BE Civil Engineering (IOE, TU) Probability and Statistics (IOE, SH 552) Question Paper 2080 Nepal

We use class mid-points $x_i$ and the class width $h = 4$.

(a) Mean, Median, Mode

Mean:

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1584}{50} = 31.68\ \text{MPa}$$

Median: $N/2 = 25$, which falls in the class 28–32 (CF jumps from 13 to 27). Here $L = 28$, $CF_{prev} = 13$, $f = 14$, $h = 4$.

$$\text{Median} = L + \frac{N/2 - CF_{prev}}{f}\,h = 28 + \frac{25 - 13}{14}\times 4 = 28 + \frac{48}{14} = 28 + 3.4286 = 31.43\ \text{MPa}$$

Mode: the modal class is 28–32 (highest frequency 14). $L = 28$, $f_1 = 14$, $f_0 = 9$, $f_2 = 12$, $h = 4$.

$$\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\,h = 28 + \frac{14 - 9}{28 - 9 - 12}\times 4 = 28 + \frac{5}{7}\times 4 = 28 + 2.857 = 30.86\ \text{MPa}$$

Mean = 31.68 MPa, Median = 31.43 MPa, Mode = 30.86 MPa.

(b) Standard deviation and coefficient of variation

$$\sigma = \sqrt{\frac{\sum f_i x_i^2}{N} - \bar{x}^2} = \sqrt{\frac{51656}{50} - (31.68)^2} = \sqrt{1033.12 - 1003.6224} = \sqrt{29.4976}$$ $$\sigma = 5.431\ \text{MPa}$$ $$CV = \frac{\sigma}{\bar{x}}\times 100\% = \frac{5.431}{31.68}\times 100\% = 17.14\%$$

Standard deviation = 5.43 MPa, Coefficient of variation = 17.14%.

(c) Karl Pearson coefficient of skewness

$$S_k = \frac{\bar{x} - \text{Mode}}{\sigma} = \frac{31.68 - 30.86}{5.431} = \frac{0.82}{5.431} = +0.151$$

The coefficient is small and positive, so the distribution is slightly positively (right) skewed — a few high-strength cubes pull the mean above the mode. $S_k \approx +0.15$.

(a) Statements

Total probability theorem: If $B_1, B_2, \dots, B_n$ are mutually exclusive and exhaustive events with $P(B_i) > 0$, then for any event $A$:

$$P(A) = \sum_{i=1}^{n} P(B_i)\,P(A\mid B_i)$$

Bayes' theorem: Under the same conditions,

$$P(B_k \mid A) = \frac{P(B_k)\,P(A\mid B_k)}{\sum_{i=1}^{n} P(B_i)\,P(A\mid B_i)}$$

(b) Probability that a bag is substandard

Let $A$ = bag is substandard. Given:

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

By total probability:

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

P(substandard) = 0.032 (3.2%).

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

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

P(S₂ | substandard) = 0.375 (37.5%).

(d) Most likely source

$$P(S_1 \mid A) = \frac{0.010}{0.032} = 0.3125, \quad P(S_2 \mid A) = \frac{0.012}{0.032} = 0.375, \quad P(S_3 \mid A) = \frac{0.010}{0.032} = 0.3125$$

The three posteriors sum to 1 ($0.3125 + 0.375 + 0.3125 = 1$). The largest posterior is for $S_2$, so the most likely source of a substandard bag is supplier $S_2$ with probability 0.375.

Let $X \sim N(\mu = 30,\ \sigma = 4)$ and $Z = \dfrac{X - \mu}{\sigma}$.

(a) P(26 < X < 34)

$$Z_1 = \frac{26 - 30}{4} = -1, \qquad Z_2 = \frac{34 - 30}{4} = +1$$ $$P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 0.8413 - (1 - 0.8413) = 0.8413 - 0.1587 = 0.6826$$

About 68.26% of cubes lie between 26 and 34 MPa.

(b) P(X < 24) — rejection

$$Z = \frac{24 - 30}{4} = -1.5$$ $$P(X < 24) = P(Z < -1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$$

6.68% of cubes are rejected.

(c) Strength S exceeded by only the top 5%

We need $P(X > S) = 0.05$, i.e. $P(X \le S) = 0.95$. The corresponding $z$-value is $z = 1.645$.

$$S = \mu + z\sigma = 30 + (1.645)(4) = 30 + 6.58 = 36.58\ \text{MPa}$$

S = 36.58 MPa — only 5% of cubes exceed this strength.

