BE Computer Engineering (Pokhara University) Probability and Statistics (PU, MTH 216) Question Paper 2079 Nepal

(a) Statement and Proof of Bayes' Theorem

Statement. Let $E_1, E_2, \dots, E_n$ be mutually exclusive and exhaustive events of a sample space $S$ (i.e. $E_i \cap E_j = \varnothing$ for $i \ne j$ and $\bigcup_{i=1}^{n} E_i = S$), each with $P(E_i) > 0$. Let $A$ be any event with $P(A) > 0$. Then for each $k$:

$$P(E_k \mid A) = \frac{P(E_k)\,P(A \mid E_k)}{\sum_{i=1}^{n} P(E_i)\,P(A \mid E_i)}$$

Proof. Since the $E_i$ partition $S$, we can write

$$A = A \cap S = A \cap \left(\bigcup_{i=1}^{n} E_i\right) = \bigcup_{i=1}^{n} (A \cap E_i).$$

The events $A \cap E_i$ are mutually exclusive, so by the axiom of additivity (theorem of total probability):

$$P(A) = \sum_{i=1}^{n} P(A \cap E_i) = \sum_{i=1}^{n} P(E_i)\,P(A \mid E_i).$$

By the definition of conditional probability and the multiplication rule:

$$P(E_k \mid A) = \frac{P(E_k \cap A)}{P(A)} = \frac{P(E_k)\,P(A \mid E_k)}{P(A)}.$$

Substituting the total-probability expression for $P(A)$ gives

$$P(E_k \mid A) = \frac{P(E_k)\,P(A \mid E_k)}{\sum_{i=1}^{n} P(E_i)\,P(A \mid E_i)}. \qquad \blacksquare$$

(b) Numerical

Let $D$ = "board is defective." Given:

$$P(M_1)=0.30,\; P(M_2)=0.45,\; P(M_3)=0.25,$$ $$P(D\mid M_1)=0.02,\; P(D\mid M_2)=0.03,\; P(D\mid M_3)=0.04.$$

(i) Probability the board is defective (total probability):

$$P(D) = (0.30)(0.02) + (0.45)(0.03) + (0.25)(0.04)$$ $$= 0.0060 + 0.0135 + 0.0100 = 0.0295.$$

So $P(D) = 0.0295$ (about $2.95\%$).

(ii) Probability it came from $M_2$ (Bayes):

$$P(M_2 \mid D) = \frac{P(M_2)P(D\mid M_2)}{P(D)} = \frac{0.0135}{0.0295} \approx 0.4576.$$

(iii) Most likely source. Compute all posteriors:

$$P(M_1\mid D) = \frac{0.0060}{0.0295} \approx 0.2034,$$ $$P(M_2\mid D) \approx 0.4576,$$ $$P(M_3\mid D) = \frac{0.0100}{0.0295} \approx 0.3390.$$

Since $P(M_2 \mid D) \approx 0.458$ is the largest posterior probability, machine $M_2$ is most likely to have produced the defective board. Although $M_2$ has only a moderate defect rate, its large share of output (45%) dominates, so it contributes the most defectives overall.

(a) Continuous random variable and pdf properties

A continuous random variable $X$ is one that can take any value in an interval (or union of intervals) of the real line; the probability that it equals any single value is zero, and probabilities are obtained by integrating a probability density function (pdf) $f(x)$ over intervals.

A valid pdf $f(x)$ must satisfy:

1. Non-negativity: $f(x) \ge 0$ for all $x$.
2. Normalization (total area = 1): $\displaystyle \int_{-\infty}^{\infty} f(x)\,dx = 1.$
3. Probability via integration: $\displaystyle P(a \le X \le b) = \int_a^b f(x)\,dx.$

(b) Numerical

$$f(x) = kx(4-x), \quad 0 \le x \le 4.$$

(i) Find $k$. Require total area $= 1$:

$$\int_0^4 kx(4-x)\,dx = k\int_0^4 (4x - x^2)\,dx = k\left[2x^2 - \tfrac{x^3}{3}\right]_0^4 = k\left(32 - \tfrac{64}{3}\right) = k\cdot\tfrac{32}{3}.$$

Setting $\tfrac{32}{3}k = 1$ gives $\boxed{k = \dfrac{3}{32}}.$

(ii) Mean and variance.

$$E[X] = \int_0^4 x\,f(x)\,dx = \tfrac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \tfrac{3}{32}\left[\tfrac{4x^3}{3} - \tfrac{x^4}{4}\right]_0^4 = \tfrac{3}{32}\left(\tfrac{256}{3} - 64\right) = \tfrac{3}{32}\cdot\tfrac{64}{3} = 2.$$

