BE Civil Engineering (IOE, TU) Probability and Statistics (IOE, SH 552) Question Paper 2078 Nepal

We work with class mid-values $x$ and frequencies $f$ ($N=\sum f = 50$).

(a) Mean:

$$\bar{x}=\frac{\sum fx}{N}=\frac{1480}{50}=29.6\text{ MPa}$$

Median: $N/2=25$. Cumulative frequencies: 6, 18, 36, 45, 50. The 25th item falls in class 28–32 (median class). With $L=28$, $cf=18$, $f_m=18$, $h=4$:

$$\text{Median}=L+\frac{N/2-cf}{f_m}\,h=28+\frac{25-18}{18}\times4=28+1.556=29.556\text{ MPa}$$

(a) Mean = 29.6 MPa; Median ≈ 29.56 MPa.

(b) Standard deviation:

$$\sigma=\sqrt{\frac{\sum f(x-\bar{x})^2}{N}}=\sqrt{\frac{1124.48}{50}}=\sqrt{22.4896}=4.742\text{ MPa}$$

Coefficient of variation:

$$CV=\frac{\sigma}{\bar{x}}\times100=\frac{4.742}{29.6}\times100=16.02\%$$

σ ≈ 4.74 MPa; CV ≈ 16.02%.

(c) Mode: modal class is 28–32 ($f=18$, highest). $L=28$, $f_1=18$, $f_0=12$, $f_2=9$, $h=4$:

$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h=28+\frac{18-12}{36-12-9}\times4=28+\frac{6}{15}\times4=28+1.6=29.6\text{ MPa}$$

Karl Pearson coefficient of skewness:

$$S_k=\frac{\bar{x}-\text{Mode}}{\sigma}=\frac{29.6-29.6}{4.742}=0$$

The distribution is essentially symmetrical ($S_k\approx0$). (Using mean−median form: $3(29.6-29.556)/4.742\approx+0.028$, a negligible positive skew.)

Let $D$ = bag is substandard (defective). Given:

$$P(S_1)=0.50,\;P(S_2)=0.30,\;P(S_3)=0.20$$ $$P(D\mid S_1)=0.02,\;P(D\mid S_2)=0.04,\;P(D\mid S_3)=0.05$$

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

$$P(D)=\sum_{i=1}^{n}P(S_i)\,P(D\mid S_i)$$

Bayes' theorem:

$$P(S_k\mid D)=\frac{P(S_k)\,P(D\mid S_k)}{\sum_{i}P(S_i)\,P(D\mid S_i)}$$

(b) First the total probability of a defective bag:

$$P(D)=0.50(0.02)+0.30(0.04)+0.20(0.05)$$ $$=0.010+0.012+0.010=0.032$$

Then:

$$P(S_2\mid D)=\frac{P(S_2)P(D\mid S_2)}{P(D)}=\frac{0.012}{0.032}=0.375$$

P(S₂ | D) = 0.375 (37.5%).

(c) $P(D)=0.032$, so:

$$P(\text{not substandard})=1-P(D)=1-0.032=0.968$$

P(not substandard) = 0.968 (96.8%).

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

(a) $P(X>4700)$:

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

≈ 0.0764 (7.64%).

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

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

≈ 0.6826 (68.26%).

(c) Capacity sufficient on 95% of days means $P(X\le C)=0.95$, so $C$ is the 95th percentile:

$$C=\mu+z_{0.95}\,\sigma=4200+1.645\times350=4200+575.75=4775.75\text{ litres}$$

Required design capacity ≈ 4776 litres (round up to be safe).

Sample statistics. $n=10$. Sum:

$$\sum x=508+495+512+489+505+498+510+492+503+500=5012$$ $$\bar{x}=\frac{5012}{10}=501.2\text{ MPa}$$

Deviations $d=x-\bar{x}$ and $d^2$:

Sample variance and SD (with $n-1=9$):

$$s^2=\frac{\sum d^2}{n-1}=\frac{541.60}{9}=60.178,\qquad s=7.757\text{ MPa}$$

(a) Test of claim. Claim: $\mu\ge500$. Set up a one-tailed (lower) test:

$$H_0:\mu=500\quad\text{vs}\quad H_1:\mu<500$$

Test statistic:

$$t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{501.2-500}{7.757/\sqrt{10}}=\frac{1.2}{2.453}=0.489$$

Rejection region (lower tail): reject $H_0$ if $t<-t_{0.05,9}=-1.833$. Since $t=0.489>-1.833$, we fail to reject $H_0$.

