AD-302 Probability and Statistics for Data Science - Jun 2023 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
   aProve that for any discrete distribution standard deviation is not less than mean deviation from mean.
   How to answer

   • State expressions for standard deviation $\sigma = \sqrt{\frac{1}{N}\sum f_i (x_i - \bar{x})^2}$ and mean deviation $\text{MD} = \frac{1}{N}\sum f_i |x_i - \bar{x}|$.
   • Apply Cauchy-Schwarz inequality $(\sum u_i v_i)^2 \le (\sum u_i^2)(\sum v_i^2)$ or consider $\text{Var}(|X - \bar{x}|) \ge 0$.
   • Show that $E[(|X - \bar{x}|)^2] \ge (E[|X - \bar{x}|])^2$.
   • Take the square root of both sides to conclude $\sigma \ge \text{MD}_{\text{mean}}$.
2. 7 Marks
   bFind the mean deviation from the mean and standard deviation of A.P. $a, a + d, a + 2d, \dots, a + 2nd$ and verify that the latter is greater than the former.
   How to answer

   • Determine number of terms $N = 2n + 1$ and mean $\bar{x} = a + nd$.
   • Calculate mean deviation from mean: $\text{MD} = \frac{1}{2n+1} \sum_{r=-n}^n |rd| = \frac{2d}{2n+1} \frac{n(n+1)}{2} = \frac{n(n+1)}{2n+1} d$.
   • Calculate standard deviation: $\sigma^2 = \frac{1}{2n+1} \sum_{r=-n}^n (rd)^2 = \frac{2d^2}{2n+1} \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{3} d^2$, so $\sigma = d\sqrt{\frac{n(n+1)}{3}}$.
   • Compare $\sigma^2$ and $(\text{MD})^2$ to show $\sigma > \text{MD}$ for $n \ge 1$.
3. 7 Marks
   a

   Calculate the first four moments of the following distribution about the mean and hence find $\beta_1$ and $\beta_2$.

   x	0	1	2	3	4	5	6	7	8
   f	1	8	28	56	70	56	28	8	1

   How to answer

   • Note $N = \sum f = 256$, mean $\bar{x} = 4$ due to perfect symmetry.
   • Calculate $\mu_1 = 0$ and $\mu_3 = 0$ directly by symmetry.
   • Compute $\mu_2 = \frac{1}{256}\sum f_i (x_i - 4)^2 = 2$.
   • Compute $\mu_4 = \frac{1}{256}\sum f_i (x_i - 4)^4 = 11.5$.
   • Calculate $\beta_1 = \frac{\mu_3^2}{\mu_2^3} = 0$ and $\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{11.5}{4} = 2.875$.
4. 7 Marks
   bIf two dice are thrown, what is the probability that the sum is i) greater than 8 ii) neither 7 Nor 11?
   How to answer

   • Identify total sample space size $n(S) = 36$.
   • For (i): Favorable sums are 9, 10, 11, 12; number of outcomes $= 4 + 3 + 2 + 1 = 10$. Thus $P(\text{sum} > 8) = \frac{10}{36} = \frac{5}{18}$.
   • For (ii): Favorable outcomes for 7 are 6, for 11 are 2, total $= 8$. $P(7 \text{ or } 11) = \frac{8}{36} = \frac{2}{9}$.
   • Therefore, $P(\text{neither } 7 \text{ nor } 11) = 1 - \frac{2}{9} = \frac{7}{9} = \frac{28}{36}$.
5. 7 Marks
   aIn a bolt factory machine A, B and C manufacture respectively 25%, 35% and 40%, of a total of their output 5, 4, 2 percent are defective bolts. A bolt is drawn at random from the product and is found a to be defective. What are the probabilities that it was manufactured by machines A, B and C.
   How to answer

   • List prior probabilities: $P(A) = 0.25$, $P(B) = 0.35$, $P(C) = 0.40$.
   • List conditional probabilities: $P(D|A) = 0.05$, $P(D|B) = 0.04$, $P(D|C) = 0.02$.
   • Calculate total probability: $P(D) = 0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345$.
   • Calculate posterior probabilities via Bayes' theorem: $P(A|D) = \frac{125}{345} = \frac{25}{69}$, $P(B|D) = \frac{140}{345} = \frac{28}{69}$, $P(C|D) = \frac{80}{345} = \frac{16}{69}$.
6. 7 Marks
   bThe diameter of an electric cable, say X, is assumed to be a continuous random variable with probability density function $f(x) = 6x(1-x), 0 \le x \le 1$. i) Check that above is probability density function ii) Determine a number b such that $P(X < b) = P(X > b)$
   How to answer

