FE Civil Probability & Statistics practice problems

The mode is simply the value that appears most often. The value 2.1 shows up three times, more than any other, so it is the mode. The choice 2.8 mm is the median once the data is sorted, not the mode. The choice 2.84 mm is the arithmetic mean of the seven values, not the mode. The choice 3.4 mm appears only twice, fewer than the three occurrences of 2.1.

1. Tally the occurrences of each value:
   $2.1 \rightarrow 3,\quad 3.4 \rightarrow 2,\quad 2.8 \rightarrow 1,\quad 4.0 \rightarrow 1$
2. The value with the highest frequency is the mode:
   $\text{Mode} = 2.1\,\text{mm}$

FE Reference Handbook 10.6, p. 64

Because the four lifts are stated to be the whole population, divide the sum of squared deviations by $N = 4$, not $n - 1 = 3$. The mean is 51. The squared deviations sum to 20, so the variance is $20/4 = 5$ and the standard deviation is $\sqrt{5} = 2.24$. The 2.58 mm choice is the sample standard deviation, which divides by $n - 1 = 3$, which is wrong here because the problem says population. The 5.0 mm choice is the variance, not the standard deviation; the square root was skipped. The 51.0 mm choice is the mean, not a measure of spread.

1. Compute the mean:
   $\mu = \frac{50 + 54 + 48 + 52}{4} = \frac{204}{4} = 51$
2. Sum the squared deviations:
   $(-1)^2 + 3^2 + (-3)^2 + 1^2 = 1 + 9 + 9 + 1 = 20$
3. Population variance (divide by $N = 4$):
   $\sigma^2 = \frac{20}{4} = 5$
4. Population standard deviation:
   $\sigma = \sqrt{5} = 2.24\,\text{mm}$

FE Reference Handbook 10.6, p. 64

The coefficient of determination $R^2 = r^2 = 0.95^2 = 0.9025$ is the fraction of variation explained. The fraction NOT explained is the complement: $1 - 0.9025 = 0.0975$. The choice 0.05 is 1 - r, mistakenly using the complement of the correlation coefficient instead of $R^2$. The choice 0.9025 is the fraction explained, not the unexplained portion. The choice 0.20 does not correspond to any correct step; it resembles a guess from misreading the slope.

1. Compute the coefficient of determination:
   $R^2 = r^2 = (0.95)^2 = 0.9025$
2. The fraction explained is 0.9025. The fraction unexplained is the complement:
   $1 - R^2 = 1 - 0.9025 = 0.0975$

FE Reference Handbook 10.6, p. 71

Here you are given $\hat{y}$ and must solve for $x$ by inverting the line: $x = (\hat{y} - a)/b = (1455 - (-120))/450 = 1575/450 = 3.50$. The 2.97 ft option drops the intercept sign and computes (1455 - 120)/450. The 3.23 ft option ignores the intercept entirely, using 1455/450. The 654,630 ft option multiplies instead of dividing by the slope, a setup error.

1. Start from the regression line and solve for $x$:
   $x = \frac{\hat{y} - a}{b}$
2. Substitute $\hat{y} = 1{,}455$, $a = -120$, $b = 450$:
   $x = \frac{1{,}455 - (-120)}{450} = \frac{1{,}575}{450}$
3. Divide:
   $x = 3.50\,\text{ft}$

FE Reference Handbook 10.6, p. 70

Ranking the three bonuses means order matters, so use permutations: $P(7,3) = 7!/(7-3)! = 7 \times 6 \times 5 = 210$. The choice 35 is the combination C(7,3), which ignores ranking. The choice 343 is $7^3$, treating each award as an independent pick with replacement. The choice 5,040 is $7!$, arranging all seven leaders rather than just three.

1. Order matters (ranked bonuses), so use permutations:
   $P(7, 3) = \frac{7!}{(7-3)!} = \frac{7!}{4!}$
2. Simplify:
   $7 \times 6 \times 5 = 210$

FE Reference Handbook 10.6, p. 65

For a binomial variable the variance is $npq$ and the standard deviation is its square root. Here $n = 80$, $p = 0.10$, $q = 0.90$, so $\sigma^2 = 80(0.10)(0.90) = 7.2$ and $\sigma = \sqrt{7.2} = 2.68$. The choice 7.2 is the variance, not the standard deviation; the square root was skipped. The choice 8.0 is the mean $np = 80(0.10)$, not the spread. The choice 2.83 is $\sqrt{np} = \sqrt{8}$, which forgets to multiply by $q$.

1. Identify parameters: $n = 80$, $p = 0.10$, $q = 1 - p = 0.90$.
2. Compute the variance:
   $\sigma^2 = npq = 80(0.10)(0.90) = 7.2$
3. Take the square root:
   $\sigma = \sqrt{7.2} = 2.68$

FE Reference Handbook 10.6, p. 68

Compute the expected value of the risky bid: $E(A) = 40{,}000(0.6) + (-10{,}000)(0.4) = 24{,}000 - 4{,}000 = 20{,}000$ dollars. Bid B is a certain 25,000 dollars. Since $25{,}000 > 20{,}000$, Bid B is preferred by 5,000 dollars. "Bid A, by 5,000 dollars" reverses the comparison, picking the risky bid. "Bid B, by 15,000 dollars" uses 15,000 dollars, which comes from treating the loss as a positive payoff so that $E(A) = 24{,}000 + 4{,}000 = 28{,}000$ and then taking the wrong difference. "They are equal" ignores the probability weighting entirely.