Conclusion: At the 5% level the data do not contradict the manufacturer's claim; there is no significant evidence that the mean strength is below 500 MPa.

(b) 95% confidence interval (two-sided, $t_{0.025,9}=2.262$):

$$\bar{x}\pm t_{0.025,9}\,\frac{s}{\sqrt{n}}=501.2\pm2.262\times2.453=501.2\pm5.549$$ $$\Rightarrow(495.65,\;506.75)\text{ MPa}$$

95% CI for μ ≈ (495.7, 506.8) MPa. The value 500 lies inside the interval, consistent with part (a).

Let $n=7$. Build the sums (use $x$ in original units):

Sums:

$$\sum x=3.85,\;\sum y=219,\;\sum x^2=2.1875,\;\sum y^2=7195,\;\sum xy=115.55$$ $$\bar{x}=0.55,\quad\bar{y}=31.2857$$

Correction sums (about the mean):

$$S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}=2.1875-\frac{3.85^2}{7}=2.1875-2.1175=0.0700$$ $$S_{yy}=\sum y^2-\frac{(\sum y)^2}{n}=7195-\frac{219^2}{7}=7195-6852.4286=342.5714$$ $$S_{xy}=\sum xy-\frac{(\sum x)(\sum y)}{n}=115.55-\frac{3.85\times219}{7}=115.55-120.45=-4.9000$$

(a) Correlation coefficient:

$$r=\frac{S_{xy}}{\sqrt{S_{xx}\,S_{yy}}}=\frac{-4.9000}{\sqrt{0.0700\times342.5714}}=\frac{-4.9000}{\sqrt{23.98}}=\frac{-4.9000}{4.8969}=-1.000$$

r ≈ −0.9994 ≈ −1.0: an almost perfect negative linear relationship — strength falls as the water–cement ratio rises (consistent with concrete technology).

(b) Regression line of $y$ on $x$: slope and intercept:

$$b=\frac{S_{xy}}{S_{xx}}=\frac{-4.9000}{0.0700}=-70.0$$ $$a=\bar{y}-b\bar{x}=31.2857-(-70.0)(0.55)=31.2857+38.5=69.7857$$ $$\boxed{\;\hat{y}=69.786-70.0\,x\;}$$

(c) Estimate at $x=0.58$:

$$\hat{y}=69.786-70.0\times0.58=69.786-40.60=29.186\text{ MPa}$$

Estimated compressive strength ≈ 29.19 MPa.

Let $X$ = number of defective bricks, $X\sim B(n=12,\;p=0.08)$, $q=0.92$.

$$P(X=k)=\binom{12}{k}(0.08)^k(0.92)^{12-k}$$

(a) $P(X=2)$: $\binom{12}{2}=66$.

$$P(X=2)=66\,(0.08)^2(0.92)^{10}=66\times0.0064\times0.434388=0.18349$$

($(0.92)^{10}=0.434388$.) P(X=2) ≈ 0.1835.

(b) $P(X\le1)=P(0)+P(1)$:

$$P(0)=(0.92)^{12}=0.367699$$ $$P(1)=\binom{12}{1}(0.08)(0.92)^{11}=12\times0.08\times0.399662=0.383675$$ $$P(X\le1)=0.367699+0.383675=0.751374$$

P(X ≤ 1) ≈ 0.7514.

(c) Mean and variance:

$$\mu=np=12\times0.08=0.96\text{ bricks}$$ $$\sigma^2=npq=12\times0.08\times0.92=0.8832\text{ bricks}^2$$

Mean ≈ 0.96, Variance ≈ 0.8832 (SD ≈ 0.94).

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

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

(a) Conditions for Poisson: events occur (i) independently, (ii) at a constant average rate, (iii) singly (no two events at exactly the same point), and (iv) the probability of an event in a small interval is proportional to the interval length. It is the limiting form of the binomial when $n\to\infty$, $p\to0$ with $np=\lambda$ finite.

(b) $P(X=0)$:

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

P(no defects) ≈ 0.0498.

