Grade: Grade 12 Subject: Mathematics Unit: Statistics Lesson: 4 of 6 SAT: ProblemSolving+DataAnalysis ACT: Math

Word Problems

Apply statistical concepts to real-world scenarios involving surveys, experiments, quality control, and data analysis.

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Word problems require you to translate real-world situations into statistical language. Success depends on identifying the key information, choosing the right statistical method, and interpreting results in context.

Problem-Solving Framework

  1. Read carefully: Identify what is being asked
  2. Extract data: List all numerical information and given conditions
  3. Identify the statistical concept: Is this about probability, confidence intervals, hypothesis testing, or descriptive statistics?
  4. Set up the problem: Write the appropriate formula or test
  5. Calculate: Show your work step by step
  6. Interpret: State your answer in the context of the original problem

Common Contexts for Statistics Word Problems

  • Medical studies and clinical trials
  • Quality control in manufacturing
  • Survey results and polling
  • Educational testing and assessment
  • Business and economics data
  • Sports statistics
  • Environmental and scientific research

Examples

Example 1: Quality Control

Problem: A factory produces light bulbs with a mean lifespan of 1200 hours and standard deviation of 100 hours. The quality control department tests a random sample of 49 bulbs. What is the probability that the sample mean lifespan is between 1180 and 1220 hours?

Show Solution
  1. Given: Population mean = 1200, sd = 100, n = 49
  2. Standard error: SE = 100 / sqrt(49) = 100 / 7 = 14.29
  3. Z-scores:
    • z1 = (1180 - 1200) / 14.29 = -1.40
    • z2 = (1220 - 1200) / 14.29 = 1.40
  4. Probability: P(-1.40 < Z < 1.40) = 0.9192 - 0.0808 = 0.8384
  5. Answer: There is an 83.84% probability that the sample mean falls between 1180 and 1220 hours.

Example 2: Medical Study

Problem: A pharmaceutical company claims a new drug reduces blood pressure by at least 10 points on average. In a clinical trial of 64 patients, the mean reduction was 8.5 points with a standard deviation of 6 points. At the 0.05 significance level, is there evidence to doubt the company's claim?

Show Solution
  1. Hypotheses:
    • H0: mean >= 10 (company's claim)
    • Ha: mean < 10 (doubt the claim)
  2. Test statistic: z = (8.5 - 10) / (6 / sqrt(64)) = -1.5 / 0.75 = -2.0
  3. Critical value: z = -1.645 (one-tailed, alpha = 0.05)
  4. Decision: Since -2.0 < -1.645, reject H0
  5. Conclusion: There is sufficient evidence at the 0.05 level to doubt the company's claim that the drug reduces blood pressure by at least 10 points on average.

Practice Problems

Solve these 10 real-world statistics problems.

Problem 1: College Admissions

A college admission test has scores normally distributed with mean 500 and standard deviation 100. If 2500 students take the test, approximately how many will score above 650?

Show Answer

z = (650 - 500) / 100 = 1.5

P(Z > 1.5) = 0.0668

Number of students = 2500 * 0.0668 = 167 students

Problem 2: Customer Satisfaction Survey

A survey of 400 customers found that 68% were satisfied with a product. Construct a 99% confidence interval for the true proportion of satisfied customers.

Show Answer

p = 0.68, n = 400, z = 2.576 for 99% CI

SE = sqrt(0.68 * 0.32 / 400) = 0.0233

CI = 0.68 +/- 2.576 * 0.0233 = 0.68 +/- 0.06

(0.62, 0.74) or 62% to 74%

Problem 3: Manufacturing Process

A machine fills cereal boxes with a target weight of 18 ounces. A sample of 36 boxes has a mean weight of 17.8 ounces with standard deviation 0.6 ounces. At alpha = 0.05, is the machine underfilling boxes?

Show Answer

H0: mean >= 18, Ha: mean < 18

z = (17.8 - 18) / (0.6 / 6) = -0.2 / 0.1 = -2.0

Critical value: -1.645

Since -2.0 < -1.645, reject H0. Evidence suggests the machine is underfilling.