So $E[X] = 2$ (thousand hours). By symmetry of $x(4-x)$ about $x=2$, this is expected.

$$E[X^2] = \tfrac{3}{32}\int_0^4 (4x^3 - x^4)\,dx = \tfrac{3}{32}\left[x^4 - \tfrac{x^5}{5}\right]_0^4 = \tfrac{3}{32}\left(256 - \tfrac{1024}{5}\right) = \tfrac{3}{32}\cdot\tfrac{256}{5} = \tfrac{24}{5} = 4.8.$$ $$\text{Var}(X) = E[X^2] - (E[X])^2 = 4.8 - 4 = 0.8.$$

So mean $= 2$, variance $= 0.8$ (thousand-hours, hours$^2$ respectively).

(iii) $P(1 \le X \le 3)$.

$$P(1\le X\le 3) = \tfrac{3}{32}\int_1^3 (4x - x^2)\,dx = \tfrac{3}{32}\left[2x^2 - \tfrac{x^3}{3}\right]_1^3.$$

At $x=3$: $18 - 9 = 9$. At $x=1$: $2 - \tfrac13 = \tfrac53$. Difference $= 9 - \tfrac53 = \tfrac{22}{3}.$

$$P = \tfrac{3}{32}\cdot\tfrac{22}{3} = \tfrac{22}{32} = \tfrac{11}{16} = 0.6875.$$

(c) Binomial vs Poisson

Poisson is the limiting case of binomial as $n\to\infty$, $p\to 0$ with $np=\lambda$ fixed.

• Binomial example: number of defective chips in a batch of $n=20$ inspected chips, each defective with probability $p$.
• Poisson example: number of packet arrivals at a router per second (rare events at a known average rate).

(a) Procedure for testing a statistical hypothesis

General steps:

1. Formulate hypotheses: state the null hypothesis $H_0$ and alternative hypothesis $H_1$.
2. Choose level of significance $\alpha$ (e.g. 0.05).
3. Select the test statistic appropriate to the parameter and sampling distribution ($z$, $t$, $\chi^2$, $F$).
4. Determine the critical region (rejection region) from $\alpha$ and whether the test is one- or two-tailed.
5. Compute the test statistic from the sample data.
6. Decision: reject $H_0$ if the statistic falls in the critical region; otherwise do not reject $H_0$.
7. Conclusion in terms of the original problem.

Definitions:

• Null hypothesis ($H_0$): the statement of no effect / no difference, assumed true until evidence contradicts it (e.g. $\mu = \mu_0$).
• Alternative hypothesis ($H_1$): the claim accepted if $H_0$ is rejected (e.g. $\mu \ne \mu_0$, $\mu < \mu_0$, or $\mu > \mu_0$).
• Type I error: rejecting $H_0$ when it is actually true; its probability is $\alpha$.
• Type II error: failing to reject $H_0$ when it is actually false; its probability is $\beta$.
• Level of significance ($\alpha$): the maximum probability of committing a Type I error, fixed in advance.
• Critical region: the set of values of the test statistic for which $H_0$ is rejected.

(b) Numerical

Given: claimed $\mu \ge 250$, $n=36$, $\bar{x}=244$, $s=18$, $\alpha=0.05$.

(i) Hypotheses (one-tailed, testing whether mean is less than claimed):

$$H_0: \mu \ge 250 \text{ MPa} \qquad H_1: \mu < 250 \text{ MPa}.$$

(ii) Test. Since $n=36$ is large, use the $z$-test (with $s$ for $\sigma$):

$$z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{244 - 250}{18/\sqrt{36}} = \frac{-6}{18/6} = \frac{-6}{3} = -2.0.$$

For a left-tailed test at $\alpha=0.05$, the critical value is $z_{0.05} = -1.645$. Rejection region: $z < -1.645$.

Since $z = -2.0 < -1.645$, the test statistic lies in the critical region, so we reject $H_0$.

(iii) Conclusion. At the 5% level of significance there is sufficient evidence that the true mean tensile strength is less than 250 MPa. The manufacturer's claim that the mean strength is at least 250 MPa is not supported by the sample data.

