BE Computer Engineering (IOE, TU) Probability and Statistics (IOE, SH 602 / ENSH 301) Question Paper 2079 Nepal

Working table

Using class mid-points $m$ and $N=\sum f = 60$:

(a) Mean, Median, Mode

Mean: $\bar{x} = \dfrac{\sum fm}{N} = \dfrac{4700}{60} = 78.33$ kg.

Median: $N/2 = 30$ falls in the class 70-80 (cumulative frequency reaches 34). With $L=70,\ cf=16,\ f=18,\ h=10$:

$$M = L + \frac{N/2 - cf}{f}\,h = 70 + \frac{30-16}{18}\times 10 = 77.78 \text{ kg}.$$

Mode: Modal class is 70-80 (highest frequency 18). With $f_1=18,\ f_0=10,\ f_2=14$:

$$Z = L + \frac{f_1-f_0}{2f_1-f_0-f_2}\,h = 70 + \frac{18-10}{36-10-14}\times 10 = 76.67 \text{ kg}.$$

(b) Standard deviation and CV

$$\sigma = \sqrt{\frac{\sum f(m-\bar{x})^2}{N}} = \sqrt{\frac{10933.3}{60}} = \sqrt{182.22} = 13.50 \text{ kg}.$$ $$\text{CV} = \frac{\sigma}{\bar{x}}\times 100 = \frac{13.50}{78.33}\times 100 = 17.23\%.$$

Comment: A CV of about 17% indicates moderate relative variability. The breaking strengths are reasonably consistent, so the production process is fairly stable, though not highly uniform.

(c) Karl Pearson's coefficient of skewness

$$S_k = \frac{\bar{x}-Z}{\sigma} = \frac{78.33-76.67}{13.50} = +0.12.$$

(Equivalently, $S_k = \dfrac{3(\bar{x}-M)}{\sigma} = \dfrac{3(78.33-77.78)}{13.50} \approx +0.12$.)

Since $S_k$ is small and positive, the distribution is slightly positively (right) skewed — nearly symmetrical with a mild tail towards higher breaking strengths.

(a) Axioms of probability and the addition theorem

Axioms (Kolmogorov). For a sample space $S$ and any event $A$:

1. Non-negativity: $P(A) \geq 0$.
2. Normalization: $P(S) = 1$.
3. Additivity: if $A_1, A_2, \dots$ are mutually exclusive events, then $P\!\left(\bigcup_i A_i\right) = \sum_i P(A_i)$.

Addition theorem. We can write the union of $A$ and $B$ as a union of disjoint pieces:

$$A \cup B = A \cup (B \cap A^c).$$

Since $A$ and $B\cap A^c$ are mutually exclusive, by Axiom 3:

$$P(A\cup B) = P(A) + P(B\cap A^c). \quad (1)$$

Also, $B = (B\cap A) \cup (B\cap A^c)$ is a disjoint split of $B$, so:

$$P(B) = P(B\cap A) + P(B\cap A^c) \;\Rightarrow\; P(B\cap A^c) = P(B) - P(A\cap B). \quad (2)$$

Substituting (2) into (1):

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

(b) Bayes' theorem application

Let $D$ be the event that a drawn item is defective. Given:

$$P(M_1)=0.25,\ P(M_2)=0.35,\ P(M_3)=0.40,$$ $$P(D|M_1)=0.05,\ P(D|M_2)=0.04,\ P(D|M_3)=0.02.$$

Total probability of a defective:

$$P(D) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02)$$ $$= 0.0125 + 0.0140 + 0.0080 = 0.0345.$$

Bayes' theorem:

$$P(M_3|D) = \frac{P(M_3)P(D|M_3)}{P(D)} = \frac{0.0080}{0.0345} = 0.2319.$$

Thus the probability the defective item was produced by machine $M_3$ is $\approx 0.232$ (23.2%).

(a) Normal distribution: definition, properties, coincidence of averages

A continuous random variable $X$ follows a normal distribution with mean $\mu$ and variance $\sigma^2$ if its probability density function is

$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad -\infty < x < \infty.$$

Important properties:

• Bell-shaped, symmetric about $x=\mu$.
• Mean = Median = Mode = $\mu$.
• Total area under the curve is 1; the curve is asymptotic to the $x$-axis.
• The $x$-axis is approached but never touched; $\mu \pm \sigma$ are points of inflexion.
• Empirical rule: about 68%, 95% and 99.7% of values lie within $\mu\pm\sigma$, $\mu\pm2\sigma$, $\mu\pm3\sigma$.
• Quartiles: $Q_1=\mu-0.6745\sigma$, $Q_3=\mu+0.6745\sigma$; mean deviation $=0.7979\sigma$.

