AL-302 Introduction to Probability and Statistics - Jun 2024 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aDefine discrete random variables. If $X$ and $Y$ are any two random variables, then show that $E(X+Y) = E(X) + E(Y)$ Provided $E(X)$ and $E(Y)$ exist.
   How to answer

   • Define discrete random variable with range/values and probability mass function (PMF).
   • State definition of expectation for joint distribution: $E(X+Y) = \sum_x \sum_y (x+y) p(x,y)$.
   • Split summation: $\sum_x \sum_y x p(x,y) + \sum_x \sum_y y p(x,y)$.
   • Use marginal probability properties $\sum_y p(x,y) = p_X(x)$ and $\sum_x p(x,y) = p_Y(y)$.
   • Conclude $E(X+Y) = E(X) + E(Y)$.
2. 7 Marks
   bIf $X$ is a continuous random variable and $Y = aX + b$. Prove that $E(Y) = aE(X) + b$ and $V(Y) = a^2 \cdot V(X)$, where $V$ stands for variance and $a, b$ are constants.
   How to answer

   • Use definition $E(Y) = \int_{-\infty}^\infty (ax+b) f(x) dx$.
   • Split into $a \int_{-\infty}^\infty x f(x) dx + b \int_{-\infty}^\infty f(x) dx = aE(X) + b$.
   • Use definition of variance $V(Y) = E[(Y - E(Y))^2]$.
   • Substitute $Y - E(Y) = (aX+b) - (aE(X)+b) = a(X - E(X))$.
   • Evaluate $V(Y) = E[a^2 (X - E(X))^2] = a^2 V(X)$.
3. 7 Marks
   aA sample of 4 items is selected at random from a box containing 12 items of which 5 are defective. Find the expected number $E$ of defective items.
   How to answer

   • Identify probability distribution (Hypergeometric distribution without replacement).
   • Write formula for expected value: $E(X) = n \cdot p = n \cdot \frac{M}{N}$.
   • Substitute values: $n = 4$, $M = 5$, $N = 12$.
   • Calculate $E(X) = 4 \times \frac{5}{12} = \frac{5}{3} \approx 1.67$.
4. 7 Marks
   bLet $f(X)$ be the probability density function of a continuous random variable $X$. Then explain Mean, Median, Mode, variance and mean deviation.
   How to answer

   • Define Mean: $\mu = E(X) = \int_{-\infty}^\infty x f(x) dx$.
   • Define Median: Value $M$ such that $\int_{-\infty}^M f(x) dx = 0.5$.
   • Define Mode: Value of $x$ where $f(x)$ is maximum, i.e., $f'(x) = 0$ and $f''(x) < 0$.
   • Define Variance: $\sigma^2 = \int_{-\infty}^\infty (x - \mu)^2 f(x) dx$.
   • Define Mean Deviation: $MD = \int_{-\infty}^\infty |x - \text{Average}| f(x) dx$ (about mean or median).
5. 14 Marks
   The probability density $f(X)$ of a continuous random variable is given by $f(x) = C e^{-|x|}, -\infty < x < \infty$. Show that $C = 1/2$ and find the mean and Variance of the distribution. Also find the probability that the variate lies between 0 and 4.
   How to answer

   • Show normalization: $\int_{-\infty}^\infty C e^{-|x|} dx = 2C \int_0^\infty e^{-x} dx = 2C(1) = 1 \implies C = 1/2$.
   • Calculate Mean: Since $x f(x) = \frac{1}{2} x e^{-|x|}$ is an odd function, $E(X) = 0$.
   • Calculate Variance: $E(X^2) = 2 \int_0^\infty \frac{1}{2} x^2 e^{-x} dx = \Gamma(3) = 2! = 2$, so $V(X) = 2 - 0 = 2$.
   • Calculate probability: $P(0 < X < 4) = \int_0^4 \frac{1}{2} e^{-x} dx = \frac{1}{2} [-e^{-x}]_0^4 = \frac{1}{2} (1 - e^{-4})$.
6. 14 Marks
   State and prove Bayes theorem. Suppose 5 men out of 100 and 25 Women out of 10,000 are color blind. A color blind person is chosen at random. What is the probability of the person being a male. (Assuming male and female to be in equal numbers).
   How to answer

