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How to use this manual: In the lab, copy the blank working table (the one with only headings) at the start of each experiment's Calculation and fill it in yourself from the given data. The Calculation section then shows the completed table and the substitution so you can check your work.

List of Practical Experiments (Official Syllabus)

  1. Calculation of moments of Uniform distribution.
  2. Calculation of skewness and kurtosis of Uniform distribution.
  3. Fitting of Exponential distribution.
  4. Gamma distribution application-oriented problems.
  5. Fitting of Normal distribution — Areas method.
  6. Fitting of Normal distribution — Ordinates method.
  7. Problems related to Standard Normal distribution.

Experiment 1 — Moments of a Uniform Distribution

1. Problem

(a) For \(X \sim U(2, 8)\), compute the first four central moments, the standard deviation and the mean deviation. (b) For \(X \sim U(0, 10)\), find \(E(X^3)\).

2. Aim

To compute the moments and related measures of a continuous Uniform distribution.

3. Formula

\[ \mu = \frac{a+b}{2}, \qquad \mu_2 = \frac{(b-a)^2}{12}, \qquad \mu_3 = 0, \qquad \mu_4 = \frac{(b-a)^4}{80} \] \[ \sigma = \sqrt{\mu_2}, \qquad \text{MD} = \frac{b-a}{4}, \qquad E(X^r) = \int_a^b \frac{x^r}{b-a}\,dx \]

Applying it:

  1. Identify the limits \(a, b\) and the range \(b - a\).
  2. Use the standard results for the mean, variance and higher central moments of \(U(a, b)\).
  3. For a raw moment such as \(E(X^3)\), integrate \(x^3 f(x)\) over the support, with \(f(x) = 1/(b-a)\).

4. Calculation

Blank working table (fill in each measure for \(U(2,8)\)):

MeasureValue
Mean \(\mu\)
\(\mu_2\) (variance)
\(\mu_3\)
\(\mu_4\)
SD \(\sigma\)
MD

(a) \(a = 2,\ b = 8,\ b - a = 6\). \(\mu = (2+8)/2 = 5\); \(\mu_2 = 6^2/12 = 3\); \(\mu_3 = 0\); \(\mu_4 = 6^4/80 = 1296/80 = 16.2\); \(\sigma = \sqrt{3} = 1.732\); \(\text{MD} = 6/4 = 1.5\).

(b) \(E(X^3) = \displaystyle\int_0^{10} \frac{x^3}{10}\,dx = \frac{1}{10}\cdot\frac{10^4}{4} = 250\).

5. Result

For \(U(2, 8)\): \(\mu = 5,\ \mu_2 = 3,\ \mu_3 = 0,\ \mu_4 = 16.2,\ \sigma = 1.732,\ \text{MD} = 1.5\). For \(U(0, 10)\): \(E(X^3) = 250\).

Experiment 2 — Skewness & Kurtosis of a Uniform Distribution

1. Problem

Using the central moments of a Uniform distribution (from Experiment 1, \(\mu_2 = 3,\ \mu_3 = 0,\ \mu_4 = 16.2\) for \(U(2,8)\)), compute the coefficients of skewness \(\beta_1\) and kurtosis \(\beta_2\).

2. Aim

To measure the skewness and kurtosis of a Uniform distribution and show that its shape is independent of the limits \(a, b\).

3. Formula

\[ \beta_1 = \frac{\mu_3^2}{\mu_2^3}, \qquad \beta_2 = \frac{\mu_4}{\mu_2^2}, \qquad \gamma_2 = \beta_2 - 3 \]

Applying it:

  1. Compute \(\beta_1 = \mu_3^2/\mu_2^3\) and \(\beta_2 = \mu_4/\mu_2^2\); then \(\gamma_2 = \beta_2 - 3\).
  2. Note that for any \(U(a, b)\) the range cancels, so \(\beta_1\) and \(\beta_2\) are constants.

4. Calculation

Blank working table (fill in the coefficients):

QuantityValue
\(\beta_1\)
\(\beta_2\)
\(\gamma_2\)

\(\beta_1 = 0^2/3^3 = 0\) (symmetric).

\(\beta_2 = 16.2/3^2 = 16.2/9 = 1.8\); \(\gamma_2 = 1.8 - 3 = -1.2\).

General check: \(\beta_2 = \dfrac{(b-a)^4/80}{[(b-a)^2/12]^2} = \dfrac{144}{80} = 1.8\) — independent of \(a, b\).

5. Result

\(\beta_1 = 0\) (perfectly symmetric) and \(\beta_2 = 1.8\) (\(< 3\)), so the Uniform distribution is strongly platykurtic (\(\gamma_2 = -1.2\)), and this shape holds for every \(U(a, b)\).

Experiment 3 — Fitting an Exponential Distribution

1. Problem

The time (minutes) between successive arrivals at a bank counter, recorded for 100 inter-arrival intervals:

Class (min)0–22–44–66–88–1010+
Frequency f4025161063

Fit an Exponential distribution.

