Unit Quiz
Instructions
This quiz covers all concepts from the Statistics unit:
- Measures of center (mean, median, mode)
- Measures of spread (range, IQR, standard deviation)
- Normal distributions and the empirical rule
- Data interpretation and analysis
Unit Quiz
Complete all 10 questions. Click each to check your answer.
Question 1: Find the mean, median, and mode of: 4, 7, 7, 9, 10, 12, 15
Answer: Mean = 9.14, Median = 9, Mode = 7
Solution: Mean = 64/7 = 9.14. Median = 4th value = 9. Mode = 7 (appears twice).
Question 2: Data: 15, 18, 22, 25, 30, 35, 40, 45. Find Q1, Q3, and IQR.
Answer: Q1 = 20, Q3 = 37.5, IQR = 17.5
Solution: Q1 = (18 + 22)/2 = 20. Q3 = (35 + 40)/2 = 37.5. IQR = 37.5 - 20 = 17.5.
Question 3: A data set has mean 60 and standard deviation 8. What percentage of data falls between 44 and 76 in a normal distribution?
Answer: About 95%
Solution: 44 = 60 - 16 = 60 - 2(8) and 76 = 60 + 16 = 60 + 2(8). Within 2 standard deviations, about 95% of data falls (empirical rule).
Question 4: The mean of 6 numbers is 15. If one number (24) is removed, what is the new mean?
Answer: 13.2
Solution: Original sum = 6 x 15 = 90. New sum = 90 - 24 = 66. New mean = 66/5 = 13.2.
Question 5: Which measure would be most affected if the value 100 was added to: 10, 12, 14, 16, 18?
Answer: Mean (most affected) and Range
Solution: Original mean = 14. New mean = 170/6 = 28.3 (changed by 14.3). Median changes from 14 to 15 (only slightly). Range changes from 8 to 90.
Question 6: Test scores: 72, 78, 82, 85, 88, 90, 95. Using the 1.5 IQR rule, is 72 an outlier?
Answer: No, 72 is not an outlier.
Solution: Q1 = 78, Q3 = 90, IQR = 12. Lower fence = 78 - 18 = 60. Since 72 > 60, it's not an outlier.
Question 7: A student's z-score on a test is -1.5. The mean was 80 and standard deviation was 10. What was the student's score?
Answer: 65
Solution: z = (x - mean)/sd. -1.5 = (x - 80)/10. -15 = x - 80. x = 65.
Question 8: Two data sets have the same mean of 50. Set A has range 10, Set B has range 40. Which has more variability?
Answer: Set B has more variability.
Solution: A larger range indicates data is more spread out. Set B (range 40) has greater variability than Set A (range 10).
Question 9: Heights are normally distributed with mean 66 inches and standard deviation 3 inches. What height is at the 84th percentile (1 sd above mean)?
Answer: 69 inches
Solution: 1 standard deviation above = 66 + 3 = 69 inches. (84th percentile is approximately 1 sd above mean in a normal distribution.)
Question 10: A survey of commute times shows: mean = 25 min, median = 20 min. What does this suggest about the distribution?
Answer: The distribution is right-skewed (positively skewed).
Solution: When mean > median, there are likely some high values (long commutes) pulling the mean up. This indicates right skew.
Next Steps
- Review any questions you missed
- Revisit earlier lessons if needed
- Move on to the next unit when ready