List of Practical Experiments (Official Syllabus)

1. Writing a Questionnaire in different situations.
2. Forming a grouped and ungrouped frequency distribution table.
3. Diagrammatic presentation of data — Bar, multiple Bar and Pie.
4. Graphical presentation of data — Histogram, frequency polygon, Ogives.
5. Computation of measures of central tendency — Mean, Median and Mode.
6. Computation of measures of dispersion — QD, MD and SD.
7. Computation of non-central, central moments, β₁ and β₂ for ungrouped data.
8. Computation of non-central, central moments, β₁, β₂ and Sheppard's corrections for grouped data.
9. Computation of Karl Pearson's and Bowley's coefficients of skewness.
10. Computation of kurtosis.

Experiment 1 — Writing a Questionnaire

1. Problem

Frame a structured questionnaire for each of the following situations: (a) a survey on the reading habits of college students, and (b) a customer-satisfaction survey for a mobile-service provider.

2. Aim

To design a clear, unbiased and well-sequenced questionnaire suitable for collecting primary data in a given research situation.

3. Formula

A questionnaire has no formula; it is built to these rules of design.

Applying it:

1. Begin with a short cover note stating the purpose and assuring confidentiality.
2. Keep questions short, simple and unambiguous; avoid technical jargon.
3. Avoid leading and double-barrelled questions.
4. Prefer closed (multiple-choice / rating) questions for easy tabulation; use mutually exclusive and exhaustive options.
5. Arrange questions in a logical flow — easy to difficult, general to specific; place personal (demographic) items at the end.
6. Pilot-test the draft on a few respondents and revise before full deployment.

4. Calculation

The working of this experiment is the drafting: each question is written, given closed options where it can be, and put in order by the rules above.

5. Result

Two structured questionnaires are framed: (a) eight questions on the reading habits of college students and (b) seven on customer satisfaction with a mobile service. Both use closed, mutually exclusive options and rating scales, so the answers can be tabulated directly, and each ends with one open question for suggestions. Before use, each is pilot-tested on a few respondents and revised; in (a), the personal items (name, age, gender, class) are better moved to the end, as rule 5 advises.

Experiment 2 — Forming a Frequency Distribution

1. Problem

(a) Ungrouped: The number of accidents recorded per day for 25 days is:
1, 2, 0, 3, 2, 1, 4, 0, 2, 1, 3, 2, 1, 0, 2, 1, 3, 2, 0, 1, 2, 4, 1, 2, 3.
Form a frequency distribution.

(b) Grouped: The marks of 30 students range from 30 to 80. Form a grouped frequency distribution using suitable class intervals.

2. Aim

To arrange raw data into ungrouped and grouped frequency distribution tables.

3. Formula

Applying it:

1. For ungrouped data, list each distinct value, place a tally mark against it for every occurrence, and count the tallies to obtain the frequency \(f\).
2. For grouped data, decide the number of classes by Sturges' rule, fix the class width \(h\), and tally each observation into its class.
3. Verify that the total of the frequencies equals the number of observations, \(\sum f = N\).

4. Calculation

Blank working table — ungrouped (fill in the tally and frequency):

Accidents (x)	Tally	Frequency (f)
0
1
2
3
4
Total

(a) Ungrouped — tallying the 25 values:

Accidents (x)	Tally	Frequency (f)
0	||||	4
1	|||| ||	7
2	|||| |||	8
3	||||	4
4	||	2
Total		25

Check: \(4 + 7 + 8 + 4 + 2 = 25 = N\) ✓.

(b) Grouped — Sturges' rule gives \(k = 1 + 3.322\,\log_{10} 30 = 1 + 3.322(1.477) \approx 5.9\) (about 6 classes). Rounding the width to a convenient \(h = 10\) over the range 30–80 yields 5 classes (frequencies obtained by tallying the marks):

Marks (class)	Frequency (f)
30–40	4
40–50	8
50–60	9
60–70	6
70–80	3
Total	30

5. Result

The ungrouped distribution has frequencies 4, 7, 8, 4, 2 (total 25). The grouped distribution has five class intervals of width 10 with frequencies 4, 8, 9, 6, 3 (total 30).

