List of Practical Experiments (Official Syllabus)

1. Calculation of moments of a univariate random variable for a given pmf.
2. Calculation of coefficient of skewness and kurtosis for a given pmf.
3. Calculation of moments of a univariate random variable for a given pdf.
4. Calculation of coefficient of skewness and kurtosis for a given pdf.
5. Problems related to jpmf, mpmf and conditional pmf and its independence.
6. Problems related to jpdf, mpdf and conditional pdf and its independence.
7. Chebyshev's inequality — application-oriented problems.

Experiment 1 — Moments from a PMF

1. Problem

A discrete random variable \(X\) has the probability mass function:

x	0	1	2	3	4
p(x)	0.10	0.20	0.30	0.25	0.15

Compute the first four raw moments \(\mu'_1, \mu'_2, \mu'_3, \mu'_4\) and the central moments \(\mu_2, \mu_3, \mu_4\).

2. Aim

To compute the raw and central moments (including the mean and variance) of a discrete random variable from its pmf.

3. Formula

Applying it:

1. Verify that \(\sum p(x) = 1\).
2. Form the columns \(x\,p(x),\ x^2 p(x),\ x^3 p(x),\ x^4 p(x)\) and total them to get the raw moments.
3. Convert the raw moments to central moments using the standard relations below.

4. Calculation

Blank working table (fill in each product column and total it):

x	p(x)	x p(x)	x² p(x)	x³ p(x)	x⁴ p(x)
0	0.10
1	0.20
2	0.30
3	0.25
4	0.15
Σ	1.00

x	p(x)	x p(x)	x² p(x)	x³ p(x)	x⁴ p(x)
0	0.10	0	0	0	0
1	0.20	0.20	0.20	0.20	0.20
2	0.30	0.60	1.20	2.40	4.80
3	0.25	0.75	2.25	6.75	20.25
4	0.15	0.60	2.40	9.60	38.40
Σ	1.00	2.15	6.05	18.95	63.65

Raw moments: \(\mu'_1 = 2.15,\ \mu'_2 = 6.05,\ \mu'_3 = 18.95,\ \mu'_4 = 63.65\).

\(\mu_2 = 6.05 - 2.15^2 = 6.05 - 4.6225 = 1.4275\).

\(\mu_3 = 18.95 - 3(6.05)(2.15) + 2(2.15)^3 = 18.95 - 39.0225 + 19.8768 = -0.1957\).

\(\mu_4 = 63.65 - 4(18.95)(2.15) + 6(6.05)(2.15)^2 - 3(2.15)^4 = 63.65 - 162.97 + 167.7968 - 64.1025 = 4.3742\).

5. Result

Mean \(= \mu'_1 = 2.15\); variance \(= \mu_2 = 1.4275\); \(\mu_3 = -0.196\); \(\mu_4 = 4.374\). The small negative \(\mu_3\) indicates a slight left skew.

Experiment 2 — Skewness & Kurtosis from a PMF

1. Problem

Using the central moments obtained in Experiment 1 (\(\mu_2 = 1.4275,\ \mu_3 = -0.1957,\ \mu_4 = 4.3742\)), compute the coefficients of skewness and kurtosis.

2. Aim

To measure the skewness and kurtosis of a discrete distribution from its central moments.

3. Formula

Applying it:

1. Compute \(\beta_1\) and the signed skewness \(\gamma_1 = \sqrt{\beta_1}\) (with the sign of \(\mu_3\)).
2. Compute \(\beta_2\) and the excess kurtosis \(\gamma_2 = \beta_2 - 3\).
3. Interpret the sign of \(\gamma_1\) (direction of skew) and the value of \(\beta_2\) (peakedness).

4. Calculation

Blank working table (enter the moment inputs and the computed coefficients):

Quantity	Value
\(\mu_2\)
\(\mu_3\)
\(\mu_4\)
\(\beta_1,\ \gamma_1\)
\(\beta_2,\ \gamma_2\)

\(\beta_1 = \dfrac{(-0.1957)^2}{1.4275^3} = \dfrac{0.03832}{2.9089} = 0.0132\); \(\quad \gamma_1 = \dfrac{-0.1957}{1.4275^{3/2}} = \dfrac{-0.1957}{1.7055} = -0.115\).

\(\beta_2 = \dfrac{4.3742}{1.4275^2} = \dfrac{4.3742}{2.0378} = 2.147\); \(\quad \gamma_2 = 2.147 - 3 = -0.853\).

5. Result

\(\gamma_1 = -0.115\) — the distribution is slightly negatively (left) skewed; \(\beta_2 = 2.147\) (\(< 3\)), so it is platykurtic (\(\gamma_2 = -0.853\)).