(d) Expected number with strength > 34 MPa in 200 cubes

$$Z = \frac{34 - 30}{4} = 1, \qquad P(X > 34) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$$ $$\text{Expected count} = 200 \times 0.1587 = 31.74 \approx 32\ \text{cubes}$$

About 32 cubes are expected to exceed 34 MPa.

(a) Sample mean and standard deviation

Data: 70, 72, 74, 71, 69, 73, 70, 72, 71, 68. Sum $= 710$, $n = 10$.

$$\bar{x} = \frac{710}{10} = 71\ \text{mm}$$

Deviations $d_i = x_i - 71$: $-1, 1, 3, 0, -2, 2, -1, 1, 0, -3$. Squares: $1, 1, 9, 0, 4, 4, 1, 1, 0, 9$; $\sum d_i^2 = 30$.

$$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} = \sqrt{\frac{30}{9}} = \sqrt{3.3333} = 1.826\ \text{mm}$$

Mean = 71 mm, s = 1.826 mm.

(b) Two-tailed t-test

Hypotheses:

$$H_0:\ \mu = 75\ \text{mm} \qquad H_1:\ \mu \neq 75\ \text{mm}$$

Test statistic:

$$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{71 - 75}{1.826/\sqrt{10}} = \frac{-4}{1.826/3.1623} = \frac{-4}{0.5774} = -6.928$$

Degrees of freedom $= n - 1 = 9$. Critical value $t_{0.025,9} = \pm 2.262$.

Since $|t| = 6.928 > 2.262$, the test statistic falls in the rejection region.

(c) Conclusion

We reject $H_0$ at the 5% level. There is strong evidence that the true mean slump differs from (is significantly less than) the claimed 75 mm; the sample mean of 71 mm is significantly lower than the manufacturer's claim. Calculated $t = -6.93$, reject $H_0$.

Let $n = 7$. Compute the required sums.

$\sum x = 3.85$, $\sum y = 236$, $\sum x^2 = 2.1875$, $\sum y^2 = 8158$, $\sum xy = 126.05$.

(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: $7(126.05) - (3.85)(236) = 882.35 - 908.60 = -26.25$.

First bracket: $7(2.1875) - (3.85)^2 = 15.3125 - 14.8225 = 0.49$.

Second bracket: $7(8158) - (236)^2 = 57106 - 55696 = 1410$.

$$r = \frac{-26.25}{\sqrt{0.49 \times 1410}} = \frac{-26.25}{\sqrt{690.9}} = \frac{-26.25}{26.286} = -0.9986$$

r = −0.999, indicating an almost perfect negative linear correlation: strength decreases as the water–cement ratio increases.

(b) Regression line of y on x

Means: $\bar{x} = 3.85/7 = 0.55$, $\bar{y} = 236/7 = 33.714$.

Slope:

$$b_{yx} = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{-26.25}{0.49} = -53.571$$

Intercept:

$$a = \bar{y} - b_{yx}\bar{x} = 33.714 - (-53.571)(0.55) = 33.714 + 29.464 = 63.179$$

Regression line: $\boxed{y = 63.18 - 53.57\,x}$

(c) Estimate at x = 0.58

$$\hat{y} = 63.179 - 53.571(0.58) = 63.179 - 31.071 = 32.11\ \text{MPa}$$

Estimated strength ≈ 32.11 MPa.

Let $X$ = number of defective joints in 6, $X \sim \text{Bin}(n = 6,\ p = 0.08)$, $q = 0.92$.

$$P(X = x) = \binom{6}{x} (0.08)^x (0.92)^{6-x}$$

(a) Conditions: fixed number of independent trials $n$; each trial has only two outcomes (success/failure); probability of success $p$ is constant across trials; trials are independent.

(b) P(X = 2)

$$P(X=2) = \binom{6}{2}(0.08)^2(0.92)^4 = 15 \times 0.0064 \times 0.71639 = 0.06877$$

P(exactly 2 defective) = 0.0688.

(c) P(X ≤ 1) = P(0) + P(1)

$$P(X=0) = (0.92)^6 = 0.60643$$ $$P(X=1) = \binom{6}{1}(0.08)(0.92)^5 = 6 \times 0.08 \times 0.65908 = 0.31636$$ $$P(X \le 1) = 0.60643 + 0.31636 = 0.92279$$

P(at most 1 defective) = 0.9228.

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

(a) PMF:

$$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}, \quad x = 0,1,2,\dots$$

For the Poisson distribution the mean and variance are equal: $E(X) = \text{Var}(X) = \lambda = 3$.

(b) P(X = 2)

$$P(X=2) = \frac{e^{-3}\,3^2}{2!} = \frac{0.049787 \times 9}{2} = \frac{0.448083}{2} = 0.22404$$

P(exactly 2) = 0.2240.

