BE Civil Engineering (IOE, TU) Probability and Statistics (IOE, SH 552) Question Paper 2079 Nepal

We use class mid-points $x_i$ and frequencies $f_i$. Total $N=\sum f_i = 60$.

(a) Mean and Median

$$\bar{x}=\frac{\sum f_ix_i}{N}=\frac{1924}{60}=32.0667\ \text{MPa}$$

Median class: $N/2 = 30$, which falls in the class 32–36 (CF just exceeds 30). Here $L=32$, $CF_{prev}=29$, $f=18$, $h=4$.

$$\text{Median}=L+\frac{N/2-CF_{prev}}{f}\times h = 32+\frac{30-29}{18}\times 4 = 32+0.2222 = 32.2222\ \text{MPa}$$

Mean = 32.07 MPa, Median = 32.22 MPa.

(b) Standard deviation and CV

$$\sigma=\sqrt{\frac{\sum f_i(x_i-\bar{x})^2}{N}}=\sqrt{\frac{1584.00}{60}}=\sqrt{26.40}=5.1381\ \text{MPa}$$ $$CV=\frac{\sigma}{\bar{x}}\times100=\frac{5.1381}{32.0667}\times100=16.02\%$$

σ ≈ 5.14 MPa, CV ≈ 16.02 %.

(c) Karl Pearson coefficient of skewness (using mode)

Modal class is the one with highest frequency = 32–36 ($f_1=18$). With $L=32$, $f_0=16$ (previous), $f_2=9$ (next), $h=4$:

$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h = 32+\frac{18-16}{2(18)-16-9}\times4 = 32+\frac{2}{11}\times4 = 32+0.7273=32.7273\ \text{MPa}$$ $$S_k=\frac{\bar{x}-\text{Mode}}{\sigma}=\frac{32.0667-32.7273}{5.1381}=\frac{-0.6606}{5.1381}=-0.1286$$

Karl Pearson skewness ≈ −0.13. The small negative value indicates the distribution is slightly negatively (left) skewed but very nearly symmetric.

Let $S_1,S_2,S_3$ be the events that a bar comes from each supplier, and $F$ the event that a bar fails the test.

Given: $P(S_1)=0.50,\ P(S_2)=0.30,\ P(S_3)=0.20$ and $P(F|S_1)=0.02,\ P(F|S_2)=0.04,\ P(F|S_3)=0.05$.

(a) Statements

Total probability: If $S_1,\dots,S_n$ are mutually exclusive and exhaustive events with $P(S_i)>0$, then for any event $F$:

$$P(F)=\sum_{i=1}^{n}P(S_i)\,P(F|S_i).$$

Bayes' theorem:

$$P(S_k|F)=\frac{P(S_k)\,P(F|S_k)}{\sum_{i=1}^{n}P(S_i)\,P(F|S_i)}.$$

(b) Probability a bar fails

$$P(F)=P(S_1)P(F|S_1)+P(S_2)P(F|S_2)+P(S_3)P(F|S_3)$$ $$=(0.50)(0.02)+(0.30)(0.04)+(0.20)(0.05)$$ $$=0.010+0.012+0.010=0.032$$

P(F) = 0.032 (3.2 %).

(c) Posterior probability for $S_2$

$$P(S_2|F)=\frac{P(S_2)P(F|S_2)}{P(F)}=\frac{0.012}{0.032}=0.375$$

P(S₂ | F) = 0.375 (37.5 %).

Let $X\sim N(\mu=4200,\ \sigma=350)$. Standardize with $z=\dfrac{X-\mu}{\sigma}$.

(a) $P(X>4700)$

$$z=\frac{4700-4200}{350}=\frac{500}{350}=1.4286\approx1.43$$ $$P(X>4700)=1-\Phi(1.43)=1-0.9236=0.0764$$

About 7.64 % of days the demand exceeds 4700 m³.

(b) $P(3850<X<4550)$

$$z_1=\frac{3850-4200}{350}=\frac{-350}{350}=-1.00,\qquad z_2=\frac{4550-4200}{350}=\frac{350}{350}=1.00$$ $$P(-1<Z<1)=\Phi(1.00)-\Phi(-1.00)=0.8413-(1-0.8413)=0.8413-0.1587=0.6826$$

About 68.26 % of days the demand lies between 3850 and 4550 m³.