Mean = Median = Mode. The curve is symmetric about $x=\mu$, so $\mu$ divides the area into two equal halves $\Rightarrow$ Median $=\mu$. Differentiating $f(x)$ and setting $f'(x)=0$ gives $-\frac{(x-\mu)}{\sigma^2}f(x)=0 \Rightarrow x=\mu$, and $f''(\mu)<0$, so $f(x)$ is maximum at $x=\mu \Rightarrow$ Mode $=\mu$. The mean of a symmetric distribution equals its centre of symmetry $\mu$. Hence mean = median = mode = $\mu$.

(b) Numerical part ($\mu=1200,\ \sigma=150$)

Standardize with $z = \dfrac{x-\mu}{\sigma}$.

(i) $P(X>1450)$: $z = \dfrac{1450-1200}{150} = 1.67$.

$$P(X>1450) = P(Z>1.67) = 0.5 - 0.4525 = 0.0475 \approx 4.75\%.$$

(ii) $P(1000<X<1300)$: $z_1 = \dfrac{1000-1200}{150} = -1.33$, $z_2 = \dfrac{1300-1200}{150} = 0.67$.

$$P = P(-1.33<Z<0.67) = 0.4082 + 0.2486 = 0.6568 \approx 65.7\%.$$

(iii) Guarantee period $g$ such that $P(X<g)=0.05$. The lower 5% point is $z=-1.645$:

$$g = \mu + z\sigma = 1200 + (-1.645)(150) = 1200 - 246.75 = 953.25 \text{ hours}.$$

The manufacturer should set the guarantee period at about 953 hours.

Sums (n = 8)

$$\sum X = 57,\ \sum Y = 380,\ \sum X^2 = 491,\ \sum Y^2 = 20786,\ \sum XY = 3187,$$ $$\bar{X} = 7.125,\quad \bar{Y} = 47.5.$$

(a) Karl Pearson's correlation coefficient

$$r = \frac{n\sum XY - \sum X\sum Y}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}$$ $$= \frac{8(3187) - (57)(380)}{\sqrt{[8(491)-57^2][8(20786)-380^2]}} = \frac{25496 - 21660}{\sqrt{(3928-3249)(166288-144400)}}$$ $$= \frac{3836}{\sqrt{679 \times 21888}} = \frac{3836}{3855.1} = 0.995.$$

There is a very strong positive linear correlation between hours studied and marks obtained.

(b) Regression equations

Regression coefficients:

$$b_{YX} = \frac{n\sum XY - \sum X\sum Y}{n\sum X^2 - (\sum X)^2} = \frac{3836}{679} = 5.649,$$ $$b_{XY} = \frac{n\sum XY - \sum X\sum Y}{n\sum Y^2 - (\sum Y)^2} = \frac{3836}{21888} = 0.1753.$$

Regression of $Y$ on $X$: $Y - \bar{Y} = b_{YX}(X-\bar{X})$

$$Y = 47.5 + 5.649(X - 7.125) \Rightarrow \boxed{Y = 7.25 + 5.649X.}$$

Regression of $X$ on $Y$: $X - \bar{X} = b_{XY}(Y-\bar{Y})$

$$X = 7.125 + 0.1753(Y - 47.5) \Rightarrow \boxed{X = -1.20 + 0.1753Y.}$$

(c) Estimate at X = 10 and verification

$$\hat{Y}|_{X=10} = 7.25 + 5.649(10) = 63.74 \approx 64 \text{ marks}.$$

Verification: $\sqrt{b_{YX}\cdot b_{XY}} = \sqrt{5.649 \times 0.1753} = \sqrt{0.9903} = 0.995 = r.$

Hence the geometric mean of the two regression coefficients equals the correlation coefficient (sign of $r$ = common sign of the coefficients, here positive).

(a) Value of $k$

Since $\sum P(X) = 1$:

$$k + 2k + 3k + 2k + k = 9k = 1 \Rightarrow \boxed{k = \tfrac{1}{9}}.$$

So the distribution is $P(0)=\tfrac19,\ P(1)=\tfrac29,\ P(2)=\tfrac39,\ P(3)=\tfrac29,\ P(4)=\tfrac19.$

(b) Expectation and variance

$$E(X) = \sum xP(x) = \frac{0+2+6+6+4}{9} = \frac{18}{9} = 2.$$ $$E(X^2) = \sum x^2 P(x) = \frac{0+2+12+18+16}{9} = \frac{48}{9} = 5.333.$$ $$\text{Var}(X) = E(X^2) - [E(X)]^2 = 5.333 - 4 = 1.333 = \tfrac{4}{3}.$$

(The distribution is symmetric about $x=2$, consistent with $E(X)=2$.)