(c) P(X > 2) = 1 − [P(0) + P(1) + P(2)]

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

P(more than 2 accidents) = 0.5768.

Given $n = 64$, $\bar{x} = 12.4$ mm, $\sigma = 0.8$ mm, $z_{0.025} = 1.96$.

(a) 95% confidence interval (z-interval, σ known)

Standard error:

$$SE = \frac{\sigma}{\sqrt{n}} = \frac{0.8}{\sqrt{64}} = \frac{0.8}{8} = 0.1\ \text{mm}$$

Margin of error:

$$E = z\cdot SE = 1.96 \times 0.1 = 0.196\ \text{mm}$$

Confidence interval:

$$\bar{x} \pm E = 12.4 \pm 0.196 = (12.204,\ 12.596)\ \text{mm}$$

95% CI = (12.204 mm, 12.596 mm).

(b) Interpretation: We are 95% confident that the true mean diameter of all steel rods lies between 12.204 mm and 12.596 mm. If sampling were repeated many times, about 95% of such intervals would contain the true mean.

(c) Required sample size for E ≤ 0.1 mm

$$n = \left(\frac{z\,\sigma}{E}\right)^2 = \left(\frac{1.96 \times 0.8}{0.1}\right)^2 = (15.68)^2 = 245.86$$

Round up: n = 246 rods.

Hypotheses

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

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

Test statistic

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

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

Decision: Since calculated $\chi^2 = 3.80 < 11.07$, we fail to reject $H_0$.

Conclusion: There is no significant evidence at the 5% level that the die is biased; the observed frequencies are consistent with a fair die. $\chi^2 = 3.80$, accept $H_0$.

(a) Value of k

Probabilities must sum to 1:

$$0.1 + 0.25 + k + 0.2 + 0.1 = 1 \implies 0.65 + k = 1 \implies k = 0.35$$

k = 0.35.

Distribution: $P(0)=0.1,\ P(1)=0.25,\ P(2)=0.35,\ P(3)=0.2,\ P(4)=0.1$.

(b) Expected value

$$E(X) = \sum x\,P(x) = 0(0.1) + 1(0.25) + 2(0.35) + 3(0.2) + 4(0.1)$$ $$= 0 + 0.25 + 0.70 + 0.60 + 0.40 = 1.95$$

E(X) = 1.95 cranes.

(c) Variance and standard deviation

$$E(X^2) = \sum x^2 P(x) = 0 + 1(0.25) + 4(0.35) + 9(0.2) + 16(0.1)$$ $$= 0.25 + 1.40 + 1.80 + 1.60 = 5.05$$ $$\text{Var}(X) = E(X^2) - [E(X)]^2 = 5.05 - (1.95)^2 = 5.05 - 3.8025 = 1.2475$$ $$\sigma = \sqrt{1.2475} = 1.117$$

Var(X) = 1.2475, standard deviation = 1.117 cranes.

(a) Two-proportion z-test

Method A: $x_1 = 45$, $n_1 = 300$ → $\hat{p}_1 = 45/300 = 0.15$. Method B: $x_2 = 28$, $n_2 = 250$ → $\hat{p}_2 = 28/250 = 0.112$.

Hypotheses:

$$H_0:\ p_1 = p_2 \qquad H_1:\ p_1 \neq p_2$$

Pooled proportion:

$$\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{45 + 28}{300 + 250} = \frac{73}{550} = 0.132727$$ $$\hat{q} = 1 - \hat{p} = 0.867273$$

Standard error:

$$SE = \sqrt{\hat{p}\hat{q}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.132727 \times 0.867273 \times \left(\frac{1}{300} + \frac{1}{250}\right)}$$ $$\frac{1}{300} + \frac{1}{250} = 0.0033333 + 0.0040000 = 0.0073333$$ $$SE = \sqrt{0.115097 \times 0.0073333} = \sqrt{0.00084405} = 0.029052$$

Test statistic:

$$z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.15 - 0.112}{0.029052} = \frac{0.038}{0.029052} = 1.308$$

Critical value $z_{0.025} = \pm 1.96$. Since $|z| = 1.308 < 1.96$, we fail to reject $H_0$.

Conclusion: There is no significant difference between the two curing methods in defective proportion at the 5% level. $z = 1.31$, accept $H_0$.

(b) Type I vs Type II error

• Type I error (α): rejecting the null hypothesis $H_0$ when it is in fact true (a "false positive"). Its probability equals the significance level $\alpha$.
• Type II error (β): failing to reject (accepting) $H_0$ when it is actually false (a "false negative"). Its probability is $\beta$, and $1 - \beta$ is the power of the test.