BE Civil Engineering (IOE, TU) Probability and Statistics (IOE, SH 552) Question Paper 2077 Nepal

Set up the working table. Let $x$ be the class mid-value, $f$ the frequency, $N=\sum f = 50$.

(a) Mean.

$$\bar x = \frac{\sum fx}{N} = \frac{5820}{50} = 116.4 \text{ litres}$$

Median. $N/2 = 25$, which falls in the cumulative frequency 28, so the median class is 100–120. Lower boundary $L=100$, c.f. before class $C=14$, class frequency $f_m=14$, width $h=20$.

$$\text{Median} = L + \frac{N/2 - C}{f_m}\,h = 100 + \frac{25-14}{14}\times 20 = 100 + \frac{11}{14}\times 20 = 100 + 15.714 = 115.714 \text{ litres}$$

(a) Answer: Mean $\approx \mathbf{116.4}$ litres, Median $\approx \mathbf{115.71}$ litres.

(b) Standard deviation.

$$\sigma = \sqrt{\frac{\sum f(x-\bar x)^2}{N}} = \sqrt{\frac{36552}{50}} = \sqrt{731.04} = 27.038 \text{ litres}$$

Coefficient of variation.

$$\text{CV} = \frac{\sigma}{\bar x}\times 100 = \frac{27.038}{116.4}\times 100 = 23.23\%$$

(b) Answer: $\sigma \approx \mathbf{27.04}$ litres, CV $\approx \mathbf{23.23\%}$.

(c) Karl Pearson coefficient of skewness.

$$S_k = \frac{\bar x - \text{Median (approx. mode via empirical relation)}}{\sigma}$$

Using the median-based form $S_k = \dfrac{3(\bar x - \text{Median})}{\sigma}$:

$$S_k = \frac{3(116.4 - 115.714)}{27.038} = \frac{3(0.686)}{27.038} = \frac{2.058}{27.038} = 0.0761$$

(c) Answer: $S_k \approx \mathbf{+0.076}$. Since $S_k>0$ but very small, the distribution is very slightly positively (right) skewed, i.e. close to symmetric.

(a) Bayes' theorem. If $E_1,E_2,\dots,E_n$ are mutually exclusive and exhaustive events with $P(E_i)>0$, and $A$ is any event with $P(A)>0$, then

$$P(E_i\mid A) = \frac{P(E_i)\,P(A\mid E_i)}{\sum_{j=1}^{n} P(E_j)\,P(A\mid E_j)}.$$

Given. $P(A)=0.50,\;P(B)=0.30,\;P(C)=0.20$. Defective rates: $P(D\mid A)=0.02,\;P(D\mid B)=0.04,\;P(D\mid C)=0.05$.

Total probability of a defective bag.

$$P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)$$ $$P(D)=(0.50)(0.02)+(0.30)(0.04)+(0.20)(0.05)$$ $$P(D)=0.010+0.012+0.010=0.032$$

(b) Posterior probabilities.

$$P(A\mid D)=\frac{0.010}{0.032}=0.3125$$ $$P(B\mid D)=\frac{0.012}{0.032}=0.3750$$ $$P(C\mid D)=\frac{0.010}{0.032}=0.3125$$

Check: $0.3125+0.3750+0.3125=1.000$ ✓

(b) Answer: $P(A\mid D)=\mathbf{0.3125}$, $P(B\mid D)=\mathbf{0.3750}$, $P(C\mid D)=\mathbf{0.3125}$.

(c) Probability of a non-defective bag.

$$P(\bar D)=1-P(D)=1-0.032=0.968$$

(c) Answer: $P(\text{non-defective}) = \mathbf{0.968}$.

Let $X\sim N(\mu=28,\sigma=3)$ and $Z=\dfrac{X-\mu}{\sigma}$ be standard normal.

(a) $P(X>32)$.

$$Z=\frac{32-28}{3}=\frac{4}{3}=1.33$$ $$P(X>32)=P(Z>1.33)=1-\Phi(1.33)=1-0.9082=0.0918$$

Answer: $\approx \mathbf{0.0918}$ (about 9.18% of cubes).

(b) $P(25<X<31)$.

$$Z_1=\frac{25-28}{3}=-1.00,\qquad Z_2=\frac{31-28}{3}=1.00$$ $$P(-1.00<Z<1.00)=\Phi(1.00)-\Phi(-1.00)=\Phi(1.00)-[1-\Phi(1.00)]=2\Phi(1.00)-1$$ $$=2(0.8413)-1=1.6826-1=0.6826$$

Answer: $\approx \mathbf{0.6826}$ (about 68.26% of cubes).