Using class midpoints $x$ and frequencies $f$ ($N=50$):

(a) Mean and Median

Mean: $\bar{x} = \dfrac{\sum fx}{N} = \dfrac{1990}{50} = 39.8.$

Median: $N/2 = 25$ lies in the class 30–40 (CF just exceeds 25 there). With $L=30$, $CF=12$ (before), $f=14$, $h=10$:

$$\text{Median} = L + \frac{N/2 - CF}{f}\times h = 30 + \frac{25-12}{14}\times 10 = 30 + 9.29 = 39.29.$$

(b) Standard deviation and CV

$$\sigma = \sqrt{\frac{\sum f(x-\bar{x})^2}{N}} = \sqrt{\frac{9280}{50}} = \sqrt{185.6} \approx 13.62.$$ $$\text{CV} = \frac{\sigma}{\bar{x}}\times 100 = \frac{13.62}{39.8}\times 100 \approx 34.2\%.$$

Comment: A coefficient of variation of about $34\%$ is fairly high, indicating that the marks are not very consistent — there is considerable dispersion of student performance about the mean.

With $n=8$, compute the sums:

$$\sum x = 56,\quad \sum y = 362,\quad \sum xy = 2962,\quad \sum x^2 = 470,\quad \sum y^2 = 18696.$$

($\bar{x}=7,\ \bar{y}=45.25$.)

(a) Least-squares line of $y$ on $x$

$$b = \frac{n\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2} = \frac{8(2962) - (56)(362)}{8(470) - 56^2} = \frac{23696 - 20272}{3760 - 3136} = \frac{3424}{624} \approx 5.487.$$ $$a = \bar{y} - b\bar{x} = 45.25 - 5.487(7) = 45.25 - 38.41 = 6.84.$$

Regression line: $\hat{y} = 6.84 + 5.487x.$

(b) Estimate for $x = 7$ hours

$$\hat{y} = 6.84 + 5.487(7) = 6.84 + 38.41 \approx 45.25.$$

The estimated test score is about 45 marks.

(c) Karl Pearson correlation coefficient

$$r = \frac{n\sum xy - \sum x\sum y}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}.$$

Denominator pieces: $n\sum x^2-(\sum x)^2 = 624$; $n\sum y^2-(\sum y)^2 = 8(18696) - 362^2 = 149568 - 131044 = 18524.$

$$r = \frac{3424}{\sqrt{624\times 18524}} = \frac{3424}{\sqrt{11558976}} = \frac{3424}{3399.85} \approx 0.993.$$

Interpretation: $r \approx 0.99$ indicates a very strong positive linear correlation — students who practise more hours tend to score substantially higher.

Let $X \sim \text{Poisson}(\lambda)$ with $\lambda = 5$ per minute. PMF: $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}.$

(a) Exactly 3 requests in a minute ($\lambda=5$)

$$P(X=3) = \frac{e^{-5}5^3}{3!} = \frac{e^{-5}\cdot 125}{6} = 20.833\,e^{-5} \approx 20.833(0.006738) \approx 0.1404.$$

(b) At most 2 requests in a minute

$$P(X\le 2) = e^{-5}\left(1 + 5 + \frac{25}{2}\right) = e^{-5}(18.5) = 18.5(0.006738) \approx 0.1247.$$

(c) More than 1 request in a 30-second interval

For a 30-second (half-minute) interval the mean is $\lambda' = 5\times 0.5 = 2.5.$

$$P(X>1) = 1 - P(X=0) - P(X=1) = 1 - e^{-2.5}(1 + 2.5) = 1 - 3.5\,e^{-2.5}.$$

With $e^{-2.5}\approx 0.082085$: $P(X>1) = 1 - 3.5(0.082085) = 1 - 0.2873 \approx 0.7127.$

So $P(X>1) \approx 0.713.$

Let $X \sim N(\mu=12.0,\ \sigma=0.04)$ mm. Standardize with $z=\dfrac{x-\mu}{\sigma}.$

(a) Probability a bearing is acceptable ($11.92 \le X \le 12.08$)

$$z_1 = \frac{11.92-12.0}{0.04} = -2,\qquad z_2 = \frac{12.08-12.0}{0.04} = +2.$$ $$P(-2 \le Z \le 2) = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9544.$$

So about 95.44% of bearings are acceptable.

(b) Number rejected out of 5000

Probability of rejection $= 1 - 0.9544 = 0.0456.$

$$\text{Expected rejects} = 5000 \times 0.0456 = 228 \text{ bearings}.$$

(c) Diameter exceeded by only 2.5% of bearings

We need $d$ with $P(X > d) = 0.025$, i.e. $P(X \le d) = 0.975$, so $z = 1.96.$

$$d = \mu + z\sigma = 12.0 + 1.96(0.04) = 12.0 + 0.0784 = 12.078 \text{ mm}.$$

Only about $2.5\%$ of bearings exceed $12.078$ mm.

