BE Civil Engineering (IOE, TU) Probability and Statistics (IOE, SH 552) Question Paper 2076 Nepal

We use class mid-values $x$, class width $h = 4$, and $N = \sum f = 50$.

(a) Mean and Median

$$\bar x = \frac{\sum fx}{N} = \frac{1572}{50} = 31.84\ \text{MPa}$$

Median class: $N/2 = 25$ lies in the 28–32 class (CF reaches 28). Here $L=28$, CF before $=13$, $f_m=15$, $h=4$.

$$\text{Median} = L + \frac{N/2 - CF}{f_m}\,h = 28 + \frac{25 - 13}{15}\times 4 = 28 + 3.20 = 31.20\ \text{MPa}$$

(b) Standard deviation and CV

$$s = \sqrt{\frac{\sum f(x-\bar x)^2}{N}} = \sqrt{\frac{1376.35}{50}} = \sqrt{27.527} = 5.247\ \text{MPa}$$ $$\text{CV} = \frac{s}{\bar x}\times 100 = \frac{5.247}{31.84}\times 100 = 16.48\%$$

(c) Karl Pearson skewness (using mode)

Modal class is 28–32 (highest $f=15$). With $L=28$, $f_1=15$, $f_0=9$, $f_2=12$, $h=4$:

$$\text{Mode} = L + \frac{f_1-f_0}{2f_1-f_0-f_2}\,h = 28 + \frac{15-9}{30-9-12}\times 4 = 28 + \frac{6}{9}\times 4 = 28 + 2.667 = 30.667\ \text{MPa}$$ $$S_k = \frac{\bar x - \text{Mode}}{s} = \frac{31.84 - 30.667}{5.247} = \frac{1.173}{5.247} = +0.224$$

Final answers: Mean = 31.84 MPa, Median = 31.20 MPa, s = 5.25 MPa, CV = 16.48%, Mode = 30.67 MPa, $S_k = +0.224$. The small positive skewness means the distribution is mildly skewed to the right (a slightly longer tail toward higher strengths).

(a) Statements

Let $B_1, B_2, \dots, B_n$ be mutually exclusive and exhaustive events with $P(B_i)>0$, and let $A$ be any event.

Total probability: $\;P(A) = \sum_{i=1}^{n} P(B_i)\,P(A\mid B_i).$

Bayes' theorem: $\;P(B_k\mid A) = \dfrac{P(B_k)\,P(A\mid B_k)}{\sum_{i=1}^{n} P(B_i)\,P(A\mid B_i)}.$

Given: $P(S_1)=0.50,\ P(S_2)=0.30,\ P(S_3)=0.20$; $P(F\mid S_1)=0.02,\ P(F\mid S_2)=0.04,\ P(F\mid S_3)=0.05$, where $F$ = "bag fails".

(b) Probability a bag fails (total probability)

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

$P(F) = 0.032$ (3.2%).

(c) Posterior probability it came from $S_2$ (Bayes)

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

$P(S_2\mid F) = 0.375$ (37.5%).

(d) Two bags with replacement, both fail

With replacement the draws are independent, each with failure probability $P(F)=0.032$:

$$P(\text{both fail}) = (0.032)^2 = 0.001024$$

$P(\text{both fail}) = 0.001024 \approx 1.02\times 10^{-3}$.

(a) Properties of the normal distribution

1. It is symmetric and bell-shaped about its mean; mean = median = mode.
2. The total area under the curve is 1, and the curve is asymptotic to the horizontal axis.
3. About 68.27%, 95.45% and 99.73% of values lie within $\mu\pm 1\sigma$, $\mu\pm 2\sigma$, $\mu\pm 3\sigma$ respectively.

Standardize with $z = \dfrac{x-\mu}{\sigma}$, $\mu=30$, $\sigma=4$.

(b) $P(X < 25)$

$$z = \frac{25-30}{4} = -1.25$$ $$P(X<25) = \Phi(-1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056$$

= 10.56% of specimens.

(c) $P(28 < X < 36)$

$$z_1 = \frac{28-30}{4} = -0.50,\qquad z_2 = \frac{36-30}{4} = 1.50$$ $$P = \Phi(1.50) - \Phi(-0.50) = 0.9332 - (1-0.6915) = 0.9332 - 0.3085 = 0.6247$$

= 62.47% of specimens.

