AL-302 Introduction to Probability and Statistics - Jun 2025 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
   aDetermine (i) $P\left(\frac{B}{A}\right)$ (ii) $P\left(\frac{A}{B^C}\right)$ if A and B are events with $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$
   How to answer

   • Calculate $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 1/3 + 1/4 - 1/2 = 1/12$
   • Calculate $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{1/12}{1/3} = 1/4$
   • Calculate $P(A \cap B^C) = P(A) - P(A \cap B) = 1/3 - 1/12 = 3/12 = 1/4$
   • Calculate $P(B^C) = 1 - P(B) = 1 - 1/4 = 3/4$
   • Calculate $P(A/B^C) = \frac{P(A \cap B^C)}{P(B^C)} = \frac{1/4}{3/4} = 1/3$
2. 7 Marks
   b

   A random variable X has the following probability function.

   $X_i$	-3	-2	-1	0	1	2	3
   $P(X_i)$	K	0.1	K	0.2	2K	0.4	2K

   Find i) K ii) Mean iii) Variance

   How to answer

   • Set sum of probabilities equal to 1: $K + 0.1 + K + 0.2 + 2K + 0.4 + 2K = 1 \implies 6K + 0.7 = 1 \implies K = 0.05$
   • Calculate Mean: $E(X) = \sum X_i P(X_i)$
   • Calculate $E(X^2) = \sum X_i^2 P(X_i)$
   • Calculate Variance: $\text{Var}(X) = E(X^2) - [E(X)]^2$
3. 7 Marks
   aDefine Moments. Explain types of moments.
   How to answer

   • Definition of moments in statistics as expected values of powers of a random variable.
   • Explain raw moments (moments about origin) with formula: $\mu'_r = E(X^r)$ or $\frac{1}{N}\sum f_i x_i^r$.
   • Explain central moments (moments about mean) with formula: $\mu_r = E[(X - \mu)^r]$.
   • Explain moments about an arbitrary point 'a': $\mu'_r(a) = E[(X - a)^r]$.
   • Briefly state relations between central and raw moments (e.g., $\mu_2 = \mu'_2 - (\mu'_1)^2$).
4. 7 Marks
   bFind the mean and standard deviation of a normal distribution in which 7% of items are under 35 and 89% are under 63.
   How to answer

   • Convert given probabilities into standard normal variable $Z$: $P(X < 35) = P\left(Z < \frac{35-\mu}{\sigma}\right) = 0.07$ and $P(X < 63) = P\left(Z < \frac{63-\mu}{\sigma}\right) = 0.89$.
   • From standard normal distribution table: $z_1 = -1.476$ (or $-1.48$) and $z_2 = 1.227$ (or $1.23$).
   • Set up equations: $\frac{35 - \mu}{\sigma} = -1.48$ and $\frac{63 - \mu}{\sigma} = 1.23$.
   • Solve simultaneously to obtain $\mu$ and $\sigma$.
5. 7 Marks
   aThe density function of a random variable X is $f(x) = e^{-x}, x \ge 0$. Find $E(x)$, $E(x^2)$ and variance of X.
   How to answer

   • Calculate $E(X) = \int_0^\infty x e^{-x} dx = 1! = 1$.
   • Calculate $E(X^2) = \int_0^\infty x^2 e^{-x} dx = 2! = 2$.
   • Calculate $\text{Var}(X) = E(X^2) - [E(X)]^2 = 2 - 1^2 = 1$.
6. 7 Marks
   bDefine gamma distribution. Find the mean and Variance of gamma distribution.
   How to answer

   • Define gamma distribution with PDF: $f(x) = \frac{\lambda^k}{\Gamma(k)} x^{k-1} e^{-\lambda x}$ for $x > 0$ (or standard form with single parameter).
   • Derive Mean: $E(X) = \int_0^\infty x f(x) dx = \frac{k}{\lambda}$.
   • Derive $E(X^2) = \int_0^\infty x^2 f(x) dx = \frac{k(k+1)}{\lambda^2}$.
   • Calculate Variance: $\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{k}{\lambda^2}$.
7. 7 Marks
   aTwo factories produce identical clocks. The production of the first factory consists of 10,000 clocks of which 100 are defective. The second factory produces 20,000 clocks of which 300 are defective. What is the probability that a particular defective clock was produced in the first factory.
   How to answer