2. Aim

To fit an Exponential distribution to grouped waiting-time data and obtain the expected frequencies.

3. Formula

\[ \hat\lambda = \frac{1}{\bar x}, \qquad P(a \le X < b) = e^{-\hat\lambda a} - e^{-\hat\lambda b}, \qquad P(X \ge a) = e^{-\hat\lambda a} \]

Applying it:

  1. Take the class mid-points and compute the mean \(\bar x\); estimate \(\hat\lambda = 1/\bar x\).
  2. Find each class probability from the Exponential CDF, and \(\hat f = N\,P\).

4. Calculation

Blank working table (compute the mid-points, probabilities and expected frequencies):

ClassMid-pointProbabilityExpected f = 100·PObserved f
0–240
2–425
4–616
6–810
8–106
10+3

Mid-points 1, 3, 5, 7, 9, 11. \(\bar x = (40 + 75 + 80 + 70 + 54 + 33)/100 = 352/100 = 3.52\); \(\hat\lambda = 1/3.52 = 0.284\).

ClassProbabilityExpected fObserved f
0–21 − e−0.568 = 0.433443.340
2–4e−0.568 − e−1.136 = 0.245624.625
4–6e−1.136 − e−1.704 = 0.139113.916
6–8e−1.704 − e−2.272 = 0.07887.910
8–10e−2.272 − e−2.840 = 0.04474.56
10+e−2.840 = 0.05845.83
Total1.000100.0100

5. Result

The fitted Exponential distribution has \(\hat\lambda = 0.284\) per minute (mean inter-arrival time 3.52 min). The expected frequencies agree reasonably with the observed, so the Exponential model is acceptable.

Experiment 4 — Gamma Distribution Applications

1. Problem

(a) Let \(X \sim \text{Gamma}(r = 3, \lambda = 2)\) (the waiting time for the third call when calls arrive at rate 2 per hour). Find \(E(X)\), \(\operatorname{Var}(X)\) and \(P(X \le 1)\).
(b) If \(X_1 \sim \text{Gamma}(2, 1)\) and \(X_2 \sim \text{Gamma}(3, 1)\) are independent, find the distribution of \(X_1 + X_2\).

2. Aim

To apply the mean/variance formulas and the additive property of the Gamma distribution.

3. Formula

\[ E(X) = \frac{r}{\lambda}, \qquad \operatorname{Var}(X) = \frac{r}{\lambda^2}, \qquad P(X \le t) = 1 - e^{-\lambda t}\sum_{k=0}^{r-1}\frac{(\lambda t)^k}{k!} \] \[ X_1 \sim \text{Gamma}(r_1, \lambda),\ X_2 \sim \text{Gamma}(r_2, \lambda) \ \Rightarrow\ X_1 + X_2 \sim \text{Gamma}(r_1 + r_2, \lambda) \]

Applying it:

  1. For \(\text{Gamma}(r, \lambda)\): mean \(= r/\lambda\), variance \(= r/\lambda^2\).
  2. For integer \(r\), the CDF has the closed form below (a finite Poisson-type sum).
  3. Use the additive property: independent Gammas with a common \(\lambda\) add their shape parameters.

4. Calculation

Blank working table (fill in the results):

QuantityValue
\(E(X)\), \(\operatorname{Var}(X)\) — part (a)
\(P(X \le 1)\) — part (a)
Distribution of \(X_1 + X_2\) — part (b)

(a) \(E(X) = 3/2 = 1.5\) h; \(\operatorname{Var}(X) = 3/4 = 0.75\). With \(f(x) = \dfrac{2^3}{2!}x^2 e^{-2x} = 4x^2 e^{-2x}\),

\(P(X \le 1) = 1 - e^{-2}\left(1 + 2 + \dfrac{2^2}{2!}\right) = 1 - e^{-2}(1 + 2 + 2) = 1 - 5e^{-2} \approx 0.3233\).

(b) Both have \(\lambda = 1\), so \(X_1 + X_2 \sim \text{Gamma}(2 + 3, 1) = \text{Gamma}(5, 1)\), with mean 5 and variance 5.

5. Result

(a) \(E(X) = 1.5\) h, \(\operatorname{Var}(X) = 0.75\), \(P(X \le 1) \approx 0.3233\). (b) \(X_1 + X_2 \sim \text{Gamma}(5, 1)\).

Experiment 5 — Fitting a Normal Distribution (Areas Method)

1. Problem

Heights (cm) of 200 students:

Class150–155155–160160–165165–170170–175175–180
f104060503010

Fit a Normal distribution by the areas method.

2. Aim

To fit a Normal distribution to grouped data by computing the area (probability) under the normal curve for each class.