Experiment 3 — Bar, Multiple Bar and Pie Diagrams

1. Problem

Bar: Production (lakh tonnes) of a factory over 5 years: 12, 15, 18, 20, 22.
Multiple bar: Imports and Exports (₹ crores) for 2021–2024: (200, 180), (240, 220), (260, 260), (290, 270).
Pie: A family's monthly budget (₹): Food 8 000, Rent 5 000, Education 4 000, Savings 3 000.

2. Aim

To represent the given data by simple bar, multiple bar and pie diagrams.

3. Formula

Applying it:

1. Simple bar: draw bars of equal width and equal spacing, with heights proportional to the values.
2. Multiple bar: for each year draw two adjoining bars (Imports, Exports) in different shades, using a common scale.
3. Pie: convert each component into an angle at the centre using the formula below, then draw the sectors.

4. Calculation

Blank working table — pie-chart angles (fill in the angle for each item):

Item	Value (₹)	Angle = (Value/Total) × 360°
Food	8 000
Rent	5 000
Education	4 000
Savings	3 000
Total	20 000

Total budget \(= 8000 + 5000 + 4000 + 3000 = 20\,000\).

Item	Value (₹)	Angle
Food	8 000	(8000/20000) × 360 = 144°
Rent	5 000	(5000/20000) × 360 = 90°
Education	4 000	(4000/20000) × 360 = 72°
Savings	3 000	(3000/20000) × 360 = 54°
Total	20 000	360°

5. Result

The bar diagram shows a steady rise in production. The multiple bar diagram shows imports exceeding exports in every year except 2023 (equal). The pie diagram is drawn with sector angles 144°, 90°, 72° and 54°, which sum to 360°.

Experiment 4 — Histogram, Frequency Polygon and Ogives

1. Problem

For the grouped marks distribution (classes 30–40, 40–50, 50–60, 60–70, 70–80 with frequencies 4, 8, 9, 6, 3; \(N = 30\)), draw the histogram, frequency polygon, and the less-than and more-than ogives.

2. Aim

To present a grouped frequency distribution graphically and to locate the median from the ogives.

3. Formula

Applying it:

1. Histogram: erect a rectangle on each class with height equal to its frequency.
2. Frequency polygon: plot frequency against the class mid-point and join the points; close the polygon at the mid-points of the empty classes on either side (25 and 85).
3. Ogives: form the less-than and more-than cumulative frequencies and plot them against the upper (resp. lower) class boundaries. Their intersection gives the median.

4. Calculation

Blank working table (fill in the mid-points and both cumulative frequencies):

Class	f	Mid-point	Less-than CF	More-than CF
30–40	4
40–50	8
50–60	9
60–70	6
70–80	3
Total	30

Class	f	Mid-point	Less-than CF	More-than CF
30–40	4	35	4	30
40–50	8	45	12	26
50–60	9	55	21	18
60–70	6	65	27	9
70–80	3	75	30	3
Total	30

• Histogram: rectangles of heights 4, 8, 9, 6, 3.
• Frequency polygon: plot (35,4), (45,8), (55,9), (65,6), (75,3) and close at (25,0), (85,0).
• Less-than ogive: plot (40,4), (50,12), (60,21), (70,27), (80,30).
• More-than ogive: plot (30,30), (40,26), (50,18), (60,9), (70,3).

5. Result

The histogram, frequency polygon and the two ogives are drawn as above. The two ogives intersect at \(N/2 = 15\), corresponding to a mark of about 53, which is the median of the distribution.

Experiment 5 — Mean, Median and Mode

1. Problem

Compute the mean, median and mode of the following grouped frequency distribution:

Class	0–10	10–20	20–30	30–40	40–50	50–60
f	5	10	20	25	15	5

2. Aim

To compute the three measures of central tendency — arithmetic mean, median and mode — for grouped data.

3. Formula

Applying it:

1. Find the class mid-point \(x\) and the product \(fx\); the mean is \(\sum fx / N\).
2. Form the cumulative frequency (CF); the median class is the one containing \(N/2\).
3. The modal class is the class with the highest frequency.