   • Define events: $F_1$ = produced by factory 1, $F_2$ = produced by factory 2, $D$ = defective clock.
   • Compute prior probabilities: $P(F_1) = 10000/30000 = 1/3$, $P(F_2) = 20000/30000 = 2/3$.
   • Compute likelihoods: $P(D|F_1) = 100/10000 = 0.01$, $P(D|F_2) = 300/20000 = 0.015$.
   • Apply Bayes' Theorem: $P(F_1|D) = \frac{P(F_1)P(D|F_1)}{P(F_1)P(D|F_1) + P(F_2)P(D|F_2)} = \frac{100}{100 + 300} = \frac{100}{400} = 0.25$ or $1/4$.
8. 7 Marks
   bExplain bivariate distributions and their properties.
   How to answer

   • Define bivariate distribution involving two random variables $(X, Y)$.
   • Explain joint PMF for discrete case $P(X=x, Y=y)$ and joint PDF $f(x,y)$ for continuous case.
   • List properties: $f(x,y) \ge 0$, $\iint f(x,y) dx dy = 1$.
   • Define marginal distributions $f_X(x)$ and $f_Y(y)$.
   • Define conditional distributions and statistical independence.
9. 7 Marks
   aIf the variance of a Poisson variate is 3, then find $P(x=0)$, $P(0 < x \le 3)$ and $P(1 \le x < 4)$.
   How to answer

   • Identify parameter $\lambda = 3$ since for a Poisson distribution, $\text{Mean} = \text{Variance} = \lambda = 3$.
   • Calculate $P(X=0) = \frac{e^{-3} 3^0}{0!} = e^{-3} \approx 0.0498$.
   • Calculate $P(0 < X \le 3) = P(X=1) + P(X=2) + P(X=3) = e^{-3}(3 + 4.5 + 4.5) = 12 e^{-3} \approx 0.5974$.
   • Calculate $P(1 \le X < 4) = P(X=1) + P(X=2) + P(X=3) = 12 e^{-3} \approx 0.5974$.
10. 7 Marks
    b

    Fit a straight line to the form $y = a + bx$ for the following data.

    X	0	5	10	15	20	25
    Y	12	15	17	22	24	30

    How to answer

    • Form normal equations: $\sum Y = na + b\sum X$ and $\sum XY = a\sum X + b\sum X^2$ with $n = 6$.
    • Compute summations: $\sum X = 75$, $\sum Y = 120$, $\sum X^2 = 1375$, $\sum XY = 1810$.
    • Substitute values: $6a + 75b = 120$ and $75a + 1375b = 1810$.
    • Solve for $a$ and $b$ ($b = 0.709$, $a = 11.14$).
    • Write the fitted equation: $y = 11.14 + 0.709x$.
11. 7 Marks
    a20% of items produced from a factory are defective. Find the probability that in a sample of 5 choosen at random. i) none is defective ii) one is defective iii) $P(1 < x < 4)$
    How to answer

    • State binomial distribution parameters: $n = 5, p = 0.2, q = 0.8$.
    • i) Calculate $P(X = 0) = ^5C_0 (0.2)^0 (0.8)^5 = (0.8)^5 = 0.32768$.
    • ii) Calculate $P(X = 1) = ^5C_1 (0.2)^1 (0.8)^4 = 5 \times 0.2 \times 0.4096 = 0.4096$.
    • iii) Calculate $P(1 < X < 4) = P(X = 2) + P(X = 3) = ^5C_2(0.2)^2(0.8)^3 + ^5C_3(0.2)^3(0.8)^2 = 10(0.04)(0.512) + 10(0.008)(0.64) = 0.2048 + 0.0512 = 0.256$.
12. 7 Marks
    b

    Calculate the coefficient of rank correlation from the below data.

