AL-302 Introduction to Probability and Statistics - Nov 2022 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. 14 Marks
   A random variable X has the following probability function:

   x	0	1	2	3	4	5	6	7
   P(x)	0	K	2K	2K	3K	$K^2$	$2K^2$	$7K^2 + K$

   i) Determine K ii) Evaluate $P(x < 6)$ and $P(0 \le x \le 4)$ iii) If $P(x \le K) > (1/2)$ find the minimum value of K iv) Determine the distribution function of X v) Mean

   How to answer

   • Set sum of all probabilities equal to 1 to form a quadratic equation in K: 10K^2 + 9K - 1 = 0, solve to find K = 1/10.
   • Calculate P(x < 6) and P(0 <= x <= 4) by summing corresponding probabilities.
   • Determine the minimum value satisfying P(X <= k) > 1/2.
   • Tabulate cumulative distribution function F(x) = P(X <= x).
   • Compute Mean E(X) = \sum x P(x).
2. 7 Marks
   aThe density function of a random variable X is $$f(x) = e^{-x} \text{ when } x \ge 0$$ Find $E(x)$, $E(x^2)$ and variance of X.
   How to answer

   • Integrate E(X) = \int_0^\infty x e^{-x} dx = 1.
   • Integrate E(X^2) = \int_0^\infty x^2 e^{-x} dx = 2.
   • Compute Var(X) = E(X^2) - [E(X)]^2 = 2 - 1 = 1.
3. 7 Marks
   bSuppose a continuous random variable X has the probability density function $f(x) = K(1-x^2)$ for $0 < x < 1$, and $f(x) = 0$ otherwise. Find: i) K ii) Mean iii) Variance
   How to answer

   • Integrate \int_0^1 K(1-x^2) dx = 1 to find K = 3/2.
   • Calculate Mean E(X) = \int_0^1 (3/2) x (1-x^2) dx = 3/8.
   • Calculate E(X^2) = \int_0^1 (3/2) x^2 (1-x^2) dx = 1/5.
   • Compute Variance = 1/5 - (3/8)^2 = 19/320.
4. 6 Marks
   aDefine Bivariate distribution. Explain their properties.
   How to answer

   • Definition of bivariate distribution for pairs of random variables (X, Y).
   • Joint probability mass function (discrete) and joint probability density function (continuous).
   • Key properties: non-negativity and total probability equaling 1.
   • Marginal distributions and independence conditions.
5. 8 Marks
   bDefine Baye's theorem. The chance that doctor A will diagnose a disease X correctly is 60%. The chance that a patient will die by his treatment after correct diagnosis is 40% and the chance of death by wrong diagnosis is 70%. A patient of doctor A, who had disease X, died. What is the chance that his disease was diagnosed correctly?
   How to answer

   • State Bayes' theorem formula: P(E_1|A) = [P(E_1)P(A|E_1)] / [P(E_1)P(A|E_1) + P(E_2)P(A|E_2)].
   • Identify events: E1 = correct diagnosis (P(E1) = 0.60), E2 = wrong diagnosis (P(E2) = 0.40).
   • Identify conditional probabilities of death D: P(D|E1) = 0.40, P(D|E2) = 0.70.
   • Compute P(E1|D) = (0.60 * 0.40) / (0.60 * 0.40 + 0.40 * 0.70) = 0.24 / 0.52 = 6/13.
6. 7 Marks
   aThe mean and variance of a binomial variable X with parameters $n$ and P are 16 and 8 respectively. Find $P(X \ge 1)$ and $P(X > 2)$.
   How to answer

   • Find q = Variance/Mean = 8/16 = 1/2, so p = 1/2.
   • Calculate n = Mean/p = 16 / (1/2) = 32.
   • Compute P(X >= 1) = 1 - P(X = 0) = 1 - (1/2)^32.
   • Compute P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)].
7. 7 Marks
   bA manufacturer knows that the condensers he makes contain on average 1% defectives. He packs them in boxes of 100. What is the probability that a box picked at random will contain 3 or more faulty condensers?
   How to answer

   • Recognize Poisson distribution approximation with parameter \lambda = np = 100 * 0.01 = 1.
   • Set up P(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)].
   • Calculate P(X = 0) = e^{-1}, P(X = 1) = e^{-1}, P(X = 2) = e^{-1} / 2.
   • Sum probabilities: P(X <= 2) = 2.5 e^{-1} = 2.5 * 0.3679 = 0.9197.
   • Result: P(X >= 3) = 1 - 0.9197 = 0.0803.
8. 7 Marks
   aIf X is a normal variate with mean 30 and standard deviation is 5. Find i) $P(26 \le X \le 40)$ ii) $P(X \ge 45)$
   How to answer