(a) Addition and multiplication theorems

Addition theorem (for any two events $A,B$):

$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$

If $A$ and $B$ are mutually exclusive, $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)$.

Multiplication theorem:

$$P(A \cap B) = P(A)\,P(B \mid A) = P(B)\,P(A \mid B).$$

If $A$ and $B$ are independent, $P(A\cap B)=P(A)\,P(B)$.

(b) System reliability

Given $P(A)=0.9$, $P(B)=0.8$, independent.

(i) Series — system works only if both components work:

$$R_{\text{series}} = P(A)\,P(B) = 0.9 \times 0.8 = 0.72.$$

(ii) Parallel — system works if at least one component works:

$$R_{\text{parallel}} = 1 - (1-0.9)(1-0.8) = 1 - (0.1)(0.2) = 1 - 0.02 = 0.98.$$

The parallel configuration ($0.98$) is far more reliable than the series configuration ($0.72$), illustrating redundancy.

Given $n=100$, $\bar{x}=102$, $s=8$, large sample so use $z$.

(a) 95% confidence interval for $\mu$

Standard error $= \dfrac{s}{\sqrt{n}} = \dfrac{8}{\sqrt{100}} = 0.8.$ For 95% confidence, $z_{0.025}=1.96.$

$$\text{CI} = \bar{x} \pm z\cdot\frac{s}{\sqrt{n}} = 102 \pm 1.96(0.8) = 102 \pm 1.568.$$ $$\boxed{(100.43,\ 103.57) \text{ ohms}}.$$

(b) Interpretation

We are 95% confident that the true mean resistance of the production lot lies between about 100.43 and 103.57 ohms. Operationally, if many such samples were taken and an interval computed from each, about 95% of those intervals would contain the true mean resistance.

(c) Required sample size for margin $\pm 1$ ohm

Margin of error $E = z\cdot\dfrac{\sigma}{\sqrt{n}} \le 1$, so

$$n \ge \left(\frac{z\,s}{E}\right)^2 = \left(\frac{1.96 \times 8}{1}\right)^2 = (15.68)^2 = 245.86.$$

Rounding up, the required sample size is $n = 246$ resistors.

This is a test of a single proportion (large sample, two-tailed).

Data: $n=400$, successes $x=240$, so sample proportion $\hat{p} = \dfrac{240}{400} = 0.6.$ Hypothesized $p_0 = 0.5$, $\alpha = 0.01.$

Hypotheses:

$$H_0: p = 0.5 \qquad H_1: p \ne 0.5 \quad (\text{two-tailed}).$$

Test statistic (under $H_0$, standard error uses $p_0$):

$$z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}} = \frac{0.6 - 0.5}{\sqrt{\dfrac{0.5 \times 0.5}{400}}} = \frac{0.1}{\sqrt{0.000625}} = \frac{0.1}{0.025} = 4.0.$$

Critical value: for a two-tailed test at $\alpha=0.01$, $z_{0.005} = \pm 2.576.$

Decision: $|z| = 4.0 > 2.576$, so the statistic falls in the rejection region — reject $H_0$.

Conclusion: At the 1% level of significance, the proportion of developers preferring statically typed languages ($0.6$) differs significantly from $0.5$; the data provide strong evidence of a genuine preference for statically typed languages.

(a) Skewness and kurtosis

Skewness measures the asymmetry of a frequency distribution about its mean. If the longer tail is on the right (mean > median), skewness is positive (right-skewed); if on the left (mean < median), it is negative (left-skewed); a symmetric distribution has zero skewness.

Kurtosis measures the peakedness (and tail weight) of a distribution relative to the normal curve. A leptokurtic distribution ($\beta_2 > 3$) is more peaked with heavier tails; a platykurtic distribution ($\beta_2 < 3$) is flatter; a mesokurtic distribution ($\beta_2 = 3$) matches the normal curve.

Together they describe the shape of a distribution: skewness tells how lopsided it is, kurtosis tells how sharp/flat its peak and how heavy its tails are.

(b) Karl Pearson coefficient of skewness

Given mean $=36$, median $=34$, $\sigma = 6$. Using the median-based formula:

$$S_k = \frac{3(\text{Mean} - \text{Median})}{\sigma} = \frac{3(36 - 34)}{6} = \frac{6}{6} = 1.0.$$

Comment: Since $S_k = +1.0 > 0$, the distribution is positively (right) skewed — it has a longer tail toward the higher values, and the value $+1$ indicates a fairly strong degree of skewness.