Experiment 3 — Moments from a PDF

1. Problem

A continuous random variable \(X\) has the density \(f(x) = 6x(1-x),\ 0 \le x \le 1\) (this is the Beta(2, 2) density). Compute the first four raw moments and the central moments \(\mu_2, \mu_3, \mu_4\).

2. Aim

To compute the moments of a continuous random variable from its pdf using integration.

3. Formula

Applying it:

1. Verify that \(\int_0^1 f(x)\,dx = 1\).
2. Obtain the general raw moment \(\mu'_r = \int_0^1 x^r f(x)\,dx\) and evaluate it for \(r = 1,2,3,4\).
3. Convert the raw moments to central moments using the same relations as in Experiment 1.

4. Calculation

Blank working table (evaluate the raw moment for each order):

r	μ′r = 6 / [(r+2)(r+3)]
1
2
3
4

r	μ′r
1	6/(3 · 4) = 0.5
2	6/(4 · 5) = 0.3
3	6/(5 · 6) = 0.2
4	6/(6 · 7) ≈ 0.1429

\(\mu_2 = 0.3 - 0.5^2 = 0.05\).

\(\mu_3 = 0.2 - 3(0.3)(0.5) + 2(0.5)^3 = 0.2 - 0.45 + 0.25 = 0\).

\(\mu_4 = 0.1429 - 4(0.2)(0.5) + 6(0.3)(0.5)^2 - 3(0.5)^4 = 0.1429 - 0.4 + 0.45 - 0.1875 = 0.00536\).

5. Result

Mean \(= 0.5\); variance \(= \mu_2 = 0.05\); \(\mu_3 = 0\) (symmetric); \(\mu_4 = 0.00536\).

Experiment 4 — Skewness & Kurtosis from a PDF

1. Problem

Using the central moments of the Beta(2, 2) density from Experiment 3 (\(\mu_2 = 0.05,\ \mu_3 = 0,\ \mu_4 = 0.00536\)), compute the coefficients of skewness and kurtosis.

2. Aim

To measure the skewness and kurtosis of a continuous distribution from its central moments.

3. Formula

Applying it:

1. Compute \(\beta_1 = \mu_3^2/\mu_2^3\) (and \(\gamma_1 = \mu_3/\mu_2^{3/2}\)).
2. Compute \(\beta_2 = \mu_4/\mu_2^2\) and \(\gamma_2 = \beta_2 - 3\).

4. Calculation

Blank working table (enter the moment inputs and the computed coefficients):

Quantity	Value
\(\mu_2\)
\(\mu_4\)
\(\beta_1\)
\(\beta_2,\ \gamma_2\)

\(\beta_1 = \dfrac{0^2}{0.05^3} = 0\) — confirming Beta(2, 2) is symmetric about 0.5.

\(\beta_2 = \dfrac{0.00536}{0.05^2} = \dfrac{0.00536}{0.0025} = 2.14\); \(\quad \gamma_2 = 2.14 - 3 = -0.86\).

5. Result

\(\beta_1 = 0\) (symmetric); \(\beta_2 = 2.14\) (\(< 3\)), so the distribution is platykurtic (\(\gamma_2 = -0.86\)). This matches the known excess kurtosis of the Beta(2, 2) distribution, \(-6/7 \approx -0.857\).

Experiment 5 — Joint, Marginal & Conditional PMF; Independence

1. Problem

The joint pmf of \((X, Y)\) is:

X \ Y	1	2	3
0	1/12	2/12	1/12
1	2/12	3/12	3/12

(a) Find the marginal pmfs of \(X\) and \(Y\). (b) Find \(P(Y = 2 \mid X = 1)\). (c) Determine whether \(X\) and \(Y\) are independent.

2. Aim

To obtain marginal and conditional pmfs from a joint pmf and to test the independence of two discrete random variables.

3. Formula

Applying it:

1. Marginal of \(X\): sum each row; marginal of \(Y\): sum each column.
2. Conditional pmf: divide the joint probability by the relevant marginal.
3. Independence holds only if \(p(x, y) = p_X(x)\,p_Y(y)\) for every cell.

4. Calculation

Blank working table (fill in the row totals \(p_X\) and column totals \(p_Y\)):

X \ Y	1	2	3	Row total pX(x)
0	1/12	2/12	1/12
1	2/12	3/12	3/12
Col total pY(y)				1

(a) Marginals: \(p_X(0) = \tfrac{1+2+1}{12} = \tfrac{4}{12} = \tfrac13\), \(p_X(1) = \tfrac{2+3+3}{12} = \tfrac{8}{12} = \tfrac23\); \(p_Y(1) = \tfrac{3}{12} = \tfrac14\), \(p_Y(2) = \tfrac{5}{12}\), \(p_Y(3) = \tfrac{4}{12} = \tfrac13\).

