Word Problems
Learning Objectives
In this lesson, you will:
- Apply statistical concepts to real-world scenarios
- Interpret data in context
- Choose appropriate measures for different situations
- Make predictions and draw conclusions from data
Practice Quiz
Work through these 10 problems. Click each question to reveal the answer and explanation.
Question 1: A student scored 78, 82, 85, 91, and 89 on five tests. What score does she need on the sixth test to have a mean of 86?
Answer: 91
Solution: Needed sum = 86 x 6 = 516. Current sum = 78 + 82 + 85 + 91 + 89 = 425. Needed = 516 - 425 = 91.
Question 2: The heights (in inches) of 8 basketball players are: 72, 74, 76, 76, 78, 78, 80, 82. What is the median height?
Answer: 77 inches
Solution: With 8 values, median is average of 4th and 5th values: (76 + 78) / 2 = 77 inches.
Question 3: A company's salaries (in thousands) are: 35, 40, 42, 45, 48, 50, 200. Why might the median be a better measure of center than the mean?
Answer: The $200K salary is an outlier that skews the mean upward.
Solution: Mean = 460/7 = 65.7K. Median = 45K. The median (45K) better represents a typical salary since the 200K outlier inflates the mean.
Question 4: Test scores have mean 75 and standard deviation 10. Maria scored 85. How many standard deviations above the mean is she?
Answer: 1 standard deviation above the mean
Solution: z = (85 - 75) / 10 = 10 / 10 = 1 standard deviation above.
Question 5: Two classes took the same test. Class A: mean 78, std dev 5. Class B: mean 78, std dev 12. Which class had more consistent scores?
Answer: Class A had more consistent scores.
Solution: A lower standard deviation means scores are closer to the mean. Class A (std dev 5) has more consistent scores than Class B (std dev 12).
Question 6: Weekly rainfall (inches) for 5 weeks: 0.5, 1.2, 0.8, 2.5, 1.0. Find the range and explain what it tells us.
Answer: Range = 2.0 inches
Solution: Range = 2.5 - 0.5 = 2.0 inches. This tells us the difference between the wettest and driest weeks was 2 inches.
Question 7: Home prices in a neighborhood: $180K, $195K, $200K, $210K, $220K, $850K. Which measure of center would a real estate agent use to attract buyers?
Answer: Median ($205K) to avoid the outlier's effect.
Solution: Mean = $309K (inflated by $850K outlier). Median = (200K + 210K)/2 = $205K. The median gives a more accurate picture of typical home prices.
Question 8: IQ scores are normally distributed with mean 100 and standard deviation 15. About what percent of people have IQs between 85 and 115?
Answer: About 68%
Solution: 85 to 115 is within 1 standard deviation of the mean (100 plus/minus 15). By the empirical rule, about 68% of data falls within 1 standard deviation.
Question 9: A store tracks daily sales: $420, $380, $450, $395, $500, $410, $425. Find the IQR and identify any outliers using the 1.5 IQR rule.
Answer: IQR = $55; no outliers
Solution: Ordered: 380, 395, 410, 420, 425, 450, 500. Q1 = 395, Q3 = 450, IQR = 55. Fences: 395 - 82.5 = 312.5 and 450 + 82.5 = 532.5. All values within fences, so no outliers.
Question 10: A survey asks how many pets households have. Results: 0, 0, 1, 1, 1, 2, 2, 3, 8. What is the mode, and why is it useful here?
Answer: Mode = 1 pet
Solution: 1 appears 3 times, more than any other value. Mode is useful because it tells us the most common number of pets (1), which is practical information for pet-related businesses.
Next Steps
- Practice identifying which measure of center is most appropriate
- Look for real data sets to analyze
- Move on to common mistakes when ready