List of Practical Experiments (Official Syllabus)

1. Fitting of Binomial distribution — Direct method.
2. Fitting of Binomial distribution — Recurrence relation method.
3. Fitting of Poisson distribution — Direct method.
4. Fitting of Poisson distribution — Recurrence relation method.
5. Fitting of Negative Binomial distribution — Direct method.
6. Fitting of Negative Binomial distribution — Recurrence relation method.
7. Fitting of Geometric distribution — Direct method.
8. Fitting of Geometric distribution — Recurrence relation method.
9. Fitting of Hypergeometric distribution.

Experiment 1 — Fitting a Binomial Distribution (Direct Method)

1. Problem

Eight coins are tossed and the number of heads \(X\) is recorded; the experiment is repeated 256 times:

x	0	1	2	3	4	5	6	7	8
f	2	6	30	52	67	56	32	10	1

Fit a Binomial distribution and obtain the expected frequencies.

2. Aim

To fit a Binomial distribution to observed data by the direct method and compare expected with observed frequencies.

3. Formula

Applying it:

1. Find \(N = \sum f\) and the mean \(\bar x = \dfrac{\sum f x}{N}\).
2. Estimate \(\hat p = \bar x / n\) (here \(n = 8\)) and \(\hat q = 1 - \hat p\).
3. Compute \(p(x)\) for each \(x\) and the expected frequency \(\hat f_x = N\,p(x)\).

4. Calculation

Blank working table (compute p(x) and the expected frequencies):

x	p(x)	Expected f = 256 p(x)	Observed f
0			2
1			6
2			30
3			52
4			67
5			56
6			32
7			10
8			1
Total	1.000	256	256

\(N = 256\); \(\sum f x = 0+6+60+156+268+280+192+70+8 = 1040\); \(\bar x = 1040/256 = 4.0625\).

\(\hat p = 4.0625/8 = 0.5078\), \(\hat q = 0.4922\).

x	p(x)	Expected f	Observed f
0	0.0034	0.88	2
1	0.0284	7.28	6
2	0.1027	26.28	30
3	0.2118	54.23	52
4	0.2732	69.93	67
5	0.2255	57.72	56
6	0.1163	29.77	32
7	0.0343	8.78	10
8	0.0044	1.13	1
Total	1.000	256.0	256

5. Result

The fitted distribution is \(B(8,\ 0.5078)\). The expected frequencies (0.9, 7.3, 26.3, 54.2, 69.9, 57.7, 29.8, 8.8, 1.1) are close to the observed values, so the Binomial model fits the data well.

Experiment 2 — Fitting a Binomial Distribution (Recurrence Method)

1. Problem

For the same coin-tossing data as Experiment 1, obtain the Binomial probabilities using the recurrence relation instead of computing each \(\binom{n}{x}\) separately.

2. Aim

To fit a Binomial distribution using the recurrence relation, which avoids repeated factorial computation.

3. Formula

Applying it:

1. Use \(\hat p = 0.5078,\ \hat q = 0.4922\) (from Experiment 1); compute the ratio \(\hat p/\hat q = 1.0317\).
2. Start from \(P(0) = \hat q^{\,n} = \hat q^{8}\).
3. Generate each successive probability by the recurrence, then multiply by \(N\) for expected frequencies.

4. Calculation

Blank working table (fill in each probability from the recurrence):

x	Multiplier (n−x)/(x+1) · p/q	P(x)
0	—
1
2
3
4

\(P(0) = 0.4922^{8} = 0.0034\).

• \(P(1) = \dfrac{8}{1}(1.0317)(0.0034) = 0.0284\)
• \(P(2) = \dfrac{7}{2}(1.0317)(0.0284) = 0.1027\)
• \(P(3) = \dfrac{6}{3}(1.0317)(0.1027) = 0.2118\)
• \(P(4) = \dfrac{5}{4}(1.0317)(0.2118) = 0.2732\)
• … continuing gives \(P(5)=0.2255,\ P(6)=0.1163,\ P(7)=0.0343,\ P(8)=0.0044\).