   • State Bayes' Theorem clearly.
   • Provide mathematical proof using $P(A_i \cap B) = P(B) P(A_i|B) = P(A_i) P(B|A_i)$ and Total Probability Theorem.
   • Set up numerical events: $M$: Male, $W$: Female, $C$: Color blind.
   • Probabilities: $P(M) = 0.5$, $P(W) = 0.5$, $P(C|M) = 5/100 = 0.05$, $P(C|W) = 25/10000 = 0.0025$.
   • Apply Bayes' formula: $P(M|C) = \frac{P(M)P(C|M)}{P(M)P(C|M) + P(W)P(C|W)} = \frac{0.5 \times 0.05}{0.5 \times 0.05 + 0.5 \times 0.0025} = \frac{0.05}{0.0525} = \frac{20}{21} \approx 0.952$.
7. 7 Marks
   aThe first four central moments of a distribution are $0, 2.5, 0.7$ and $1.75$. Calculate $\beta_1, \beta_2$.
   How to answer

   • Write formula for skewness coefficient: $\beta_1 = \frac{\mu_3^2}{\mu_2^3}$.
   • Calculate $\beta_1 = \frac{(0.7)^2}{(2.5)^3} = \frac{0.49}{15.625} \approx 0.03136$.
   • Write formula for kurtosis coefficient: $\beta_2 = \frac{\mu_4}{\mu_2^2}$.
   • Calculate $\beta_2 = \frac{1.75}{(2.5)^2} = \frac{1.75}{6.25} = 0.28$.
8. 7 Marks
   bThe mean and variance of a binomial variable $X$ with parameters $n$ and $p$ are 16 and 8. Find : $P(x \ge 1)$ and $P(x > 2)$.
   How to answer

   • Find $q$: $\frac{npq}{np} = q = \frac{8}{16} = 0.5$.
   • Find $p = 1 - q = 0.5$ and $n = \frac{16}{0.5} = 32$.
   • Calculate $P(X=0) = \binom{32}{0} (0.5)^{32} = (0.5)^{32}$. Thus $P(X \ge 1) = 1 - (0.5)^{32}$.
   • Calculate $P(X=1) = 32(0.5)^{32}$ and $P(X=2) = \binom{32}{2}(0.5)^{32} = 496(0.5)^{32}$.
   • Calculate $P(X > 2) = 1 - (1 + 32 + 496)(0.5)^{32} = 1 - 529 \times (0.5)^{32}$.
9. 7 Marks
   aIf $X$ is normally distributed with mean 2 and variance 0.1, then find $P(|x - 2| \ge 0.01)$.
   How to answer

   • Given $\mu = 2$, $\sigma = \sqrt{0.1} \approx 0.3162$.
   • Transform: $P(|X - 2| \ge 0.01) = P\left(|Z| \ge \frac{0.01}{0.3162}\right) = P(|Z| \ge 0.0316)$.
   • Express in terms of standard normal distribution: $P(|Z| \ge 0.0316) = 2 [1 - \Phi(0.0316)]$.
   • Compute or look up value: $\Phi(0.0316) \approx 0.5126$, yielding probability $\approx 2(1 - 0.5126) = 0.9748$.
10. 7 Marks
    bExplain comparison between Correlation and Regression.
    How to answer