4. Calculation

Blank working table (fill in x, fx and CF):

Class	f	x	fx	CF
0–10	5
10–20	10
20–30	20
30–40	25
40–50	15
50–60	5
Total	80

Class	f	x	fx	CF
0–10	5	5	25	5
10–20	10	15	150	15
20–30	20	25	500	35
30–40	25	35	875	60
40–50	15	45	675	75
50–60	5	55	275	80
Total	80		2 500

Mean \(= 2500/80 = 31.25\).

Median: \(N/2 = 40\) ⇒ median class 30–40 with \(L=30,\ C=35,\ f=25,\ h=10\):
Median \(= 30 + \dfrac{40-35}{25}\times 10 = 32\).

Mode: modal class 30–40 with \(f_1=25,\ f_0=20,\ f_2=15,\ h=10\):
Mode \(= 30 + \dfrac{25-20}{50-20-15}\times 10 = 30 + \dfrac{50}{15} \approx 33.33\).

Empirical check: \(3\,\text{Median} - 2\,\text{Mean} = 96 - 62.5 = 33.5 \approx 33.33\) ✓.

5. Result

Mean = 31.25, Median = 32, Mode ≈ 33.33. Since Mean < Median < Mode, the distribution is slightly negatively skewed.

Experiment 6 — Quartile Deviation, Mean Deviation and Standard Deviation

1. Problem

For the same distribution as Experiment 5, compute the quartile deviation (QD), the mean deviation about the mean (MD), the standard deviation (SD) and the coefficient of variation (CV).

2. Aim

To compute the absolute and relative measures of dispersion for grouped data.

3. Formula

Applying it:

1. Locate \(Q_1\) and \(Q_3\) from the CF (classes containing \(N/4\) and \(3N/4\)); then find QD.
2. Compute \(|x-\bar{x}|\) and \(f|x-\bar{x}|\); the MD is \(\sum f|x-\bar{x}|/N\).
3. Compute \(fx^2\); the SD follows from the variance formula; CV expresses SD as a percentage of the mean.

4. Calculation

Blank working table (fill in the deviation and squared columns; \(\bar{x}=31.25\)):

x	f	|x − 31.25|	f|x − 31.25|	fx²
5	5
15	10
25	20
35	25
45	15
55	5
Total	80

QD: \(N/4 = 20\) ⇒ \(Q_1\) class 20–30 (\(L=20,\ C=15,\ f=20,\ h=10\)): \(Q_1 = 20 + \dfrac{20-15}{20}\times 10 = 22.5\).
\(3N/4 = 60\) ⇒ \(Q_3\) class 30–40 (\(L=30,\ C=35,\ f=25,\ h=10\)): \(Q_3 = 30 + \dfrac{60-35}{25}\times 10 = 40\).
QD \(= (40 - 22.5)/2 = \mathbf{8.75}\).

x	f	|x − 31.25|	f|x − 31.25|	fx²
5	5	26.25	131.25	125
15	10	16.25	162.50	2 250
25	20	6.25	125.00	12 500
35	25	3.75	93.75	30 625
45	15	13.75	206.25	30 375
55	5	23.75	118.75	15 125
Total	80		837.50	91 000

MD \(= 837.5/80 = \mathbf{10.47}\).

SD: \(\sigma^2 = 91000/80 - 31.25^2 = 1137.5 - 976.5625 = 160.9375\); \(\sigma = \sqrt{160.9375} \approx \mathbf{12.69}\).

CV \(= 12.69/31.25 \times 100 \approx 40.6\%\).

5. Result

QD = 8.75, MD (about mean) = 10.47, SD = 12.69, CV ≈ 40.6 %.

Experiment 7 — Moments, β₁ and β₂ (Ungrouped Data)

1. Problem

For the data 1, 3, 5, 7, 9, compute the first four central moments and the coefficients \(\beta_1\) and \(\beta_2\).

2. Aim

To compute the central moments of an ungrouped data set and the moment-based coefficients of skewness (\(\beta_1\)) and kurtosis (\(\beta_2\)).