(d) Characteristic strength $f_{ck}$ with 5% below

We need $P(X < f_{ck}) = 0.05$, so $z = -1.645$ (5% lies in the lower tail).

$$f_{ck} = \mu + z\sigma = 30 + (-1.645)(4) = 30 - 6.58 = 23.42\ \text{MPa}$$

$f_{ck} = 23.42$ MPa.

Let $n=7$. Build the working table.

$\sum x = 385,\ \sum y = 233,\ \sum x^2 = 21875,\ \sum y^2 = 8003,\ \sum xy = 12400.$

(a) Correlation coefficient

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

Numerator $= 7(12400) - (385)(233) = 86800 - 89705 = -2905.$

$n\sum x^2 - (\sum x)^2 = 7(21875) - 385^2 = 153125 - 148225 = 4900.$

$n\sum y^2 - (\sum y)^2 = 7(8003) - 233^2 = 56021 - 54289 = 1732.$

$$r = \frac{-2905}{\sqrt{4900\times 1732}} = \frac{-2905}{\sqrt{8\,486\,800}} = \frac{-2905}{2913.21} = -0.9972$$

$r = -0.997$. A very strong negative linear relationship: as the water-cement ratio increases, compressive strength decreases.

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

$$b_{yx} = \frac{n\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2} = \frac{-2905}{4900} = -0.5929$$

$\bar x = 385/7 = 55,\ \bar y = 233/7 = 33.2857.$

$$a = \bar y - b_{yx}\bar x = 33.2857 - (-0.5929)(55) = 33.2857 + 32.6071 = 65.893$$

Regression line: $\;\hat y = 65.893 - 0.5929\,x.$

(c) Estimate at $x = 58\%$

$$\hat y = 65.893 - 0.5929(58) = 65.893 - 34.388 = 31.51\ \text{MPa}$$

Estimated compressive strength ≈ 31.51 MPa.

Sample size $n = 10$. Data sum $= 33+34+36+32+35+31+34+33+35+32 = 335$.

$$\bar x = \frac{335}{10} = 33.5\ \text{MPa}$$

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

Sample variance (with $n-1$ divisor):

$$s^2 = \frac{\sum d^2}{n-1} = \frac{22.50}{9} = 2.50,\qquad s = 1.581\ \text{MPa}$$

(a) Hypotheses (claim: mean is at least 35 MPa)

$$H_0:\ \mu = 35\ \text{MPa} \qquad H_1:\ \mu < 35\ \text{MPa (left-tailed)}$$

(b) Test statistic

$$t = \frac{\bar x - \mu_0}{s/\sqrt{n}} = \frac{33.5 - 35}{1.581/\sqrt{10}} = \frac{-1.5}{1.581/3.1623} = \frac{-1.5}{0.5} = -3.0$$

Decision rule (left-tailed, 5%): reject $H_0$ if $t < -t_{0.05,9} = -1.833$. Since $t = -3.0 < -1.833$, we reject $H_0$.

Conclusion: There is sufficient evidence at the 5% level that the true mean strength is less than 35 MPa; the manufacturer's claim is not supported.

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

$$\bar x \pm t_{0.025,9}\frac{s}{\sqrt n} = 33.5 \pm 2.262\times 0.5 = 33.5 \pm 1.131$$

95% CI = (32.37 MPa, 34.63 MPa). Note that 35 MPa lies outside this interval, consistent with rejecting the claim.

Let $X$ = number of sub-standard bars in the sample. Then $X \sim \text{Binomial}(n=12,\ p=0.08)$, $q = 0.92$.

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

(a) Exactly 2

$$P(X=2) = \binom{12}{2}(0.08)^2(0.92)^{10} = 66 \times 0.0064 \times 0.434388 = 0.18348$$

($0.92^{10} = 0.434388$.)