(c) Capacity exceeded on at most 2.5 % of days

We need $C$ such that $P(X>C)=0.025$, i.e. $P(X\le C)=0.975$, so $z_{0.975}=1.96$.

$$C=\mu+z\sigma=4200+1.96\times350=4200+686=4886\ \text{m}^3$$

Required design capacity = 4886 m³.

Let $n=6$. Build the working table (using deviations from means to keep arithmetic clean).

$\bar{x}=\dfrac{300+320+340+360+380+400}{6}=\dfrac{2100}{6}=350$

$\bar{y}=\dfrac{24+27+28+32+33+36}{6}=\dfrac{180}{6}=30$

(a) Correlation coefficient

$$r=\frac{\sum uv}{\sqrt{\sum u^2}\,\sqrt{\sum v^2}}=\frac{820}{\sqrt{7000}\,\sqrt{98}}=\frac{820}{83.666\times9.8995}=\frac{820}{828.25}=0.9900$$

r ≈ 0.990, a very strong positive correlation.

(b) Regression line of $Y$ on $X$

Slope: $b_{yx}=\dfrac{\sum uv}{\sum u^2}=\dfrac{820}{7000}=0.11714$

Intercept: $a=\bar{y}-b_{yx}\bar{x}=30-0.11714\times350=30-41.00=-11.00$

$$\boxed{\hat{Y}=-11.00+0.11714\,X}$$

(c) Estimate at $X=350$

$$\hat{Y}=-11.00+0.11714\times350=-11.00+41.00=30.00\ \text{MPa}$$

Estimated strength ≈ 30.0 MPa (as expected, the regression line passes through $(\bar{x},\bar{y})=(350,30)$).

Given: $n=16$, $\bar{x}=1476$ N/mm², $s=48$ N/mm², claimed $\mu_0=1500$ N/mm². Population SD unknown and $n<30$, so use the one-sample $t$-test.

Standard error: $SE=\dfrac{s}{\sqrt{n}}=\dfrac{48}{\sqrt{16}}=\dfrac{48}{4}=12$ N/mm².

(a) Hypothesis test (lower-tailed)

The claim is $\mu\ge1500$. We test whether data contradict it (i.e. true mean is less).

• $H_0:\mu=1500$ N/mm²   vs   $H_1:\mu<1500$ N/mm²
• Test statistic:

$$t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{1476-1500}{12}=\frac{-24}{12}=-2.00$$

• Degrees of freedom $=n-1=15$. Critical value (one-tailed, $\alpha=0.05$): $-t_{0.05,15}=-1.753$.
• Decision: Since $t=-2.00 < -1.753$, it falls in the rejection region → reject $H_0$.

Conclusion: At the 5% level there is sufficient evidence that the mean tensile strength is less than 1500 N/mm²; the data contradict the supplier's claim.

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

$$\bar{x}\pm t_{0.025,15}\cdot\frac{s}{\sqrt{n}}=1476\pm2.131\times12=1476\pm25.572$$ $$(1450.43,\ 1501.57)\ \text{N/mm}^2$$

95% CI ≈ (1450.4, 1501.6) N/mm². (Note the two-tailed 95% interval barely includes 1500, which is consistent with the borderline one-tailed rejection.)

Let $X$ = number of defective rivets in a sample of $n=10$, with $p=0.08$, $q=1-p=0.92$. Then $X\sim B(10,0.08)$ and $P(X=k)=\binom{10}{k}(0.08)^k(0.92)^{10-k}$.

(a) Conditions for binomial distribution

1. Fixed number of independent trials $n$.
2. Each trial has only two outcomes (success/failure).
3. Constant probability of success $p$ across trials.
4. Trials are mutually independent.

(b) Exactly 2 defective

$$P(X=2)=\binom{10}{2}(0.08)^2(0.92)^{8}=45\times0.0064\times0.51322=0.14780$$

(where $0.92^{8}=0.51322$). P(X = 2) ≈ 0.1478.