    • Define Correlation (degree of association) and Regression (functional relationship/prediction).
    • Symmetry: Correlation is symmetric ($r_{xy} = r_{yx}$); regression is non-symmetric ($b_{yx} \ne b_{xy}$).
    • Units: Correlation is dimensionless; regression coefficients have physical units.
    • Range: $r$ lies between -1 and 1; regression coefficients can take any real value (subject to $b_{yx} b_{xy} \le 1$).
    • Present comparison in a clear tabular format.
11. 7 Marks
    aFit a parabola for the following data. $$ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 1 & 1.8 & 1.3 & 2.5 & 6.3 \\ \hline \end{array} $$
    How to answer

    • State second-degree equation: $y = a + bx + cx^2$.
    • Set up normal equations using principle of least squares.
    • Compute table of sums: $N=5$, $\sum x=10$, $\sum x^2=30$, $\sum x^3=100$, $\sum x^4=354$, $\sum y=12.9$, $\sum xy=37.1$, $\sum x^2 y=131.7$.
    • Form simultaneous linear equations in $a, b, c$.
    • Solve the system to obtain values of $a, b, c$ and write the final fitted parabola equation.
12. 7 Marks
    bIn a big city 325 men out of 600 men were found to be smokers. Does this information support the conclusion that the majority of men in this city are smokers.
    How to answer

    • Formulate hypotheses: $H_0: P = 0.5$ vs $H_1: P > 0.5$ (one-tailed test).
    • Sample proportion: $p = 325/600 \approx 0.5417$.
    • Calculate standard error: $SE = \sqrt{\frac{0.5 \times 0.5}{600}} = \sqrt{\frac{0.25}{600}} \approx 0.0204$.
    • Calculate test statistic: $z = \frac{0.5417 - 0.5}{0.0204} \approx 2.04$.
    • Compare with critical value $z_{0.05} = 1.645$ (or $z_{0.01} = 2.33$). Since $2.04 > 1.645$, reject $H_0$ at 5% significance level and conclude that data supports the majority being smokers.
13. 7 Marks
    aThe table below give the number of air craft accidents that occurred during the various days of the week. Test whether the accidents are uniformly distributed over the week. $$ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline \text{Days} & \text{Mon.} & \text{Tue.} & \text{Wed.} & \text{Thu.} & \text{Fri.} & \text{Sat.} & \text{Sun.} & \text{Total no. accidents} \\ \hline \text{No. of accidents} & 14 & 18 & 12 & 11 & 15 & 14 & 14 & 98 \\ \hline \end{array} $$
    How to answer

    • State $H_0$: Accidents are uniformly distributed over the week.
    • Compute expected frequency for each of the 7 days: $E_i = 98/7 = 14$.
    • Calculate $(O_i - E_i)^2 / E_i$: $\frac{(14-14)^2}{14} + \frac{(18-14)^2}{14} + \frac{(12-14)^2}{14} + \frac{(11-14)^2}{14} + \frac{(15-14)^2}{14} + \frac{(14-14)^2}{14} + \frac{(14-14)^2}{14} = \frac{0 + 16 + 4 + 9 + 1 + 0 + 0}{14} = \frac{30}{14} \approx 2.143$.
    • Degrees of freedom: $df = 7 - 1 = 6$.
    • Critical value of $\chi^2$ at $df=6$ and $\alpha=0.05$ is $12.59$.
    • Conclusion: Since $\chi^2_{cal} = 2.143 < 12.59$, accept $H_0$ (accidents are uniformly distributed).
14. 7 Marks
    bShow that the maximum value of rank Correlation coefficient is 1.
    How to answer

    • State Spearman's rank correlation formula: $\rho = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}$.
    • Note that $d_i = R_{1i} - R_{2i}$ represents difference between ranks.
    • Since $d_i^2 \ge 0$ for all $i$, $\sum d_i^2 \ge 0$.
    • The minimum possible value of $\sum d_i^2$ is 0, which occurs when ranks are identical ($R_{1i} = R_{2i}$ for all $i$).
    • Substitute $\sum d_i^2 = 0$ to get $\rho_{\max} = 1 - 0 = 1$.