5. Result

The recurrence reproduces exactly the probabilities of Experiment 1 (0.0034, 0.0284, 0.1027, 0.2118, 0.2732, …) with far less arithmetic, confirming the direct-method fit.

Experiment 3 — Fitting a Poisson Distribution (Direct Method)

1. Problem

The number of accidents per day at a junction over 100 days:

x	0	1	2	3	4	5
f	46	38	12	3	1	0

Fit a Poisson distribution.

2. Aim

To fit a Poisson distribution to observed count data by the direct method.

3. Formula

Applying it:

1. Find \(N = \sum f\) and \(\bar x = \dfrac{\sum f x}{N}\); take \(\hat\lambda = \bar x\).
2. Compute \(p(x) = e^{-\hat\lambda}\hat\lambda^{x}/x!\) and \(\hat f_x = N\,p(x)\).

4. Calculation

Blank working table (compute p(x) and the expected frequencies):

x	p(x)	Expected f = 100 p(x)	Observed f
0			46
1			38
2			12
3			3
4			1
5+			0
Total	1.000	100	100

\(N = 100\); \(\sum f x = 0+38+24+9+4+0 = 75\); \(\hat\lambda = 0.75\); \(e^{-0.75} = 0.4724\).

x	p(x)	Expected f	Observed f
0	0.4724	47.24	46
1	0.3543	35.43	38
2	0.1329	13.29	12
3	0.0332	3.32	3
4	0.0062	0.62	1
5+	0.0010	0.10	0
Total	1.000	100.0	100

5. Result

The fitted distribution is Poisson with \(\hat\lambda = 0.75\). Expected and observed frequencies agree closely, so the data follow a Poisson law.

Experiment 4 — Fitting a Poisson Distribution (Recurrence Method)

1. Problem

For the same accident data as Experiment 3, obtain the Poisson probabilities using the recurrence relation.

2. Aim

To fit a Poisson distribution using the recurrence relation.

3. Formula

Applying it:

1. Take \(\hat\lambda = 0.75\) and \(P(0) = e^{-0.75}\).
2. Generate each successive probability by the recurrence and multiply by \(N\).

4. Calculation

Blank working table (fill in each probability from the recurrence):

x	Multiplier λ/(x+1)	P(x)
0	—
1
2
3
4

• \(P(0) = e^{-0.75} = 0.4724\)
• \(P(1) = (0.75/1)(0.4724) = 0.3543\)
• \(P(2) = (0.75/2)(0.3543) = 0.1329\)
• \(P(3) = (0.75/3)(0.1329) = 0.0332\)
• \(P(4) = (0.75/4)(0.0332) = 0.0062\)

5. Result

The recurrence gives the same probabilities as the direct method (0.4724, 0.3543, 0.1329, 0.0332, 0.0062), confirming the Poisson fit with \(\hat\lambda = 0.75\).

Experiment 5 — Fitting a Negative Binomial Distribution (Direct Method)

1. Problem

The number of claims \(X\) filed per policy-holder in a year, recorded over 200 policy-holders:

x	0	1	2	3	4	5	6	7	8	9	10
f	25	38	40	31	23	16	11	7	4	3	2

Fit a Negative Binomial distribution (\(X\) = number of failures before the \(r\)-th success).

2. Aim

To fit a Negative Binomial distribution by the method of moments, appropriate when the data are over-dispersed (variance > mean).

3. Formula

Applying it:

1. Compute the sample mean \(\bar x\) and variance \(s^2\). If \(s^2 > \bar x\), the NB model is appropriate.
2. Method of moments: \(\hat p = \bar x / s^2\) and \(\hat r = \bar x^2 / (s^2 - \bar x)\); round \(\hat r\) to the nearest positive integer and re-estimate \(\hat p = r / (r + \bar x)\).
3. Compute \(p(x)\) and \(\hat f_x = N\,p(x)\).

