dispersion · frequency tables · level: challenging

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Test scores for 40 students are recorded in this grouped frequency table:

Class Interval	Frequency (f)
[50 – 60)	4
[60 – 70)	10
[70 – 80)	14
[80 – 90)	8
[90 – 100]	4

Estimate (a) the mean and (b) the variance using midpoints.

1

Find midpoints \(m_i\) for each class:

Interval	Midpoint \(m\)	\(f\)	\(f \cdot m\)	\(f \cdot m^2\)
[50–60)	55	4	220	12 100
[60–70)	65	10	650	42 250
[70–80)	75	14	1050	78 750
[80–90)	85	8	680	57 800
[90–100]	95	4	380	36 100
Σ	—	40	2980	227 000

2

(a) Estimated mean:

\[\bar{x} = \frac{\sum f_i m_i}{\sum f_i} = \frac{2980}{40} = 74.5\]

3

(b) Variance using the computational formula:

\[\sigma^2 = \frac{\sum f_i m_i^2}{n} - \bar{x}^2 = \frac{227000}{40} - (74.5)^2\]

\[\sigma^2 = 5675 - 5550.25 = 124.75\]

Tip: The formula \(\sigma^2 = \frac{\sum f m^2}{n} - \bar{x}^2\) is equivalent to the deviation formula and avoids rounding errors from computing each \((m - \bar{x})^2\) separately.

✓ Estimated mean = 74.5 | Variance \(\approx\) 124.75 | SD \(\approx\) 11.17

Q 10

Coefficient of Variation — Comparing Datasets topic: relative dispersion · level: challenging

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Two investment portfolios show the following annual returns (%) over 5 years:
Portfolio A: 8, 10, 7, 12, 9
Portfolio B: 2, 18, −3, 25, 12

Calculate the coefficient of variation (CV) for each and determine which is relatively more stable.

1

The coefficient of variation formula:

\[\text{CV} = \frac{\sigma}{\bar{x}} \times 100\%\]

2

Portfolio A — Mean:

\[\bar{x}_A = \frac{8+10+7+12+9}{5} = \frac{46}{5} = 9.2\%\]

Deviations squared: \((8-9.2)^2 + (10-9.2)^2 + (7-9.2)^2 + (12-9.2)^2 + (9-9.2)^2\) \(= 1.44 + 0.64 + 4.84 + 7.84 + 0.04 = 14.80\)

\[\sigma_A = \sqrt{\frac{14.80}{5}} = \sqrt{2.96} \approx 1.72\%\]

\[\text{CV}_A = \frac{1.72}{9.2} \times 100 \approx 18.7\%\]

3

Portfolio B — Mean:

\[\bar{x}_B = \frac{2+18+(-3)+25+12}{5} = \frac{54}{5} = 10.8\%\]

Deviations squared: \((2-10.8)^2+(18-10.8)^2+(-3-10.8)^2+(25-10.8)^2+(12-10.8)^2\) \(= 77.44 + 51.84 + 190.44 + 201.64 + 1.44 = 522.80\)

\[\sigma_B = \sqrt{\frac{522.80}{5}} = \sqrt{104.56} \approx 10.23\%\]

\[\text{CV}_B = \frac{10.23}{10.8} \times 100 \approx 94.7\%\]

4

Compare:

Portfolio A: CV ≈ 18.7% — very stable relative to its mean
Portfolio B: CV ≈ 94.7% — extremely volatile relative to its mean

Although B has a higher average return (10.8% vs 9.2%), the CV shows it carries nearly 5× more relative risk. The CV is essential when comparing datasets with different means.

✓ CV(A) ≈ 18.7% | CV(B) ≈ 94.7% | Portfolio A is far more stable