   • For (i): Show $f(x) \ge 0$ for $0 \le x \le 1$. Evaluate $\int_0^1 6(x - x^2)dx = 6 [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = 6(\frac{1}{6}) = 1$, proving it is a valid pdf.
   • For (ii): Note $P(X < b) = P(X > b) \implies P(X < b) = 0.5$, which means $b$ is the median.
   • Set $\int_0^b 6(x - x^2) dx = 3b^2 - 2b^3 = 0.5 \implies 4b^3 - 6b^2 + 1 = 0$.
   • Since $f(x)$ is symmetric about $x = 0.5$ on $[0, 1]$, $b = 0.5$ (or solve $(2b-1)(2b^2-2b-1)=0$ to get $b=1/2$).
7. 7 Marks
   a

   Fit a parabola of second degree to the following data:

   X:	0	1	2	3	4
   Y:	1	1.8	1.3	2.5	6.3

   How to answer

   • Assume parabola equation $Y = a + bX + cX^2$.
   • Construct table of sums: $N=5$, $\sum X = 10$, $\sum X^2 = 30$, $\sum X^3 = 100$, $\sum X^4 = 354$, $\sum Y = 12.9$, $\sum XY = 39.1$, $\sum X^2Y = 139.5$.
   • Write normal equations: $5a + 10b + 30c = 12.9$, $10a + 30b + 100c = 39.1$, $30a + 100b + 354c = 139.5$.
   • Solve the linear system for $a$, $b$, and $c$, and state the final fitted parabola.
8. 7 Marks
   b

   Obtain the rank correlation coefficient for the following 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 $X$ values (noting ties: 75 appears 2 times, 64 appears 3 times).
   • Assign ranks to $Y$ values (noting ties: 68 appears 2 times).
   • Calculate differences $d = R_X - R_Y$ and $d^2$ for each pair.
   • Apply the tied-ranks correction factor: $CF = \frac{2(4-1)}{12} + \frac{3(9-1)}{12} + \frac{2(4-1)}{12} = 0.5 + 2 + 0.5 = 3$.
   • Compute rank correlation coefficient $\rho = 1 - \frac{6(\sum d^2 + CF)}{N(N^2-1)}$ with $N=10$.
9. 7 Marks
   a

   Calculate the correlation Coefficient for the following heights (in inches) of Father (X) and their Sons (Y):

   X:	65	66	67	67	68	69	70	72
   Y:	67	68	65	68	72	72	69	71

   How to answer

   • Compute $N = 8$, $\sum X = 544$, $\bar{X} = 68$, $\sum Y = 552$, $\bar{Y} = 69$.
   • Calculate deviations $u = X - 68$ and $v = Y - 69$.
   • Compute $\sum u^2$, $\sum v^2$, and $\sum uv$.
   • Substitute into $r = \frac{\sum uv}{\sqrt{\sum u^2 \sum v^2}}$ and calculate the value of $r$.
10. 7 Marks
    bCan $Y = 5 + 2.8 X$ and $X = 3 - 0.5Y$ be the estimated regression equation of Y on X and X on Y respectively ? Explain your answer with suitable theoretical arguments.
    How to answer

    • Identify the regression coefficients from the given equations: $b_{yx} = 2.8$ and $b_{xy} = -0.5$.
    • Check sign consistency property: Both regression coefficients $b_{yx}$ and $b_{xy}$ must have the same sign as $r$. Here, one is positive ($2.8$) and the other is negative ($-0.5$), which is impossible.
    • Check magnitude: $r^2 = b_{yx} b_{xy} = 2.8 \times (-0.5) = -1.4 < 0$, which is impossible since $r^2$ must be real and $0 \le r^2 \le 1$.
    • Conclude that these cannot be the estimated regression equations.
11. 7 Marks
    a

    The following figure show the distribution of digits in numbers chosen at random from a telephonic directory:

    Digits:	0	1	2	3	4	5	6	7	8	9	Total
    Freq.	1026	1107	997	966	1075	933	1107	972	964	853	10000

    Test whether the digits may be taken to occur equally frequently in the directory.

    How to answer

    • Formulate null hypothesis $H_0$: Digits occur equally frequently ($p_i = 1/10$).
    • Formulate alternative hypothesis $H_1$: Digits do not occur equally frequently.
    • Expected frequency for each digit: $E_i = \frac{10000}{10} = 1000$.
    • Calculate $\chi^2 = \sum \frac{(O_i - 1000)^2}{1000} = \frac{1}{1000}[26^2 + 107^2 + (-3)^2 + (-34)^2 + 75^2 + (-67)^2 + 107^2 + (-28)^2 + (-36)^2 + (-147)^2] = 58.542$.
    • Degrees of freedom $\text{df} = 10 - 1 = 9$. Critical value at $\alpha = 0.05$ is $16.92$.
    • Since $\chi_{\text{cal}}^2 > \chi_{\text{tab}}^2$, reject $H_0$; the digits do not occur equally frequently.
12. 7 Marks
    bIt is believed that the precision (as measured by the variance) of an instrument is no more than 0.16. Write down the null and alternative hypothesis from testing this belief. Carry out the test at 1% level, given 11 measurements of the same subjects on the instrument: 2.5, 2.3, 2.4, 2.3, 2.5, 2.7, 2.5, 2.6, 2.6, 2.7, 2.5
    How to answer