(c) At most 1 defective

$$P(X\le1)=P(0)+P(1)$$ $$P(0)=(0.92)^{10}=0.43439$$ $$P(1)=\binom{10}{1}(0.08)^1(0.92)^9=10\times0.08\times0.47220=0.37776$$ $$P(X\le1)=0.43439+0.37776=0.81215$$

P(X ≤ 1) ≈ 0.8122.

Let $X$ = number of cracks per km, $X\sim\text{Poisson}(\lambda=3)$, so $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}=\dfrac{e^{-3}3^k}{k!}$.

(a) Exactly 4 cracks

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

P(X = 4) ≈ 0.1680.

(b) At least 2 cracks

$$P(X\ge2)=1-P(0)-P(1)$$ $$P(0)=e^{-3}=0.049787$$ $$P(1)=\frac{e^{-3}\,3^1}{1!}=3\times0.049787=0.149361$$ $$P(X\ge2)=1-0.049787-0.149361=1-0.199148=0.800852$$

P(X ≥ 2) ≈ 0.8009.

(a) Value of $k$

The probabilities must sum to 1:

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

So the distribution is $P(0)=0.1,\ P(1)=0.3,\ P(2)=0.3,\ P(3)=0.2,\ P(4)=0.1$.

(b) Expected value

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

E(X) = 1.9.

(c) Variance

$$E(X^2)=\sum x^2P(x)=0+1(0.3)+4(0.3)+9(0.2)+16(0.1)$$ $$=0+0.3+1.2+1.8+1.6=4.9$$ $$\operatorname{Var}(X)=E(X^2)-[E(X)]^2=4.9-(1.9)^2=4.9-3.61=1.29$$

Var(X) = 1.29 (standard deviation $=\sqrt{1.29}\approx1.136$).

(a) Central Limit Theorem (CLT)

If random samples of size $n$ are drawn from a population with mean $\mu$ and finite variance $\sigma^2$, then as $n$ becomes large the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu$ and standard error $\sigma/\sqrt{n}$, regardless of the shape of the parent population:

$$\bar{X}\ \dot\sim\ N\!\left(\mu,\ \frac{\sigma^2}{n}\right).$$

(b) Probability that sample mean exceeds 84

Given $\mu=80$, $\sigma=15$, $n=36$. Standard error:

$$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{36}}=\frac{15}{6}=2.5\ \text{units}$$

Standardize:

$$z=\frac{\bar{x}-\mu}{\sigma_{\bar{X}}}=\frac{84-80}{2.5}=\frac{4}{2.5}=1.60$$ $$P(\bar{X}>84)=1-\Phi(1.60)=1-0.9452=0.0548$$

P(\u0058̄ > 84) ≈ 0.0548 (5.48 %).

Hypotheses

• $H_0$: the die is fair (each face equally likely, $p=1/6$).
• $H_1$: the die is not fair.

Expected frequency under $H_0$: $E_i=120\times\frac{1}{6}=20$ for each face.

Chi-square statistic $\chi^2=\sum\dfrac{(O_i-E_i)^2}{E_i}$:

$$\chi^2_{calc}=3.80$$

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

Decision: Since $\chi^2_{calc}=3.80 < 11.07$, we fail to reject $H_0$.

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

(a) Point estimate of proportion

$$\hat{p}=\frac{x}{n}=\frac{260}{400}=0.65$$

Estimated proportion = 0.65 (65 %).

(b) 95% confidence interval

Standard error of the proportion:

$$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.65\times0.35}{400}}=\sqrt{\frac{0.2275}{400}}=\sqrt{0.00056875}=0.023849$$

Margin of error: $E=z_{0.025}\times SE=1.96\times0.023849=0.046744$.

$$\hat{p}\pm E=0.65\pm0.04674\ \Rightarrow\ (0.6033,\ 0.6967)$$

95% CI ≈ (0.603, 0.697), i.e. about (60.3 %, 69.7 %).

(c) Interpretation

We are 95% confident that the true proportion of all households in the municipality with piped water access lies between 60.3% and 69.7%.