(b) Conditional: \(P(Y = 2 \mid X = 1) = \dfrac{p(1,2)}{p_X(1)} = \dfrac{3/12}{8/12} = \dfrac38\).

(c) Independence check: \(p_X(0)\,p_Y(1) = \tfrac13 \cdot \tfrac14 = \tfrac1{12} = p(0,1)\) ✓, but \(p_X(1)\,p_Y(3) = \tfrac23 \cdot \tfrac13 = \tfrac29 \ne \tfrac{3}{12} = \tfrac14 = p(1,3)\).

5. Result

Marginals: \(p_X = (\tfrac13, \tfrac23)\) and \(p_Y = (\tfrac14, \tfrac5{12}, \tfrac13)\); \(P(Y=2 \mid X=1) = \tfrac38\). Since the factorisation fails for at least one cell, \(X\) and \(Y\) are not independent.

Experiment 6 — Joint, Marginal & Conditional PDF; Independence

1. Problem

(a) \(f(x, y) = 4xy,\ 0 \le x, y \le 1\): find the marginal densities, the conditional density \(f_{Y|X}(y \mid x)\), and check independence.
(b) \(f(x, y) = x + y,\ 0 \le x, y \le 1\): examine independence.

2. Aim

To obtain marginal and conditional densities from a joint pdf and to test the independence of two continuous random variables.

3. Formula

Applying it:

1. Marginal of \(X\): integrate the joint density over \(y\); marginal of \(Y\): integrate over \(x\).
2. Conditional density: divide the joint density by the marginal.
3. Independence holds iff \(f(x, y) = f_X(x)\,f_Y(y)\) throughout the support.

4. Calculation

Blank working table (write each density expression):

Quantity	(a) f = 4xy	(b) f = x + y
\(f_X(x)\)
\(f_Y(y)\)
\(f_X(x)\,f_Y(y)\)
Independent?

(a) \(f_X(x) = \int_0^1 4xy\,dy = 4x \cdot \tfrac12 = 2x\) and, by symmetry, \(f_Y(y) = 2y\) (\(0 \le x, y \le 1\)).
Conditional: \(f_{Y|X}(y \mid x) = \dfrac{4xy}{2x} = 2y = f_Y(y)\). Also \(f_X(x)\,f_Y(y) = 2x \cdot 2y = 4xy = f(x, y)\) ✓.

(b) \(f_X(x) = \int_0^1 (x + y)\,dy = x + \tfrac12\) and \(f_Y(y) = y + \tfrac12\).
Product \((x + \tfrac12)(y + \tfrac12) \ne x + y\).

5. Result

(a) Since \(f(x, y) = f_X(x)\,f_Y(y)\), \(X\) and \(Y\) are independent. (b) The factorisation fails, so \(X\) and \(Y\) are not independent.

Experiment 7 — Chebyshev's Inequality Applications

1. Problem

(a) The mean and SD of the marks of 1 000 students are 60 and 8. Find a lower bound for the number of students whose marks lie between 44 and 76.
(b) For i.i.d. variables with variance \(\sigma^2 = 4\), determine the sample size \(n\) so that the sample mean is within 0.5 of \(\mu\) with probability at least 0.95.

2. Aim

To apply Chebyshev's inequality to bound probabilities and to determine a required sample size.

3. Formula

Applying it:

1. Express the given interval as \(\mu \pm k\sigma\) to identify \(k\).
2. Apply Chebyshev's inequality; for a sample mean use \(\operatorname{Var}(\bar X) = \sigma^2/n\).
3. Convert the probability bound into the required count or sample size.

4. Calculation

Blank working table (record the given quantities):

Quantity	(a)	(b)
Mean \(\mu\)	60	—
SD \(\sigma\) / variance	8	\(\sigma^2 = 4\)
Tolerance		0.5
\(k\) / required bound		0.05

(a) The interval \([44, 76]\) is \(60 \pm 16 = \mu \pm 2\sigma\), so \(k = 2\).
\(P(|X - 60| < 16) \ge 1 - \tfrac{1}{2^2} = 1 - \tfrac14 = 0.75\).
Expected count \(\ge 0.75 \times 1000 = 750\).

(b) \(\operatorname{Var}(\bar X) = \sigma^2/n = 4/n\). By Chebyshev, \(P(|\bar X - \mu| \ge 0.5) \le \dfrac{4/n}{0.5^2} = \dfrac{16}{n}\).
Require \(\dfrac{16}{n} \le 0.05 \Rightarrow n \ge 320\).

5. Result

(a) At least 750 of the 1 000 students score between 44 and 76. (b) A sample size of \(n \ge \mathbf{320}\) is required.

Lab Record Format (to be followed for every experiment)

1. Problem — the data / distribution 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 a brief interpretation.