    X	68	64	75	50	64	80	75	40	55	64
    Y	62	58	68	45	81	60	68	48	50	70

    How to answer

    • Assign ranks to series $X$ (noting tied values for 64 repeated 3 times, 75 repeated 2 times).
    • Assign ranks to series $Y$ (noting tied values for 68 repeated 2 times).
    • Calculate differences $d_i = R_{X_i} - R_{Y_i}$ and $d_i^2$.
    • Apply tie correction factor $\frac{m(m^2-1)}{12}$ for each tied rank group.
    • Substitute values into Spearman's formula and calculate $r_s$.
13. 7 Marks
    aThe average marks scored by 32 boys in 72 with a standard deviation of 8. While that for 36 girls in 70 with a standard deviation of 6. does this indicate that the boys perform better than girls at level of significance 0.05? Explain.
    How to answer

    • State null hypothesis $H_0: \mu_1 = \mu_2$ and alternative hypothesis $H_1: \mu_1 > \mu_2$ (right-tailed test).
    • Calculate test statistic: $Z = \frac{72 - 70}{\sqrt{\frac{8^2}{32} + \frac{6^2}{36}}} = \frac{2}{\sqrt{\frac{64}{32} + \frac{36}{36}}} = \frac{2}{\sqrt{2 + 1}} = \frac{2}{\sqrt{3}} \approx 1.155$.
    • Find critical value at $\alpha = 0.05$ for one-tailed test: $Z_{0.05} = 1.645$.
    • Compare $|Z| = 1.155 < 1.645$: fail to reject $H_0$.
    • Conclusion: The data does not provide sufficient evidence at 0.05 significance level to conclude that boys perform better than girls.
14. 7 Marks
    bExperience had show that 20% of a manufactured product is of the top quality. In one day's production of 400 articles only 50 are of top quality. Test the hypothesis at 0.05 level.
    How to answer

    • State null hypothesis $H_0: P = 0.20$ and alternative hypothesis $H_1: P \ne 0.20$ (two-tailed test).
    • Calculate sample proportion: $p = 50 / 400 = 0.125$.
    • Calculate standard error: $SE = \sqrt{\frac{P(1-P)}{n}} = \sqrt{\frac{0.20 \times 0.80}{400}} = \sqrt{\frac{0.16}{400}} = \frac{0.4}{20} = 0.02$.
    • Compute test statistic: $Z = \frac{0.125 - 0.20}{0.02} = \frac{-0.075}{0.02} = -3.75$.
    • Critical value for two-tailed test at $\alpha = 0.05$ is $Z_{\text{crit}} = \pm 1.96$.
    • Since $|Z| = 3.75 > 1.96$, reject $H_0$ at the 0.05 level of significance.
15. 7 Marks
    aExplain partial and multiple correlation coefficients.
    How to answer

    • Define partial correlation as the degree of association between two variables while eliminating the effect of other variables.
    • Provide formula for first-order partial correlation: $r_{12.3} = \frac{r_{12} - r_{13}r_{23}}{\sqrt{(1-r_{13}^2)(1-r_{23}^2)}}$.
    • Define multiple correlation as the degree of association between one dependent variable and the combined influence of two or more independent variables.
    • Provide formula for multiple correlation: $R_{1.23} = \sqrt{\frac{r_{12}^2 + r_{13}^2 - 2 r_{12} r_{13} r_{23}}{1 - r_{23}^2}}$.
    • Discuss their ranges and significance ($0 \le R \le 1$ and $-1 \le r \le 1$).
16. 7 Marks
    bDefine chi-square distribution. Explain it's properties.
    How to answer

    • Define chi-square distribution as the distribution of sum of squares of $n$ independent standard normal variables: $\chi^2 = \sum_{i=1}^n Z_i^2$ with $n$ degrees of freedom.
    • Write PDF of chi-square distribution: $f(x) = \frac{1}{2^{n/2} \Gamma(n/2)} x^{n/2 - 1} e^{-x/2}, x > 0$.
    • List key properties: Mean $= n$, Variance $= 2n$, Mode $= n - 2$ (for $n \ge 2$).
    • State additive (reproductive) property of independent chi-square variables.
    • Explain asymptotic normality as $n \to \infty$.