3. Formula

\[ z = \frac{x - \bar x}{s}, \qquad P(\text{class}) = \Phi(z_{\text{upper}}) - \Phi(z_{\text{lower}}), \qquad \hat f = N\,P(\text{class}) \]

Applying it:

  1. Compute the mean \(\bar x\) and standard deviation \(s\) from the grouped data.
  2. Convert each class boundary to a standard value \(z = (x - \bar x)/s\) and read \(\Phi(z)\).
  3. Class probability = difference of successive \(\Phi(z)\). Take the first class as "\(\le\) upper boundary" and the last as "\(\ge\) lower boundary" so the tails are included and the expected frequencies sum to \(N\).
  4. Expected frequency \(= N \times\) (class probability).

4. Calculation

Blank working table (fill in z, Φ(z) and the class probabilities/expected frequencies):

Upper boundaryzΦ(z)
155
160
165
170
175

Mid-points 152.5, …, 177.5. \(\bar x = 32\,900/200 = 164.5\); \(\sum f(x - \bar x)^2 = 7700\), so \(s = \sqrt{7700/200} = \sqrt{38.5} = 6.2\).

Upper boundaryz = (x − 164.5)/6.2Φ(z)
155−1.530.0630
160−0.730.2327
1650.080.5319
1700.890.8133
1751.690.9545
ClassP(class)Expected = 200·PObserved
≤ 1550.063012.610
155–1600.169733.940
160–1650.299259.860
165–1700.281456.350
170–1750.141228.230
≥ 1750.04559.110
Total1.000200.0200

5. Result

The fitted Normal distribution has \(\hat\mu = 164.5,\ \hat\sigma = 6.2\). The expected frequencies (12.6, 33.9, 59.8, 56.3, 28.2, 9.1) sum to 200 and follow the observed pattern, so the heights are approximately Normal.

Experiment 6 — Fitting a Normal Distribution (Ordinates Method)

1. Problem

For the same height data (with \(\hat\mu = 164.5,\ \hat\sigma = 6.2,\ N = 200,\ h = 5\)), fit a Normal distribution by the ordinates method.

2. Aim

To fit a Normal distribution using the ordinate (density) at each class mid-point, and to compare it with the areas method.

3. Formula

\[ \phi(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}, \qquad \hat f_i \approx \frac{N h}{\hat\sigma}\,\phi\!\left(\frac{x_i - \hat\mu}{\hat\sigma}\right) \]

Applying it:

  1. For each mid-point \(x_i\), compute \(z = (x_i - \hat\mu)/\hat\sigma\) and the standard normal ordinate \(\phi(z)\).
  2. Multiply by the constant \(Nh/\hat\sigma\) to obtain the expected frequency.

4. Calculation

Blank working table (fill in z, φ(z) and the expected frequencies; \(Nh/\hat\sigma = 161.29\)):

Mid xzφ(z)Expected f = 161.29·φ(z)
152.5
157.5
162.5
167.5
172.5
177.5

Coefficient \(Nh/\hat\sigma = 200 \times 5 / 6.2 = 161.29\).

Mid xzφ(z)Expected f
152.5−1.940.06089.81
157.5−1.130.210733.99
162.5−0.320.379061.13
167.50.480.355557.34
172.51.290.173628.00
177.52.100.04407.10

5. Result

The ordinate-method expected frequencies (9.8, 34.0, 61.1, 57.3, 28.0, 7.1) are very close to those from the areas method, confirming the Normal fit with \(\hat\mu = 164.5,\ \hat\sigma = 6.2\).

Experiment 7 — Standard Normal Distribution Problems

1. Problem

(a) Marks of 1000 students follow \(N(60, 100)\). How many score between 50 and 75?
(b) IQ in a population follows \(N(100, 225)\). Find the IQ exceeded by only 5 % of the population.

2. Aim

To solve probability and percentile problems by standardising to the standard normal variable \(Z\).

3. Formula

\[ z = \frac{x - \mu}{\sigma}, \qquad P(x_1 < X < x_2) = \Phi(z_2) - \Phi(z_1), \qquad x = \mu + z\sigma \]

Applying it:

  1. Standardise: \(z = (x - \mu)/\sigma\) (here \(\sigma = \sqrt{\text{variance}}\)).
  2. For a probability, use \(\Phi\) values; for a percentile, invert \(\Phi\) to get \(z\), then \(x = \mu + z\sigma\).

4. Calculation

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

Partz value(s)Probability / IQ
(a)
(b)

(a) \(\sigma = 10\); \(z_1 = (50-60)/10 = -1\), \(z_2 = (75-60)/10 = 1.5\). \(P = \Phi(1.5) - \Phi(-1) = 0.9332 - 0.1587 = 0.7745\). Number \(= 0.7745 \times 1000 = 774.5 \approx 775\).

(b) \(\sigma = 15\); require \(P(X > x) = 0.05\), i.e. \(\Phi(z) = 0.95 \Rightarrow z = 1.645\). \(x = 100 + 1.645 \times 15 = 124.7\).

5. Result

(a) About 775 students score between 50 and 75. (b) An IQ of about 125 marks the top 5 % of the population.

Lab Record Format (to be followed for every experiment)

  1. Problem — the distribution / 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 — parameter estimation, the completed table and the substitution.
  5. Result — the final value(s) with a brief interpretation.
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