   • Standardize: Z = (X - 30) / 5.
   • For (i): z1 = (26 - 30)/5 = -0.8, z2 = (40 - 30)/5 = 2.0; P(-0.8 <= Z <= 2.0) = \Phi(2.0) - \Phi(-0.8).
   • For (ii): z = (45 - 30)/5 = 3.0; P(Z >= 3.0) = 1 - \Phi(3.0).
   • Evaluate using standard normal distribution table values.
9. 7 Marks
   b

   Find Karl pearson's coefficient of correlation from the following data:

   Wages	100	101	102	102	100	99	97	98	96	95
   Cost of living	98	99	99	97	95	92	95	94	90	91

   How to answer

   • Calculate mean of X (\bar{X}) and mean of Y (\bar{Y}) (or use assumed mean method).
   • Construct table with columns for u = X - A, v = Y - B, u^2, v^2, and uv.
   • Calculate \sum u, \sum v, \sum u^2, \sum v^2, and \sum uv.
   • Substitute into Karl Pearson's correlation coefficient formula and obtain r.
10. 14 Marks
    By the method of least squares fit a parabola of the form $y = a + bx + cx^2$ for the following data.

    X:	2	4	6	8	10
    Y:	3.07	12.85	31.47	57.38	91.29

    How to answer

    • Write down the three normal equations for fitting a second-degree parabola y = a + bx + cx^2.
    • Create calculation table for sums: \sum x, \sum x^2, \sum x^3, \sum x^4, \sum y, \sum xy, \sum x^2 y.
    • Set up the system of 3 linear equations in a, b, and c.
    • Solve the linear system simultaneously for a, b, and c.
    • Write the fitted equation: y = a + bx + cx^2.
11. 8 Marks
    aIt is claimed that a random sample of 49 tyres has mean life of 15200 km. This sampled was drawn from a population whose mean is 15150 km's and a standard deviation of 1200 km. Test the significance at 0.05 level.
    How to answer

    • State null hypothesis H0: \mu = 15150 and alternative hypothesis H1: \mu \ne 15150.
    • Calculate test statistic: Z = (15200 - 15150) / (1200 / \sqrt{49}) = 50 / (1200 / 7) = 350 / 1200 = 0.292.
    • Compare |Z| with critical value at 5% level of significance (Z_critical = 1.96).
    • Since |Z| < 1.96, accept null hypothesis H0; difference is not significant.
12. 6 Marks
    bDefine i) Correlation coefficient ii) Chebyshev's inequality
    How to answer

    • Definition of Correlation coefficient: measure of linear relationship between two random variables, formula r = Cov(X,Y)/(SD(X)*SD(Y)), and range [-1, 1].
    • Definition of Chebyshev's inequality: bounds probability that a random variable differs from its mean by more than k standard deviations, stating P(|X - \mu| >= k\sigma) <= 1/k^2.
13. 7 Marks
    aA sample of 26 bulbs gives a mean life of 900 hours with a standard deviation of 20 hours. The manufacturer claims that the mean life of bulbs is 1000 hours. Is the sample not up to the standard. (5% L.O.S.)
    How to answer

    • State null hypothesis H0: \mu = 1000 and alternative hypothesis H1: \mu < 1000 (one-tailed).
    • Calculate test statistic t = (900 - 1000) / (20 / \sqrt{25}) = -100 / 4 = -25.
    • Degrees of freedom df = n - 1 = 25.
    • Critical t value at 5% level of significance for 25 df is approx 1.708.
    • Since |-25| > 1.708, reject null hypothesis H0; sample is significantly below standard.
14. 7 Marks
    bIn one sample of 10 observations, the sum of the squares of the deviations of the sample values from mean was 120 and in the other sample of 12 observations, it was 314. Test whether the difference is significant at 5% level?
    How to answer

    • State null hypothesis H0: \sigma_1^2 = \sigma_2^2 against H1: \sigma_1^2 \ne \sigma_2^2.
    • Calculate sample variances: s_1^2 = 120 / 9 = 13.33 and s_2^2 = 314 / 11 = 28.55.
    • Calculate test statistic F = s_2^2 / s_1^2 = 28.55 / 13.33 = 2.14.
    • Find critical value F_{0.05} with degrees of freedom (v1 = 11, v2 = 9) (approx 3.10).
    • Since calculated F < critical F, fail to reject H0; the difference in variances is not significant.