(c) 5th percentile. We need $x_0$ such that $P(X<x_0)=0.05$. The lower 5% point of $Z$ is $z=-1.645$ (since $\Phi(1.645)=0.95$).

$$x_0=\mu+z\sigma=28+(-1.645)(3)=28-4.935=23.065\text{ MPa}$$

Answer: minimum acceptable strength $\approx \mathbf{23.07\text{ MPa}}$.

Working table. $n=6$.

Means: $\bar x = 108/6 = 18$, $\bar y = 143/6 = 23.8333$.

(a) Correlation coefficient.

$$S_{xy}=\sum xy-\frac{(\sum x)(\sum y)}{n}=3025-\frac{(108)(143)}{6}=3025-2574=451$$ $$S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}=2704-\frac{108^2}{6}=2704-1944=760$$ $$S_{yy}=\sum y^2-\frac{(\sum y)^2}{n}=3697-\frac{143^2}{6}=3697-3408.1667=288.8333$$ $$r=\frac{S_{xy}}{\sqrt{S_{xx}\,S_{yy}}}=\frac{451}{\sqrt{760\times288.8333}}=\frac{451}{\sqrt{219513.33}}=\frac{451}{468.523}=0.9626$$

Answer: $r\approx \mathbf{0.963}$ — a strong positive linear correlation between curing age and strength.

(b) Regression line of $y$ on $x$. Slope

$$b_{yx}=\frac{S_{xy}}{S_{xx}}=\frac{451}{760}=0.59342$$

Intercept

$$a=\bar y-b_{yx}\bar x=23.8333-(0.59342)(18)=23.8333-10.6816=13.1517$$ $$\boxed{\;\hat y = 13.152 + 0.5934\,x\;}$$

(c) Estimate at $x=40$ days.

$$\hat y = 13.152 + 0.5934(40)=13.152+23.737=36.889\text{ MPa}$$

Answer: estimated strength $\approx \mathbf{36.89\text{ MPa}}$ at 40 days. (Note: this is an extrapolation beyond the observed range.)

Sample size $n=10$, degrees of freedom $\nu=n-1=9$.

(a) Hypotheses. The claim is $\mu\ge 500$; we test against the rods being weaker:

$$H_0:\mu = 500\text{ MPa (claim holds)}\qquad H_1:\mu < 500\text{ MPa (left-tailed)}$$

(b) Test statistic. Since $\sigma$ is unknown and $n$ is small, use the $t$-test:

$$t=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{488-500}{16/\sqrt{10}}=\frac{-12}{16/3.1623}=\frac{-12}{5.0596}=-2.372$$

Decision rule (left-tailed, $\alpha=0.05$, df 9): reject $H_0$ if $t < -t_{0.05,9}=-1.833$.

Since $t=-2.372 < -1.833$, the test statistic falls in the rejection region.

Conclusion: Reject $H_0$. At the 5% level there is significant evidence that the mean tensile strength is less than 500 MPa, so the data contradict the manufacturer's claim.

(c) 95% two-sided confidence interval.

$$\text{CI}=\bar x \pm t_{0.025,9}\,\frac{s}{\sqrt n}=488 \pm 2.262\times \frac{16}{\sqrt{10}}=488 \pm 2.262\times 5.0596$$ $$=488 \pm 11.445$$ $$\Rightarrow (476.56,\;499.45)\text{ MPa}$$

Answer: 95% CI $\approx \mathbf{(476.56,\;499.45)\text{ MPa}}$. Note that 500 MPa lies outside this interval, consistent with rejecting the claim in (b).

Let $X$ = number of defective joints, $X\sim \text{Binomial}(n=8,\;p=0.2)$, $q=1-p=0.8$.

$$P(X=k)=\binom{8}{k}(0.2)^k(0.8)^{8-k}$$

(a) Exactly 2 defective.

$$P(X=2)=\binom{8}{2}(0.2)^2(0.8)^6=28\times0.04\times0.262144=0.29360$$

Answer: $P(X=2)\approx \mathbf{0.2936}$.

(b) At most 1 defective.

$$P(X=0)=\binom{8}{0}(0.2)^0(0.8)^8=1\times1\times0.16777=0.16777$$ $$P(X=1)=\binom{8}{1}(0.2)^1(0.8)^7=8\times0.2\times0.2097152=0.33554$$ $$P(X\le1)=0.16777+0.33554=0.50332$$

Answer: $P(X\le1)\approx \mathbf{0.5033}$.

Let $X$ = number of cracks per km, $X\sim\text{Poisson}(\lambda=3)$.

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

(a) Exactly 4 cracks.

$$P(X=4)=\frac{e^{-3}3^4}{4!}=\frac{0.049787\times81}{24}=\frac{4.03275}{24}=0.16803$$

Answer: $P(X=4)\approx \mathbf{0.1680}$.

(b) No cracks.

$$P(X=0)=\frac{e^{-3}3^0}{0!}=e^{-3}=0.049787$$

Answer: $P(X=0)\approx \mathbf{0.0498}$.