1. Expected value of Bid A (keep the loss negative):
   $E(A) = 40{,}000(0.6) + (-10{,}000)(0.4)$
2. Evaluate each term:
   $24{,}000 + (-4{,}000) = 20{,}000 \text{ dollars}$
3. Bid B is certain:
   $E(B) = 25{,}000 \text{ dollars}$
4. Compare and take the difference:
   $E(B) - E(A) = 25{,}000 - 20{,}000 = 5{,}000 \text{ dollars (Bid B preferred)}$

FE Reference Handbook 10.6, p. 67

Expected value is linear, so $E(U) = 1.2\,E(S) + 1.6\,E(T)$ regardless of independence. That gives $1.2(90) + 1.6(35) = 108 + 56 = 164$. The 125 option adds the means without applying the load factors. The 200 option applies a factor of 1.6 to both means. The 150 option applies a factor of 1.2 to both means. Only the 164 option uses the correct individual factors.

1. Apply linearity of expectation:
   $E(U) = 1.2\,E(S) + 1.6\,E(T)$
2. Substitute the means:
   $E(U) = 1.2(90) + 1.6(35) = 108 + 56$
3. Sum:
   $E(U) = 164\,\text{kN}$

FE Reference Handbook 10.6, p. 67

Because $\sigma$ is known, use the z-interval at 90% confidence with $z = 1.645$. The standard error is $0.8/\sqrt{64} = 0.8/8 = 0.10$, so the margin is $1.645 \times 0.10 = 0.164$, and the interval is $19.6 \pm 0.16 = (19.44, 19.76)$. The (19.40, 19.80) choice uses $z = 1.96$ (95% value) instead of the given 90% value. The (18.28, 20.92) choice forgets to divide by $\sqrt{n}$, using $1.645 \times 0.8 = 1.32$. The (19.58, 19.62) choice divides by $n = 64$ instead of $\sqrt{64} = 8$.

1. Identify: $\bar{x} = 19.6$, $\sigma = 0.8$, $n = 64$, $z_{\alpha/2} = 1.645$.
2. Compute the standard error:
   $\frac{\sigma}{\sqrt{n}} = \frac{0.8}{\sqrt{64}} = \frac{0.8}{8} = 0.10$
3. Margin of error:
   $E = 1.645 \times 0.10 = 0.164$
4. Confidence interval:
   $19.6 \pm 0.164 = (19.44,\; 19.76)$

FE Reference Handbook 10.6, p. 75

The margin of error is $E = z\,\sigma/\sqrt{n}$, so $E$ is proportional to $1/\sqrt{n}$. To halve $E$ you need $\sqrt{n}$ to double, which means $n$ must increase by a factor of $2^2 = 4$. So $n$ goes from 25 to 100. The "double the sample size to $n = 50$" option only reduces $E$ by a factor of $1/\sqrt{2} \approx 0.71$, not one half. The "halve the sample size to $n \approx 13$" option is backwards: shrinking $n$ widens the interval. The "increase by 50% to $n \approx 38$" option underestimates the required increase because it ignores the square-root relationship.

1. Margin of error depends on $n$ through:
   $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \;\propto\; \frac{1}{\sqrt{n}}$
2. To halve $E$, set the new margin equal to half the old:
   $\frac{1}{\sqrt{n_{\text{new}}}} = \frac{1}{2}\cdot\frac{1}{\sqrt{n_{\text{old}}}}$
3. Solve for the sample-size ratio:
   $\sqrt{n_{\text{new}}} = 2\sqrt{n_{\text{old}}} \;\Rightarrow\; n_{\text{new}} = 4\,n_{\text{old}}$
4. Apply to $n_{\text{old}} = 25$:
   $n_{\text{new}} = 4(25) = 100$

FE Reference Handbook 10.6, p. 75

Compute the z-statistic: standard error is $14/\sqrt{49} = 14/7 = 2.0$, so $z = (106 - 100)/2.0 = 3.0$. Since $|3.0| > 1.96$, reject $H_0$: the slump differs from the target. The "fail to reject" choice computes $z = 3.0$ correctly but then draws the wrong decision; 3.0 exceeds 1.96, so you reject. The choices using $z = 0.43$ get it from dividing by $\sigma$ itself (6/14) instead of the standard error $\sigma/\sqrt{n}$.

1. Set up a two-tailed test: $H_0: \mu = 100$, $H_1: \mu \neq 100$.
2. Compute the standard error:
   $\frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2.0$
3. Compute the test statistic:
   $z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{106 - 100}{2.0} = 3.0$
4. Compare: $|z| = 3.0 > 1.96$. Reject $H_0$.

FE Reference Handbook 10.6, p. 74

Rejecting $H_0$ when $H_0$ is actually true is a Type I error (a false alarm), and its probability is the significance level $\alpha$, chosen before the test. The "Type II error, controlled by $\beta$" choice describes a Type II error (failing to reject a false $H_0$), which is the opposite situation. The "Type I error, controlled by $\beta$" and "Type II error, controlled by $\alpha$" choices pair the error type with the wrong controlling probability: Type I goes with $\alpha$, and Type II goes with $\beta$.

1. Rejecting $H_0$ while $H_0$ is in fact true is a Type I error.
2. The probability of a Type I error is the significance level, chosen before testing:
   $P(\text{Type I}) = \alpha$
3. For contrast, failing to reject a false $H_0$ is a Type II error with probability $\beta$.

FE Reference Handbook 10.6, p. 73