3. Formula

Applying it:

1. Compute the mean; take deviations \(d = x - \bar{x}\).
2. Form the columns \(d, d^2, d^3, d^4\) and their sums.
3. Divide each sum by \(n\) to obtain the central moments \(\mu_2, \mu_3, \mu_4\), then compute \(\beta_1\) and \(\beta_2\).

4. Calculation

Blank working table (fill in the deviation-power columns; \(\bar{x}=5\)):

x	d = x − 5	d²	d³	d⁴
1
3
5
7
9
Σ

Mean \(= 25/5 = 5\).

x	d = x − 5	d²	d³	d⁴
1	−4	16	−64	256
3	−2	4	−8	16
5	0	0	0	0
7	2	4	8	16
9	4	16	64	256
Σ	0	40	0	544

\(\mu_1 = 0,\quad \mu_2 = 40/5 = 8,\quad \mu_3 = 0,\quad \mu_4 = 544/5 = 108.8\).

\(\beta_1 = \dfrac{0^2}{8^3} = 0\); \(\quad \beta_2 = \dfrac{108.8}{8^2} = \dfrac{108.8}{64} = 1.7\).

5. Result

\(\mu_2 = 8,\ \mu_3 = 0,\ \mu_4 = 108.8\); \(\beta_1 = 0\) (the distribution is symmetric) and \(\beta_2 = 1.7\) (< 3, so platykurtic).

Experiment 8 — Moments and Sheppard's Correction (Grouped Data)

1. Problem

For the grouped distribution below, compute the moments about \(A = 25\), the central moments, \(\beta_1,\ \beta_2\), and apply Sheppard's corrections (\(h = 10\)).

Class	0–10	10–20	20–30	30–40	40–50
f	5	10	20	10	5

2. Aim

To compute the moments of a grouped distribution and to correct them for grouping error using Sheppard's corrections.

3. Formula

Applying it:

1. Take the mid-point \(x\), deviations \(d = x - A\), and form \(fd, fd^2, fd^3, fd^4\).
2. Obtain the moments about \(A\): \(\mu'_r = \sum fd^{\,r}/N\).
3. Because \(\mu'_1 = 0\) here (\(A\) equals the mean), the central moments equal the moments about \(A\).
4. Apply Sheppard's corrections for the class width \(h\).

4. Calculation

Blank working table (fill in the deviation-power columns; \(A = 25\)):

Class	f	x	d = x − 25	fd	fd²	fd³	fd⁴
0–10	5
10–20	10
20–30	20
30–40	10
40–50	5
Σ	50

Class	f	x	d = x − 25	fd	fd²	fd³	fd⁴
0–10	5	5	−20	−100	2 000	−40 000	800 000
10–20	10	15	−10	−100	1 000	−10 000	100 000
20–30	20	25	0	0	0	0	0
30–40	10	35	10	100	1 000	10 000	100 000
40–50	5	45	20	100	2 000	40 000	800 000
Σ	50			0	6 000	0	1 800 000

Moments about \(A = 25\): \(\mu'_1 = 0,\ \mu'_2 = 6000/50 = 120,\ \mu'_3 = 0,\ \mu'_4 = 1\,800\,000/50 = 36\,000\).

Since \(\mu'_1 = 0\), the central moments equal these: \(\mu_2 = 120,\ \mu_3 = 0,\ \mu_4 = 36\,000\).

Uncorrected coefficients: \(\beta_1 = \dfrac{\mu_3^2}{\mu_2^3} = 0\) and \(\beta_2 = \dfrac{36\,000}{120^2} = \dfrac{36\,000}{14\,400} = 2.5\).

Sheppard's corrections with \(h = 10\):

• \(\mu_2^{*} = 120 - \dfrac{10^2}{12} = 120 - 8.33 = 111.67\).
• \(\mu_3^{*} = 0\) (no correction).
• \(\mu_4^{*} = 36\,000 - \dfrac{10^2}{2}(120) + \dfrac{7(10^4)}{240} = 36\,000 - 6\,000 + 291.67 = 30\,291.67\).