4. Calculation

Blank working table (compute p(x) and the expected frequencies):

x	p(x)	Expected f = 200 p(x)	Observed f
0			25
1			38
2			40
3			31
4			23
5			16
6			11
7			7
8			4
9			3
10			2

\(N = 200\); \(\sum f x = 577\), so \(\bar x = 577/200 = 2.885\).

\(\sum f x^2 = 2683\), so \(s^2 = 2683/200 - 2.885^2 = 13.415 - 8.323 = 5.092\). Since \(s^2 = 5.09 > \bar x = 2.89\), the data are over-dispersed and NB is appropriate.

Method of moments: \(\hat p = 2.885/5.092 = 0.567\), \(\hat r = 2.885^2/(5.092 - 2.885) = 8.323/2.207 = 3.77\). Rounding, \(r = 4\); re-estimate \(p = 4/(4 + 2.885) = 0.581\), \(q = 0.419\).

x	p(x)	Expected f	Observed f
0	0.1139	22.79	25
1	0.1910	38.19	38
2	0.2000	40.01	40
3	0.1676	33.53	31
4	0.1229	24.59	23
5	0.0824	16.48	16
6	0.0518	10.36	11
7	0.0310	6.20	7
8	0.0179	3.57	4
9	0.0100	2.00	3
10	0.0054	1.09	2

(The remaining probability, \(P(X \ge 11) \approx 0.006\), accounts for the small difference from 200.)

5. Result

The fitted distribution is Negative Binomial with \(r = 4,\ p = 0.581\). The expected frequencies track the observed values well, confirming that the over-dispersed claim data follow a Negative Binomial law.

Experiment 6 — Fitting a Negative Binomial Distribution (Recurrence Method)

1. Problem

For the claim data of Experiment 5 (fitted \(r = 4,\ p = 0.581,\ q = 0.419\)), obtain the Negative Binomial probabilities using the recurrence relation.

2. Aim

To fit a Negative Binomial distribution using the recurrence relation.

3. Formula

Applying it:

1. Start from \(P(0) = p^{\,r}\).
2. Generate each successive probability by the recurrence and multiply by \(N\).

4. Calculation

Blank working table (fill in each probability from the recurrence):

x	Multiplier (r+x)q/(x+1)	P(x)
0	—
1
2
3

• \(P(0) = 0.581^{4} = 0.1139\)
• \(P(1) = \dfrac{(4+0)(0.419)}{1}(0.1139) = 0.1910\)
• \(P(2) = \dfrac{(4+1)(0.419)}{2}(0.1910) = 0.2000\)
• \(P(3) = \dfrac{(4+2)(0.419)}{3}(0.2000) = 0.1676\)

5. Result

The recurrence reproduces the direct-method probabilities of Experiment 5 (0.1139, 0.1910, 0.2000, 0.1676, …), confirming the Negative Binomial fit with \(r = 4,\ p = 0.581\).

Experiment 7 — Fitting a Geometric Distribution (Direct Method)

1. Problem

The number of failures \(X\) before the first success of a marketing call, over 100 call-sequences:

x	0	1	2	3	4	5+
f	40	24	14	10	7	5

Fit a Geometric distribution.

2. Aim

To fit a Geometric distribution (number of failures before the first success) to observed data.

3. Formula

Applying it:

1. Find \(N\) and \(\bar x = \dfrac{\sum f x}{N}\).
2. For this form, mean \(= q/p\); hence \(\hat p = \dfrac{1}{1 + \bar x}\), \(\hat q = 1 - \hat p\).
3. Compute \(p(x) = \hat q^{\,x}\hat p\) and \(\hat f_x = N\,p(x)\).