Problem 4: Election Polling

A pollster wants to estimate voter support with a margin of error of 2% at 95% confidence. What is the minimum sample size needed?

Show Answer

Using p = 0.5 for maximum variability:

n = (1.96)^2 * 0.5 * 0.5 / (0.02)^2

n = 0.9604 / 0.0004 = 2401

Problem 5: Test Score Comparison

Students in Class A (n=30) have a mean test score of 78 with sd = 8. Students in Class B (n=25) have a mean of 82 with sd = 10. Calculate the 95% confidence interval for the difference in means (B - A).

Show Answer

Difference = 82 - 78 = 4

SE = sqrt(8^2/30 + 10^2/25) = sqrt(2.133 + 4) = sqrt(6.133) = 2.48

CI = 4 +/- 1.96 * 2.48 = 4 +/- 4.86

(-0.86, 8.86)

Note: Since the interval contains 0, the difference is not statistically significant.

Problem 6: Website Traffic

A website averages 1000 visitors per day with standard deviation 150. Over the next 36 days, what is the probability that the average daily visitors will exceed 1050?

Show Answer

SE = 150 / sqrt(36) = 25

z = (1050 - 1000) / 25 = 2.0

P(Z > 2.0) = 1 - 0.9772 = 0.0228 or 2.28%

Problem 7: Restaurant Wait Times

A restaurant claims average wait time is 20 minutes. A food critic visits 16 times and records a mean wait of 24 minutes with sd = 6 minutes. At alpha = 0.01, test if wait times exceed the claim.

Show Answer

H0: mean <= 20, Ha: mean > 20

t = (24 - 20) / (6 / 4) = 4 / 1.5 = 2.67

t-critical (df=15, one-tailed, alpha=0.01) = 2.602

Since 2.67 > 2.602, reject H0. Wait times significantly exceed 20 minutes.

Problem 8: Product Defects

A company claims only 3% of their products are defective. In a sample of 500 products, 22 are defective. At alpha = 0.05, is the defect rate higher than claimed?

Show Answer

p-hat = 22/500 = 0.044, p0 = 0.03

z = (0.044 - 0.03) / sqrt(0.03 * 0.97 / 500)

z = 0.014 / 0.00763 = 1.83

Critical value: 1.645

Since 1.83 > 1.645, reject H0. Defect rate is higher than 3%.

Problem 9: Salary Analysis

Employee salaries at a company are normally distributed with mean $65,000 and standard deviation $8,000. What salary represents the 75th percentile?

Show Answer

z for 75th percentile = 0.674

Salary = 65000 + 0.674 * 8000

Salary = 65000 + 5392 = $70,392

Problem 10: Sports Performance

A basketball player's free throw percentage is 75%. In the next 40 free throw attempts, what is the probability she makes at least 35?

Show Answer

Using normal approximation: mean = 40 * 0.75 = 30, sd = sqrt(40 * 0.75 * 0.25) = 2.74

P(X >= 35) with continuity correction: P(X >= 34.5)

z = (34.5 - 30) / 2.74 = 1.64

P(Z >= 1.64) = 1 - 0.9495 = 0.0505 or about 5%

Check Your Understanding

Reflect on these questions:

  1. How do you decide between a one-tailed and two-tailed test based on a word problem?
  2. What key phrases indicate you need to construct a confidence interval?
  3. When should you use a proportion test versus a mean test?
  4. How do you identify the appropriate significance level from a problem description?
Show Key Answers
  1. One-tailed: phrases like "more than," "less than," "exceeds," "at least," "at most"; Two-tailed: phrases like "different from," "changed," "not equal to"
  2. Phrases like "estimate," "within," "range of values," "how confident," "interval"
  3. Proportion test: dealing with percentages, rates, or categories; Mean test: dealing with measurements, averages, or continuous data
  4. Look for explicit alpha values or phrases like "95% confidence" (alpha = 0.05), "99% confidence" (alpha = 0.01)

Next Steps

  • Practice creating your own word problems from real data
  • Review the Common Mistakes lesson to avoid typical errors
  • Prepare for the Unit Quiz by reviewing all lesson topics
  • Look for statistics in news articles and practice interpreting them