(c) $P(X \geq 2)$

$$P(X\geq 2) = P(2)+P(3)+P(4) = \frac{3+2+1}{9} = \frac{6}{9} = \frac{2}{3} \approx 0.667.$$

Let $X$ = number of defective items in a sample of $n=10$, with $p=0.04$ (so $q=0.96$). Then $X \sim B(10, 0.04)$ and

$$P(X=x) = \binom{10}{x}(0.04)^x(0.96)^{10-x}.$$

(a) Exactly 2 defective:

$$P(X=2) = \binom{10}{2}(0.04)^2(0.96)^8 = 45(0.0016)(0.7214) = 0.0519.$$

(b) At most 1 defective:

$$P(X\leq 1) = P(0)+P(1) = (0.96)^{10} + 10(0.04)(0.96)^9$$ $$= 0.6648 + 0.2770 = 0.9418.$$

(c) At least 1 defective:

$$P(X\geq 1) = 1 - P(X=0) = 1 - (0.96)^{10} = 1 - 0.6648 = 0.3352.$$

Mean and variance:

$$\text{Mean} = np = 10\times 0.04 = 0.4,\qquad \text{Variance} = npq = 10\times 0.04\times 0.96 = 0.384.$$

Let $X$ = number of calls per minute, $X \sim \text{Poisson}(\lambda = 3)$, with

$$P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!} = \frac{e^{-3}3^x}{x!}.$$

(a) No call received:

$$P(X=0) = e^{-3} = 0.0498.$$

(b) More than 2 calls:

$$P(X>2) = 1 - [P(0)+P(1)+P(2)] = 1 - e^{-3}\left[1 + 3 + \tfrac{9}{2}\right]$$ $$= 1 - e^{-3}(8.5) = 1 - 0.0498\times 8.5 = 1 - 0.4232 = 0.5768.$$

(c) Poisson as a limit of the binomial. The binomial $B(n,p)$ tends to a Poisson distribution with $\lambda = np$ when:

• the number of trials $n$ is large ($n \to \infty$);
• the probability of success $p$ is very small ($p \to 0$, i.e. the event is rare);
• the product $np = \lambda$ remains finite and constant (moderate).

Under these conditions $\binom{n}{x}p^x q^{n-x} \to \dfrac{e^{-\lambda}\lambda^x}{x!}$. In practice the approximation is good when $n \geq 20$ and $p \leq 0.05$ (often stated as $n\ge 50,\ np<5$). A key feature is that mean = variance = $\lambda$ for the Poisson.

(a) 95% confidence interval for the mean

Given $n=100,\ \bar{x}=1570,\ s=120$. Since $n$ is large, use the normal (z) interval with $z_{0.025}=1.96$. Standard error:

$$SE = \frac{s}{\sqrt{n}} = \frac{120}{\sqrt{100}} = 12.$$

Confidence interval:

$$\bar{x} \pm z\cdot SE = 1570 \pm 1.96(12) = 1570 \pm 23.52.$$ $$\boxed{1546.48 \text{ hours} < \mu < 1593.52 \text{ hours}.}$$

We are 95% confident that the true mean life of the bulbs lies between about 1546.5 and 1593.5 hours.

(b) Point vs interval estimate; properties of a good estimator

• Point estimate: a single value of a sample statistic used to estimate the unknown population parameter (e.g. $\bar{x}=1570$ for $\mu$). It gives no measure of uncertainty.
• Interval estimate: a range of values, together with a confidence level, within which the parameter is expected to lie (e.g. $1546.5$ to $1593.5$ hours at 95%). It conveys the precision/reliability of the estimate.

Properties of a good estimator:

1. Unbiasedness — $E(\hat{\theta}) = \theta$.
2. Consistency — $\hat{\theta} \to \theta$ as $n \to \infty$.
3. Efficiency — smallest variance among unbiased estimators.
4. Sufficiency — uses all the information about $\theta$ contained in the sample.

(a) Z-test for the mean

Given: claimed mean $\mu_0 = 580$, $n=64$, $\bar{x}=568$, $s=40$.