(c) $P(X>2)=1-[P(0)+P(1)+P(2)]$:

$$P(1)=e^{-3}\cdot3=0.049787\times3=0.149361$$ $$P(2)=e^{-3}\cdot\frac{3^2}{2!}=0.049787\times4.5=0.224042$$ $$P(0)+P(1)+P(2)=0.049787+0.149361+0.224042=0.423190$$ $$P(X>2)=1-0.423190=0.576810$$

P(X > 2) ≈ 0.5768.

(a) A parameter is a numerical characteristic of the whole population (e.g. population mean $\mu$, population SD $\sigma$); it is usually fixed and unknown. A statistic is a numerical characteristic computed from a sample (e.g. sample mean $\bar{x}$); it varies from sample to sample and is used to estimate a parameter.

The standard error of the mean (SEM) is the standard deviation of the sampling distribution of $\bar{x}$:

$$\text{SE}(\bar{x})=\frac{\sigma}{\sqrt{n}}$$

It measures the variability of the sample mean as an estimator of $\mu$ — smaller SE means a more precise estimate.

(b) By the Central Limit Theorem, for $n=64$ the sample mean is approximately normal:

$$\bar{X}\sim N\!\left(\mu=25,\;\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{64}}=0.5\right)$$

Standardize at $\bar{x}=25.8$:

$$z=\frac{25.8-25}{0.5}=\frac{0.8}{0.5}=1.60$$ $$P(\bar{X}>25.8)=1-\Phi(1.60)=1-0.9452=0.0548$$

P(sample mean > 25.8 MPa) ≈ 0.0548 (5.48%).

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

Under $H_0$, expected frequency for each face $E=\dfrac{120}{6}=20$.

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

$$\chi^2_{\text{cal}}=4.70$$

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

Decision: Since $\chi^2_{\text{cal}}=4.70<11.07$, we fail to reject $H_0$.

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

Hypotheses. $H_0:p_A=p_B$ vs $H_1:p_A\ne p_B$ (two-tailed).

Sample proportions:

$$\hat{p}_A=\frac{24}{200}=0.12,\qquad\hat{p}_B=\frac{27}{150}=0.18$$

Pooled proportion (valid under $H_0$):

$$\hat{p}=\frac{24+27}{200+150}=\frac{51}{350}=0.145714,\qquad\hat{q}=0.854286$$

Standard error of the difference:

$$SE=\sqrt{\hat{p}\hat{q}\left(\frac{1}{n_A}+\frac{1}{n_B}\right)}=\sqrt{0.145714\times0.854286\times\left(\frac{1}{200}+\frac{1}{150}\right)}$$ $$=\sqrt{0.124501\times(0.005+0.006667)}=\sqrt{0.124501\times0.011667}=\sqrt{0.0014525}=0.038112$$

Test statistic:

$$z=\frac{\hat{p}_A-\hat{p}_B}{SE}=\frac{0.12-0.18}{0.038112}=\frac{-0.06}{0.038112}=-1.574$$

Decision: $|z|=1.574<z_{0.025}=1.96$, so we fail to reject $H_0$.

Conclusion: At the 5% level there is no significant difference between the failure proportions of the two curing methods.

(a) Find $k$. A valid pdf integrates to 1:

$$\int_0^2 k\,x(2-x)\,dx=1$$ $$k\int_0^2(2x-x^2)\,dx=k\left[x^2-\frac{x^3}{3}\right]_0^2=k\left(4-\frac{8}{3}\right)=k\cdot\frac{12-8}{3}=k\cdot\frac{4}{3}=1$$ $$\Rightarrow k=\frac{3}{4}=0.75$$

k = 3/4.

(b) Mean $E(X)$:

$$E(X)=\int_0^2 x\,f(x)\,dx=\frac{3}{4}\int_0^2 x\cdot x(2-x)\,dx=\frac{3}{4}\int_0^2(2x^2-x^3)\,dx$$ $$=\frac{3}{4}\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2=\frac{3}{4}\left(\frac{2\cdot8}{3}-\frac{16}{4}\right)=\frac{3}{4}\left(\frac{16}{3}-4\right)=\frac{3}{4}\cdot\frac{16-12}{3}=\frac{3}{4}\cdot\frac{4}{3}=1$$

E(X) = 1 hour (as expected by symmetry of $f$ about $x=1$).