MATH 170 Exhaustive Study Guide: Spring 2026 Exams 2 and 3

Administrative Guidelines and Academic Honesty Policy

The MATH 170 examinations (Exam 2 and Exam 3, Spring 2026) are conducted under a strict Academic Honesty Policy. Students must pledge that they have neither given nor received unauthorized assistance. Proper identification is required on all detached sheets. Permitted tools include Desmos (on a laptop or in Test Mode on a phone), Excel, or a TI-84 calculator. Collaboration is strictly prohibited.

Time Management and Penalties:

• A specific "stop time" is designated on the board.
• For every minute or fraction thereof exceeded past the stop time, a penalty of $2$ points is deducted from the final score.

Statistical Formula References and Computational Commands

Probability and Distribution Formulas

• Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$
• Expected Value (Discrete): $E(X) = \sum x p(x)$
• Variance (Definition): $\sigma^2 = \sum [(x - \mu)^2 p(x)]$
• Variance (Computational): $\sigma^2 = E(X^2) - E(X)^2$
• Interquartile Range: $IQR = Q_3 - Q_1$
• Complement Rule: $P(A^c) = 1 - P(A)$
• General Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
• Limit for Outliers: $Q_1 - 1.5 \times IQR$ and $Q_3 + 1.5 \times IQR$
• Independence Multiplication Rule: If $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$
• General Multiplication Rule: $P(A \cap B) = P(A)P(B|A)$

Sampling Distributions and Binomial Variables

• Binomial Mean: $\mu = np$
• Binomial Standard Deviation: $\sigma = \sqrt{np(1-p)}$
• Distribution of Sample Means: $\mu_{\bar{x}} = \mu$ and $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
• Distribution of Sample Proportions: $\mu_{\hat{p}} = p$ and $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

Inference: Test Statistics and Confidence Intervals

• Z-score for Sample Mean: $Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$
• T-score for Sample Mean: $t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$
• Z-score for Proportion: $Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$
• Confidence Interval for Means (T-distribution): $\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$
• Confidence Interval for Proportions (Z-distribution): $\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
• Sample Size for Mean: $n = \left( \frac{z_{\alpha/2} \cdot \sigma}{E} \right)^2$
• Sample Size for Proportion: $n = \left( \frac{z_{\alpha/2}}{E} \right)^2 \hat{p}(1-\hat{p})$

Software Operation Commands

• Desmos:   - Binomial Distribution: binomialdist(n, p)   - Normal Distribution: normaldist(\mu, \sigma)   - Inverse Normal: normaldist(\mu, \sigma).inversecdf(p)   - T Distribution: tdist(df)   - Inverse T: tdist(df).inversecdf(area)
• Excel:   - Binomial (PMF): binomdist(x, n, p, FALSE)   - Binomial (CDF): binomdist(x, n, p, TRUE)   - Normal (CDF): normdist(x, \mu, \sigma, TRUE)   - Inverse Normal: norminv(p, \mu, \sigma)
• TI-84 Calculator:   - Binomial (PMF/CDF): binompdf(n, p, x) / binomcdf(n, p, x)   - Normal (CDF): normalcdf(lower, upper, \mu, \sigma)   - Inverse Normal: invNorm(p, \mu, \sigma)   - T Distribution: tcdf(lower, upper, df)   - Inverse T: invT(area, df)

Exam 2: Probability and Sampling Distributions

Problem 1: Normal Distribution Applied to Apple Weights

Apples sold at a store have weights approximately normally distributed with a mean $\mu = 183\,g$ and a standard deviation $\sigma = 13\,g$.

• (a) Z-scores: Calculate z-scores for $x = 160\,g$ and $x = 190\,g$.   - $Z_{160} = \frac{160 - 183}{13} \approx -1.769$   - $Z_{190} = \frac{190 - 183}{13} \approx 0.538$
• (b) Probability Range: Calculate the probability that a randomly selected apple weighs between $160\,g$ and $190\,g$.
• (c) Percentiles: Determine the weight that marks the $90th$ percentile ($0.90$ area to the left), which represents the cutoff for the top $10\%$ heaviest apples.
• (d) Central Limit Theorem: For a sample size $n = 38$, calculate the probability that the sample average weight $P(\bar{x} > 185)$ exceeds $185\,g$.

Problem 2: Binomial Distribution (Kyle Schwarber Batting)

Batting average $p = 0.231$ ($23.1\%$ chance of a hit). Consider a series with $n = 16$ turns at bat.

• (a) Cumulative Probability: Determine the probability of scoring at most $2$ hits ($P(X \le 2)$).
• (b) Descriptive Statistics:   - Expected average number of hits: $\mu = np = 16 \times 0.231 = 3.696$   - Standard deviation: $\sigma = \sqrt{np(1-p)}$
• (c) Normal Approximation Check: Evaluate if a normal distribution can approximate part (a). This requires checking if $np > 10$ and $n(1-p) > 10$.

Problem 3: Expected Value of a Lottery Ticket

A lottery ticket costs $\$10$. Result possibilities are as follows:

• Win $\$500$ with $P = 0.003$
• Win $\$100$ with $P = 0.04$
• Win $\$0$ with $P = ?$
• (a) Winning Ticket Probability: Sum of win probabilities ($0.003 + 0.04 = 0.043$).
• (b) Losing Ticket Probability: Complement of winning ($1 - 0.043 = 0.957$).
• (c) Expected Gain/Loss: Calculate $E(X) = \sum x p(x)$ where $x$ is the net gain (e.g., $500 - 10$, $100 - 10$, or $0 - 10$).
• (d) Bonus: Calculate the standard deviation $\sigma$ for this random variable.