Hypotheses:

$$H_0: \mu = 580 \text{ N/mm}^2 \quad\text{vs}\quad H_1: \mu \neq 580 \text{ N/mm}^2 \ (\text{two-tailed}).$$

Test statistic (large sample $z$-test):

$$z = \frac{\bar{x}-\mu_0}{s/\sqrt{n}} = \frac{568-580}{40/\sqrt{64}} = \frac{-12}{5} = -2.40.$$

Critical value: at 5% (two-tailed), $z_{0.025} = \pm 1.96$.

Decision: $|z| = 2.40 > 1.96$, so we reject $H_0$.

Conclusion: At the 5% level there is sufficient evidence against the company's claim; the true mean tensile strength differs significantly from (and appears lower than) 580 N/mm².

(b) Hypotheses and error types

• $H_0$: $\mu = 580$ (the wire meets the claimed mean strength).
• $H_1$: $\mu \neq 580$ (the mean strength is not 580).
• Type I error ($\alpha$): rejecting $H_0$ when it is actually true — concluding the strength differs from 580 when in fact it is 580 (wrongly rejecting a good claim). Here $\alpha = 0.05$.
• Type II error ($\beta$): failing to reject $H_0$ when it is false — concluding the wire meets 580 N/mm² when in reality it does not.

Chi-square goodness-of-fit test

Hypotheses: $H_0$: the die is unbiased (each face equally likely) vs $H_1$: the die is biased.

Under $H_0$ each face has probability $\tfrac16$, so the expected frequency for each is

$$E = 120 \times \tfrac16 = 20.$$

Test statistic:

$$\chi^2 = \sum \frac{(O-E)^2}{E} = 4.50.$$

Degrees of freedom: $df = k - 1 = 6 - 1 = 5.$

Critical value: $\chi^2_{0.05,\,5} = 11.07.$

Decision: Since calculated $\chi^2 = 4.50 < 11.07$, we do not reject $H_0$.

Conclusion: At the 5% level of significance, the observed frequencies are consistent with a fair die — there is no significant evidence that the die is biased.

(Any three of the following.)

(a) Moments, skewness and kurtosis

The $r$-th central moment is $\mu_r = E[(X-\bar{X})^r]$. The first four describe the distribution: $\mu_1=0$, $\mu_2=\sigma^2$ (variance), $\mu_3$ measures asymmetry, $\mu_4$ measures peakedness. Skewness measures the lack of symmetry: $\beta_1 = \mu_3^2/\mu_2^3$ (or $\gamma_1=\mu_3/\sigma^3$); positive $\Rightarrow$ right tail, negative $\Rightarrow$ left tail, zero $\Rightarrow$ symmetric. Kurtosis measures peakedness relative to the normal: $\beta_2 = \mu_4/\mu_2^2$; $\beta_2=3$ (mesokurtic/normal), $>3$ leptokurtic (sharp peak), $<3$ platykurtic (flat).

(b) Mutually exclusive vs independent events

• Mutually exclusive: events that cannot occur together, $A\cap B=\varnothing$, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$. Example: getting a head and a tail on a single coin toss.
• Independent: the occurrence of one does not affect the probability of the other, $P(A\cap B)=P(A)P(B)$. Example: results of two separate coin tosses.
• Note: two events with non-zero probabilities cannot be both mutually exclusive and independent (if mutually exclusive, knowing $A$ occurred forces $P(B|A)=0\ne P(B)$).

(c) PMF vs PDF

• Probability mass function (pmf): for a discrete random variable, $p(x)=P(X=x)$, with $p(x)\ge 0$ and $\sum_x p(x)=1$. Probabilities attach to individual points.
• Probability density function (pdf): for a continuous random variable, $f(x)\ge 0$ with $\int_{-\infty}^{\infty} f(x)\,dx = 1$. Here $P(X=x)=0$; probability is an area: $P(a<X<b)=\int_a^b f(x)\,dx$.

(d) Central Limit Theorem (CLT)

For a population with mean $\mu$ and finite variance $\sigma^2$, the sampling distribution of the sample mean $\bar{X}$ of a random sample of size $n$ tends to a normal distribution $N\!\left(\mu, \sigma^2/n\right)$ as $n$ becomes large (typically $n\ge 30$), irrespective of the shape of the parent population. Significance: it justifies using normal-based ($z$) methods for confidence intervals and hypothesis tests about means even for non-normal populations, and explains why many statistics are approximately normally distributed.