Central Tendency — Advanced Practice

Advanced practice on central tendency: mean, median, and mode in grouped data, relative frequencies, word problems, and missing values.

20 questions

Question 1
5.00 pts

1. The table shows exam scores and their frequencies:

ScoreFrequency6028031001

What is the mean?

Explanation:

Solution:

Weighted sum: 2×60 + 3×80 + 1×100 = 120 + 240 + 100 = 460.
Total students: 2 + 3 + 1 = 6.
Mean = 460 ÷ 6 ≈ 76.67.

Question 2
5.00 pts

2. A final course score is calculated as: 40% from assignment average + 60% from exam.

A student earned an assignment average of 80 and an exam score of 70. What is the final score?

Explanation:

Solution:

Final = 0.4 × 80 + 0.6 × 70 = 32 + 42 = 74.

Question 3
5.00 pts

3. Group A has 10 students with a mean score of 70. Group B has 20 students with a mean score of 80. What is the mean of all 30 students together?

Explanation:

Solution:

Group A total: 10×70 = 700. Group B total: 20×80 = 1600.
Combined: 700 + 1600 = 2300. Mean = 2300 ÷ 30 ≈ 76.67.

Question 4
5.00 pts

4. The mean of four numbers is 12. Three of them are 10, 15, and 9. What is the fourth number?

Explanation:

Solution:

Total sum = 12 × 4 = 48. Known sum: 10 + 15 + 9 = 34. Fourth number = 48 − 34 = 14.

Question 5
5.00 pts

5. The class mean is 70. A 5-point bonus is added to every score. What will the new mean be?

Explanation:

Solution:

Adding the same constant to every value increases the mean by that amount. New mean = 70 + 5 = 75. The median and mode also increase by 5.

Question 6
5.00 pts

6. The mean student weight is 50 kg. Convert units from kg to grams (× 1000). What is the new mean in grams?

Explanation:

Solution:

Multiplying all values by 1000 multiplies the mean by 1000. New mean = 50 × 1000 = 50,000 grams.

Question 7
5.00 pts

7. Data: 10, 12, 13, 14, 80 — what is true?

Explanation:

Solution:

80 is an extreme value that greatly increases the mean, but the median (middle value after sorting) stays at 13. The median is "protected" against extreme values.

Question 8
5.00 pts

8. Frequency table: score 60 — 2 students, score 70 — 3 students, score 80 — 1 student. What is the median?

Explanation:

Solution:

Expanded list: 60, 60, 70, 70, 70, 80. With 6 students, the 3rd and 4th values are both 70. Median = 70.

Question 9
5.00 pts

9. Score 50 appeared 4 times, 60 appeared 2 times, 70 appeared 4 times. What is the mode?

Explanation:

Solution:

Both 50 and 70 have frequency 4 — the highest frequency. Mode = 50 and 70 (bimodal distribution).

Question 10
5.00 pts

10. Final course score: 30% from Exam A, 70% from Exam B.

A student scored 60 in Exam A and 90 in Exam B. What is the final score?

Explanation:

Solution:

0.3 × 60 = 18; 0.7 × 90 = 63. Total = 18 + 63 = 81.

Question 11
5.00 pts

11. A score table divided into class intervals:

  • 50–60: 2 students
  • 60–70: 3 students
  • 70–80: 5 students

Assuming each student scores at the midpoint of their interval, what is the approximate mean?

Explanation:

Solution:

Midpoints: 55, 65, 75.
Weighted sum: 2×55 + 3×65 + 5×75 = 110 + 195 + 375 = 680.
Total students: 10. Mean = 680 ÷ 10 = 68 ≈ 69.

Question 12
5.00 pts

12. What is the median of: 4, 9, 2, 7, 3, 10?

Explanation:

Solution:

Sort: 2, 3, 4, 7, 9, 10. With 6 values (even count), the two middle values are 4 and 7. Median = (4 + 7) ÷ 2 = 5.5.

Question 13
5.00 pts

13. Data: 5, 7, 9, 11, 13 — the mean is 9. Replace 5 with 15. What is the new mean?

Explanation:

Solution:

Original sum = 45. New sum: 15+7+9+11+13 = 55. New mean = 55 ÷ 5 = 11. Adding 10 to one value increases the total by 10, so the mean rises by 10÷5 = 2.

Question 14
5.00 pts

14. The class mean for 20 students is 75. Five new students join, each scoring 80. What is the new mean?

Explanation:

Solution:

Original total: 20 × 75 = 1500. New students: 5 × 80 = 400. Combined total: 1900. New mean = 1900 ÷ 25 = 76.

Question 15
5.00 pts

15. Most employees in a company earn between 8,000 and 12,000, but the CEO earns 80,000. Which central measure better describes the "typical employee salary"?

Explanation:

Solution:

The CEO's salary is an extreme outlier that inflates the mean. The median describes the middle employee's earnings — a better representation of the typical salary.

Question 16
5.00 pts

16. A survey asked students how many courses they take each semester:

1 course — 5 students; 2 courses — 12 students; 3 courses — 8 students; 4 courses — 5 students.

What is the modal number of courses?

Explanation:

Solution:

The value with the highest frequency is 2 courses (12 students). Mode = 2.

Question 17
5.00 pts

17. A constant of 10 is added to every value in a dataset. What happens to the mean, median, and mode?

Explanation:

Solution:

Shifting all data points by the same constant shifts all measures of central tendency by that constant. Mean, median, and mode all increase by 10.

Question 18
5.00 pts

18. Every value in a dataset is multiplied by 3. What happens to the mean, median, and mode?

Explanation:

Solution:

Multiplying all values by a constant multiplies all measures of central tendency by that constant. Mean, median, and mode are all multiplied by 3.

Question 19
5.00 pts

19. 40% of students have a mean of 70 and 60% have a mean of 90. What is the overall class mean?

Explanation:

Solution:

Weighted mean: 0.4 × 70 + 0.6 × 90 = 28 + 54 = 82.

Question 20
5.00 pts

20. The median of a dataset is 50. What can be concluded with certainty?

Explanation:

Solution:

The median is the positional midpoint: at least half the values are ≤ 50 and at least half are ≥ 50. This says nothing about the mean or mode.