Problem 4: Sampling Distributions of Software Salaries

Company salaries have $\mu = \$80,000$ and $\sigma = \$15,000$. The population distribution shape is unknown.

• (a) Sample Size Constraint: For $n = 20$, can we determine $P(\bar{x} > \$90,000)$? No, because the population is not normal and the sample size is less than $30$, failing the Central Limit Theorem requirements.
• (b) Large Sample Probability: For $n = 40$, calculate $P(\bar{x} > \$83,000)$.
• (c) Z-score for Sample Mean: If a sample of $n = 40$ has $\bar{x} = \$76,000$, find the Z-score relative to the sampling distribution.

Problem 5: Sampling Distribution of Proportions

Car ownership among college students is $57\%$ ($p = 0.57$). Sample size $n = 90$.

• (a) Mean and Standard Deviation:   - $\mu_{\hat{p}} = 0.57$   - $\sigma_{\hat{p}} = \sqrt{\frac{0.57(1-0.57)}{90}}$
• (b) Probability: Use the normal distribution to find the probability that sample proportion $\hat{p} \le 0.55$. Explain compliance with requirements ($np \ge 10, n(1-p) \ge 10$).

Problem 6: Contingency Tables and Independence

Data on final grades (F, D, C, B, A) based on course delivery (Online vs. In-person):

• (a) Conditional Probability: $P(F|O) = \frac{4}{16} = 0.25$.
• (b) Independence Test: Check if $P(F \cap O) = P(F) \times P(O)$.   - $P(F \cap O) = \frac{4}{40} = 0.1$   - $P(F) = \frac{7}{40} = 0.175$   - $P(O) = \frac{16}{40} = 0.4$   - $0.175 \times 0.4 = 0.07$. Since $0.1 \neq 0.07$, the events are not independent.

Exam 3: Inferential Statistics (Inference and Hypothesis Testing)

Problem 1: Confidence Interval for Proportions (Candidate Vote)

A simple random sample of $n = 83$ voters shows $x = 34$ will vote for a candidate. Construct a $90\%$ confidence interval.

• (a) Requirement Check: CLT requires $n\hat{p} \ge 10$ and $n(1-\hat{p}) \ge 10$. Student Memphis Woodard notes "SRS" and "CLT: $np \ge 10, n(1-p) \ge 10$".
• (b) Margin of Error Calculation:   - $\hat{p} = \frac{34}{83} = 0.4096$   - Critical value $z_{\alpha/2}$ for $90\%$ is $1.645$.   - $E = z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 1.645 \sqrt{\frac{0.4096(0.5904)}{83}} \approx 0.05397$.
• (c) Interval Calculation:   - Lower bound: $0.4096 - 0.05397 = 0.35563$   - Upper bound: $0.4096 + 0.05397 = 0.46357$   - Interval: $(0.3556, 0.4636)$.
• (d) Interpretation: "We are $90\%$ confident that the proportion of voters who will vote for the candidate is between $0.35563$ and $0.46357$."

Problem 2: Hypothesis Test for Means (Skittles Count)

Sally tests if a pack of Skittles contains less than the advertised average of $59$ pieces. $\alpha = 0.10$.

• (a) Hypotheses:   - $H_0: \mu = 59$   - $H_A: \mu < 59$
• (b) Degrees of Freedom: For $n = 60$, $df = n - 1 = 60 - 1 = 59$.
• (c) Conclusion: P-value is $0.08$. Since the P-value $(0.08) < \alpha (0.10)$, we reject the null hypothesis. The student indicates this was a "Correct!" conclusion.

Problem 3: Confidence Interval Interpretation

If an $80\%$ confidence interval for a population mean is $(24.5, 28.0)$, the chance that the true population mean does NOT fall in the interval is $100\% - 80\% = 20\%$.

Problem 4: Hypothesis Test (Internet Download Speed)

Provider claims $\mu = 100\,Mbps$. You believe it is lower. Sample $n = 40$, $\bar{x} = 95\,Mbps$, $s = 15\,Mbps$. $\alpha = 0.05$.

• (a) Hypotheses:   - $H_0: \mu = 100$   - $H_A: \mu < 100$
• (b) Test Statistic: $t = \frac{95 - 100}{\frac{15}{\sqrt{40}}} = \frac{-5}{2.3717} \approx -2.1081$.
• (c) P-value and Decision: Use tdist with $df = 39$. P-value is less than $0.05$. Decision: Reject $H_0$.
• (d) Sentence Conclusion: There is sufficient evidence to conclude the average speed is less than $100\,Mbps$.

Problem 5: Sample Size for Proportions (Congenital Disease)

Team wants a $98\%$ confidence interval with a margin of error $E \le 2\%$.

• (a) Critical Value: For $98\%$, confidence level area in tails is $0.02$. Area in one tail is $0.01$. $z_{\alpha/2} \approx 2.326$.
• (b) Sample Size Calculation: Use $p = 0.5$ for maximum variability when no estimate is given. $n = \left( \frac{2.326}{0.02} \right)^2 (0.5)(0.5)$.

Problem 6: Error Types in Hypothesis Testing

James tests if more than $50\%$ of people prefer dogs over cats ($H_0: p \le 0.50$, $H_A: p > 0.50$) at $\alpha = 0.10$.

• (a) Type I Error: Concluding that more than $50\%$ of people prefer dogs when, in reality, the proportion is $50\%$ or less.
• (b) Probability of Type I Error: The probability of making a Type I error is equal to the significance level, $\alpha$, which is $0.10$ or $10\%$.