\(\beta_1 = 0\); \(\quad \beta_2 = \dfrac{30\,291.67}{111.67^2} = \dfrac{30\,291.67}{12\,470.18} \approx 2.43\).

5. Result

Uncorrected: \(\mu_2 = 120,\ \mu_4 = 36\,000,\ \beta_2 = 2.5\). After Sheppard's correction: \(\mu_2^{*} = 111.67,\ \mu_4^{*} = 30\,291.67\), giving \(\beta_1 = 0\) and \(\beta_2 \approx 2.43\) (both < 3, platykurtic).

Experiment 9 — Karl Pearson's and Bowley's Coefficients of Skewness

1. Problem

Using the results of Experiments 5 and 6 (Mean = 31.25, Median = 32, Mode = 33.33, SD = 12.69, \(Q_1 = 22.5,\ Q_3 = 40\)), compute Karl Pearson's and Bowley's coefficients of skewness.

2. Aim

To measure the direction and degree of skewness of a distribution using the Karl Pearson and Bowley coefficients.

3. Formula

Applying it:

1. Karl Pearson's coefficient uses the mean, mode (or median) and SD.
2. When the mode is ill-defined, use the empirical relation \(\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}\), giving the second form below.
3. Bowley's coefficient is based on the three quartiles.

4. Calculation

Blank working table (enter the values you will substitute):

Quantity	Value
Mean \(\bar{x}\)
Median \(Q_2\)
Mode
SD \(\sigma\)
\(Q_1\)
\(Q_3\)

Karl Pearson: \(S_k = \dfrac{31.25 - 33.33}{12.69} = -0.164\) (mode-based).
Empirical form: \(S_k = \dfrac{3(31.25 - 32)}{12.69} = -0.177\). Both agree in sign and magnitude.

Bowley: \(S_B = \dfrac{40 + 22.5 - 2(32)}{40 - 22.5} = \dfrac{-1.5}{17.5} = -0.086\).

5. Result

Karl Pearson's coefficient ≈ −0.16 and Bowley's coefficient ≈ −0.086. Both are negative, so the distribution is mildly negatively skewed.

Experiment 10 — Kurtosis

1. Problem

(a) Using the moments from Experiment 7 (\(\mu_2 = 8,\ \mu_4 = 108.8\)), find \(\beta_2\) and \(\gamma_2\) and classify the peakedness.
(b) For a distribution with \(\mu_2 = 4\) and \(\mu_4 = 75\), find \(\beta_2\) and \(\gamma_2\).

2. Aim

To measure the peakedness (kurtosis) of a distribution and classify it as platykurtic, mesokurtic or leptokurtic.

3. Formula

Applying it:

1. Compute \(\beta_2 = \mu_4/\mu_2^{2}\) and the excess \(\gamma_2 = \beta_2 - 3\).
2. Classify: \(\beta_2 < 3\) platykurtic, \(\beta_2 = 3\) mesokurtic (normal), \(\beta_2 > 3\) leptokurtic.

4. Calculation

Blank working table (fill in the values and results):

Case	μ₂	μ₄	β₂ = μ₄/μ₂²	γ₂ = β₂ − 3	Type
(a)	8	108.8
(b)	4	75

Case	μ₂	μ₄	β₂ = μ₄/μ₂²	γ₂ = β₂ − 3	Type
(a)	8	108.8	108.8/64 = 1.70	−1.30	Platykurtic
(b)	4	75	75/16 = 4.69	1.69	Leptokurtic

5. Result

(a) \(\beta_2 = 1.70,\ \gamma_2 = -1.30\) ⇒ the distribution is platykurtic (flatter than normal).
(b) \(\beta_2 = 4.69,\ \gamma_2 = 1.69\) ⇒ the distribution is leptokurtic (more peaked than normal).

Lab Record Format (to be followed for every experiment)

1. Problem — the data and what is to be found.
2. Aim — the objective of the experiment.
3. Formula — the formula(e) used, then the steps that apply them.
4. Calculation — the filled working table and the substitution of values.
5. Result — the final value(s) with proper units and a brief interpretation.