4. Calculation

Blank working table (compute p(x) and the expected frequencies):

x	p(x)	Expected f = 100 p(x)	Observed f
0			40
1			24
2			14
3			10
4			7
5+			5

\(N = 100\); \(\sum f x = 0+24+28+30+28+25 = 135\); \(\bar x = 1.35\).

\(\hat p = 1/(1 + 1.35) = 1/2.35 = 0.4255\), \(\hat q = 0.5745\).

x	p(x)	Expected f	Observed f
0	0.4255	42.55	40
1	0.2445	24.45	24
2	0.1404	14.04	14
3	0.0807	8.07	10
4	0.0463	4.63	7
5+	0.0626	6.26	5

5. Result

The fitted Geometric distribution has \(\hat p = 0.4255\). The fit is reasonable, though the observed right tail (\(x = 3, 4\)) is a little heavier than the Geometric predicts.

Experiment 8 — Fitting a Geometric Distribution (Recurrence Method)

1. Problem

For the marketing-call data of Experiment 7 (\(\hat p = 0.4255,\ \hat q = 0.5745\)), obtain the Geometric probabilities using the recurrence relation.

2. Aim

To fit a Geometric distribution using the recurrence relation.

3. Formula

Applying it:

1. Start from \(P(0) = \hat p\).
2. Each successive probability is simply \(\hat q\) times the previous one.

4. Calculation

Blank working table (fill in each probability from the recurrence):

x	P(x) = q · P(x−1)
0
1
2
3

• \(P(0) = 0.4255\)
• \(P(1) = 0.5745 \times 0.4255 = 0.2445\)
• \(P(2) = 0.5745 \times 0.2445 = 0.1404\)
• \(P(3) = 0.5745 \times 0.1404 = 0.0807\)

5. Result

The recurrence gives the same probabilities as the direct method (0.4255, 0.2445, 0.1404, 0.0807, …), confirming the Geometric fit with \(\hat p = 0.4255\).

Experiment 9 — Fitting a Hypergeometric Distribution

1. Problem

Lots of size \(N_L = 50\) each contain \(M = 10\) defective items. From each lot a sample of \(n = 5\) items is drawn without replacement and the number of defectives \(X\) is observed, over 500 lots:

x	0	1	2	3	4	5
f	155	225	95	20	5	0

Fit a Hypergeometric distribution.

2. Aim

To fit a Hypergeometric distribution to sampling-without-replacement data whose parameters are known from the sampling scheme.

3. Formula

Applying it:

1. The parameters \(N_L = 50,\ M = 10,\ n = 5\) are fixed by the problem.
2. Compute \(p(x)\) from the Hypergeometric formula and the expected frequency \(\hat f_x = 500\,p(x)\).

4. Calculation

Blank working table (compute p(x) and the expected frequencies):

x	p(x)	Expected f = 500 p(x)	Observed f
0			155
1			225
2			95
3			20
4			5
5			0
Total	1.000	500	500

x	p(x)	Expected f	Observed f
0	0.3105	155.3	155
1	0.4313	215.7	225
2	0.2098	104.9	95
3	0.0442	22.1	20
4	0.0040	2.0	5
5	0.0001	0.05	0
Total	1.000	500.0	500

Recurrence (optional): \(P(x+1) = \dfrac{(M - x)(n - x)}{(x + 1)(N_L - M - n + x + 1)}\,P(x)\).

5. Result

The fitted Hypergeometric distribution (\(N_L = 50,\ M = 10,\ n = 5\)) gives expected frequencies in strong agreement with the observed ones, so the Hypergeometric is the correct model for this without-replacement sampling.

Lab Record Format (to be followed for every experiment)

1. Problem — the observed frequency data and the distribution to be fitted.
2. Aim — the objective of the experiment.
3. Formula — the formula(e) used, then the steps that apply them.
4. Calculation — parameter estimation, the completed table of expected frequencies, and the substitution.
5. Result — the fitted distribution and a comment on the goodness of fit.