    • State hypotheses: $H_0: \sigma^2 \le 0.16$ vs $H_1: \sigma^2 > 0.16$ (right-tailed test).
    • Calculate sample mean: $\bar{x} = \frac{27.6}{11} \approx 2.509$.
    • Compute $\sum (x_i - \bar{x})^2 \approx 0.1891$.
    • Calculate test statistic: $\chi^2 = \frac{\sum (x_i - \bar{x})^2}{\sigma_0^2} = \frac{0.1891}{0.16} \approx 1.182$.
    • Critical value for $\text{df} = 10$ at $\alpha = 0.01$ is $\chi_{0.01, 10}^2 = 23.21$.
    • Since $\chi_{\text{cal}}^2 = 1.182 < 23.21$, fail to reject $H_0$. The belief that variance is no more than 0.16 is accepted.
13. 7 Marks
    aA random sample of 10 boys had the following I.Q. : 70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Do these data support the assumption of a population mean I.Q. of 100? Find a reasonable range in which most of the mean I Q values of samples of 10 boys lie.
    How to answer

    • Set hypotheses: $H_0: \mu = 100$ vs $H_1: \mu \ne 100$.
    • Calculate sample mean: $\sum x = 972 \implies \bar{x} = 97.2$.
    • Calculate sample variance: $\sum (x - \bar{x})^2 = 1833.6 \implies s^2 = \frac{1833.6}{9} = 203.73$, so $s \approx 14.27$.
    • Calculate test statistic: $t = \frac{97.2 - 100}{14.27 / \sqrt{10}} = \frac{-2.8}{4.514} \approx -0.62$.
    • Compare with critical value $t_{0.05, 9} = 2.262$; since $|t| < 2.262$, accept $H_0$ (data support mean I.Q. of 100).
    • Compute 95% confidence interval: $\bar{x} \pm 2.262 \times 4.514 = 97.2 \pm 10.21 \implies [86.99, 107.41]$.
14. 7 Marks
    bA box contain 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random. Find the probability that among the balls drawn there is a at least one bowl of each colour.
    How to answer

    • Total number of balls $= 6 + 4 + 5 = 15$. Total ways to draw 4 balls $= \binom{15}{4} = 1365$.
    • To have at least one ball of each color among 4 balls, the distribution must be: (2 Red, 1 White, 1 Black), or (1 Red, 2 White, 1 Black), or (1 Red, 1 White, 2 Black).
    • Ways for (2R, 1W, 1B) $= \binom{6}{2} \binom{4}{1} \binom{5}{1} = 15 \times 4 \times 5 = 300$.
    • Ways for (1R, 2W, 1B) $= \binom{6}{1} \binom{4}{2} \binom{5}{1} = 6 \times 6 \times 5 = 180$.
    • Ways for (1R, 1W, 2B) $= \binom{6}{1} \binom{4}{1} \binom{5}{2} = 6 \times 4 \times 10 = 240$.
    • Total favorable ways $= 300 + 180 + 240 = 720$.
    • Required probability $= \frac{720}{1365} = \frac{48}{91} \approx 0.5275$.
15. 7 Marks
    aIn a book of 520 pages, 390 typo-graphical errors occur. Assuming poisson law for the number of per page, find the probability that a random sample of 5 pages with contain no error.
    How to answer

    • Calculate mean rate of errors per page: $\lambda = \frac{390}{520} = 0.75$.
    • For a sample of 5 pages, parameter $\lambda' = 5 \times 0.75 = 3.75$ errors.
    • Use Poisson probability formula for $k = 0$: $P(X = 0) = e^{-3.75}$.
    • Evaluate $e^{-3.75} \approx 0.0235$.
16. 7 Marks
    bTwo independent variable random X and Y are both normally distributed with means 1 and 2 and standard deviation 3 and 4 respectively. If $Z = X - Y$, write the probability density function of Z. Also state the median, S.D. and mean of the distribution of Z. find $Prob. \{Z + 1 \le 0\}$.
    How to answer

    • Determine distribution of $Z$: Since $X$ and $Y$ are independent normal variables, $Z = X - Y$ is also normal.
    • Calculate mean $\mu_Z = 1 - 2 = -1$.
    • Calculate variance $\sigma_Z^2 = 3^2 + 4^2 = 25 \implies \sigma_Z = 5$.
    • Write pdf: $f(z) = \frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(z + 1)^2}{50}\right)$ for $-\infty < z < \infty$.
    • State properties: Mean $= -1$, Median $= -1$ (by symmetry), S.D. $= 5$.
    • Calculate $P(Z + 1 \le 0) = P(Z \le -1) = P\left(\frac{Z - (-1)}{5} \le \frac{-1 - (-1)}{5}\right) = P(W \le 0) = 0.5$.