(c) More than 2 cracks.

$$P(X>2)=1-[P(0)+P(1)+P(2)]$$ $$P(1)=\frac{e^{-3}3^1}{1!}=0.149361,\qquad P(2)=\frac{e^{-3}3^2}{2!}=\frac{0.049787\times9}{2}=0.224042$$ $$P(X>2)=1-(0.049787+0.149361+0.224042)=1-0.423190=0.576810$$

Answer: $P(X>2)\approx \mathbf{0.5768}$.

(a) Find $k$. Probabilities must sum to 1:

$$0.1+k+0.3+0.2+0.1=1 \;\Rightarrow\; k=1-0.7=0.3$$

Answer: $k=\mathbf{0.3}$.

Complete distribution: $P(0)=0.1,\;P(1)=0.3,\;P(2)=0.3,\;P(3)=0.2,\;P(4)=0.1$.

(b) Expectation.

$$E(X)=\sum x\,P(x)=0(0.1)+1(0.3)+2(0.3)+3(0.2)+4(0.1)$$ $$=0+0.3+0.6+0.6+0.4=1.9$$

Second moment.

$$E(X^2)=\sum x^2 P(x)=0+1(0.3)+4(0.3)+9(0.2)+16(0.1)$$ $$=0+0.3+1.2+1.8+1.6=4.9$$

Variance.

$$\text{Var}(X)=E(X^2)-[E(X)]^2=4.9-(1.9)^2=4.9-3.61=1.29$$

Answer: $E(X)=\mathbf{1.9}$, $\text{Var}(X)=\mathbf{1.29}$.

(a) Central Limit Theorem. If $X_1,X_2,\dots,X_n$ are independent, identically distributed random variables each with mean $\mu$ and finite variance $\sigma^2$, then for large $n$ the distribution of the sample mean $\bar X$ is approximately normal:

$$\bar X \;\approx\; N\!\left(\mu,\;\frac{\sigma^2}{n}\right),$$

regardless of the shape of the parent population. Equivalently $Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}$ tends to the standard normal distribution as $n\to\infty$.

(b) Here $\mu=50$, $\sigma=1.2$, $n=36$, so the standard error is

$$\sigma_{\bar X}=\frac{\sigma}{\sqrt n}=\frac{1.2}{\sqrt{36}}=\frac{1.2}{6}=0.2\text{ kg}.$$

Standardize the limits:

$$Z_1=\frac{49.7-50}{0.2}=\frac{-0.3}{0.2}=-1.5,\qquad Z_2=\frac{50.3-50}{0.2}=\frac{0.3}{0.2}=1.5$$ $$P(49.7<\bar X<50.3)=P(-1.5<Z<1.5)=2\Phi(1.5)-1=2(0.9332)-1=0.8664$$

Answer: $P(49.7<\bar X<50.3)\approx \mathbf{0.8664}$ (about 86.64%).

Hypotheses.

$$H_0:\text{the die is fair (each face equally likely)}\qquad H_1:\text{the die is not fair}$$

Expected frequency. If fair, each face has probability $1/6$, so the expected count is

$$E_i=120\times\tfrac{1}{6}=20\text{ for every face.}$$

Compute $\chi^2$.

$$\chi^2_{\text{cal}}=\sum\frac{(O_i-E_i)^2}{E_i}=3.80$$

Degrees of freedom $=6-1=5$. Critical value $\chi^2_{0.05,5}=11.07$.

Decision. Since $\chi^2_{\text{cal}}=3.80 < 11.07$, we do not reject $H_0$.

Conclusion: At the 5% level there is no significant evidence that the die is unfair; the die may be regarded as fair.

Let $S$ = event that the student passed Surveying, $H$ = passed Hydraulics. Total $=100$.

$$P(S)=\frac{60}{100}=0.60,\quad P(H)=\frac{45}{100}=0.45,\quad P(S\cap H)=\frac{25}{100}=0.25.$$

(a) Passed at least one (union). By the addition rule:

$$P(S\cup H)=P(S)+P(H)-P(S\cap H)=0.60+0.45-0.25=0.80$$

Answer: $\mathbf{0.80}$.

(b) Passed neither.

$$P(\overline{S\cup H})=1-P(S\cup H)=1-0.80=0.20$$

Answer: $\mathbf{0.20}$.

(c) Passed exactly one subject. This is the union minus the intersection (those in exactly one set):

$$P(\text{exactly one})=P(S\cup H)-P(S\cap H)=0.80-0.25=0.55$$

Equivalently, only Surveying $=0.60-0.25=0.35$ and only Hydraulics $=0.45-0.25=0.20$, giving $0.35+0.20=0.55$. Answer: $\mathbf{0.55}$.