$P(X=2) = 0.1835.$

(b) At most 1

$$P(X\le 1) = P(0) + P(1)$$ $$P(0) = (0.92)^{12} = 0.367696$$ $$P(1) = \binom{12}{1}(0.08)(0.92)^{11} = 12\times 0.08\times 0.399670 = 0.383683$$

($0.92^{11} = 0.399670$, $0.92^{12} = 0.367696$.)

$$P(X\le 1) = 0.367696 + 0.383683 = 0.751379$$

$P(X\le 1) = 0.7514.$

(c) Mean and variance

$$\mu = np = 12\times 0.08 = 0.96\ \text{bars}$$ $$\sigma^2 = npq = 12\times 0.08\times 0.92 = 0.8832\ \text{bars}^2$$

Mean = 0.96, Variance = 0.8832.

Let $X$ = number of accidents per month, $X \sim \text{Poisson}(\lambda = 1.5)$.

$$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!},\qquad e^{-1.5} = 0.223130$$

(a) No accidents in a month

$$P(X=0) = e^{-1.5} = 0.2231$$

$P(X=0) = 0.2231.$

(b) At least 2 accidents in a month

$$P(X\ge 2) = 1 - P(0) - P(1)$$ $$P(1) = e^{-1.5}(1.5) = 0.223130\times 1.5 = 0.334695$$ $$P(X\ge 2) = 1 - 0.223130 - 0.334695 = 0.442175$$

$P(X\ge 2) = 0.4422.$

(c) Exactly 3 in two months For a 2-month period the mean is $\lambda' = 1.5\times 2 = 3.0$, with $e^{-3.0} = 0.049787$.

$$P(X=3) = \frac{e^{-3.0}(3.0)^3}{3!} = \frac{0.049787\times 27}{6} = \frac{1.344249}{6} = 0.224042$$

$P(X=3) = 0.2240.$

(a) Value of $k$

Probabilities must sum to 1:

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

$k = 0.3.$

Complete table:

(b) Expected value

$$E(X) = \sum x P(x) = 1.9$$

$E(X) = 1.9.$

(c) Variance

$$E(X^2) = \sum x^2 P(x) = 4.9$$ $$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 4.9 - (1.9)^2 = 4.9 - 3.61 = 1.29$$

$\operatorname{Var}(X) = 1.29$ (standard deviation $= 1.136$).

(a) Central Limit Theorem

For a random sample of size $n$ drawn from a population with mean $\mu$ and finite standard deviation $\sigma$, the distribution of the sample mean $\bar X$ approaches a normal distribution with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$ as $n$ becomes large (in practice $n \ge 30$), regardless of the shape of the parent population.

(b) Standard error of the mean

$$\text{SE} = \frac{\sigma}{\sqrt n} = \frac{6}{\sqrt{36}} = \frac{6}{6} = 1\ \text{mm}$$

SE = 1 mm.

(c) $P(\bar X > 101.5)$

$$z = \frac{\bar x - \mu}{\text{SE}} = \frac{101.5 - 100}{1} = 1.50$$ $$P(\bar X > 101.5) = 1 - \Phi(1.50) = 1 - 0.9332 = 0.0668$$

$P(\bar X > 101.5) = 0.0668$ (6.68%).

Hypotheses

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

Expected frequency under $H_0$: each face $E = 120/6 = 20$.

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

$$\chi^2_{\text{cal}} = 3.40$$

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

Since $\chi^2_{\text{cal}} = 3.40 < 11.07$, we do not reject $H_0$.

Conclusion: At the 5% level of significance, the data provide no evidence against fairness; the die may be regarded as unbiased.

(a) Point estimate of the proportion

$$\hat p = \frac{x}{n} = \frac{60}{400} = 0.15$$

$\hat p = 0.15$ (15%). Then $\hat q = 1-\hat p = 0.85$.

(b) Standard error of $\hat p$

$$\text{SE} = \sqrt{\frac{\hat p\,\hat q}{n}} = \sqrt{\frac{0.15\times 0.85}{400}} = \sqrt{\frac{0.1275}{400}} = \sqrt{0.00031875} = 0.017854$$

SE = 0.01785.

(c) 95% confidence interval

$$\hat p \pm z_{0.025}\,\text{SE} = 0.15 \pm 1.96\times 0.017854 = 0.15 \pm 0.035$$

Margin of error $= 1.96\times 0.017854 = 0.034994 \approx 0.035$.

95% CI = (0.115, 0.185), i.e. between about 11.5% and 18.5% of all columns are expected to have surface cracks.