Home .. Text Next

Back to Problems

1. John and Jane have five children whose ages are 2, 4, 7, 9 and 13. What are the mean, median, and modal ages of these children?

The sum of the five numbers is 35, divided by 5 gives 7, the mean. The median is also seven and there is no mode.

2. Jamie has five classes. She knows that the average number of students in her classes is 16. She remembers that in her MWF classes, her classes have 12, 14, and 20 students. She remembers that her first class on TTH has 10 students, but cannot remember how many are in the second class. Can you help her? Explain.

5*16 = 80. The totals must sum to 80. 80 = 12 + 14 + 20 + 10 + x. Solving for x, 80 = 56 + x; x = 80 - 56 = 24.

3. In cross country the score of the team is found by adding up the finishing place of the first five runners on the team. The average runner on the local team finished 10th. The first runner finished 4th, the second 6th, the third 9th, the fourth 11th. Where did the fifth runner place? Explain.

10 = 50/5 = (4 + 6 + 9 + 11 + x)/5; 50 = 30+x; x= 20.

4. John has purchased five CDs. He knows that the average price he paid was \$12, and he can find receipts for \$9.00, \$12.00, \$15.00 and \$15.00. What amount must be on the receipt that he cannot find? Explain.

The differences from the average are -\$3, \$0, +3 and +3. These differences must sum to zero, so the missing one is -\$3. Thus, the missing receipt is \$9.00.

5. Joan has collected data on gasoline prices from six area stations: \$1.19 \$1.21 \$1.37 \$1.23 \$1.25 \$1.25. Compute the mean, median, and mode of these data.

The mode is 1.25 because there are two of them. The data have to be ordered from lowest to highest to find the median. When ordered the middle numbers are \$1.23 and \$1.25, so the median is \$1.24, the average of these two. The sum is \$7.50; dividing by six gives \$1.25.

6. A professor has sixteen students in a class. He gives three A's, five B's, six C's, and two D's. What is the modal mark? What is the median mark? What is the mean mark? (Use A = 4.0, B = 3.0, etc.)

The median is 2.5; half are A or B, the other half are C or D. C or 2.00 is the modal mark. The GPA for the class is (12+15+12+2)/16 = 41/16 = 2.5625.

7. Here is grading data from a past semester, showing the number of grades given and their percentage of the total:

A 1430 (29%); A- 810 (13%); B+ 498 (10%); B 618 (13%); B- 361 (7%); C+ 203 (4%); C 287 (6%); C- 158 (3%); D+ 72 (2%); D 124 (3%); F 106 (2%); All other (P, N, W, Z, I) 395 (7%)

What was the modal grade given at this school in this semester? What was the median grade given at this school in this semester?

The modal or most common grade was an A. 42% of grades were A- or above and 52% were B+ and above. Therefore the median grade, the grade that half are above and half below, was B+.

8. A measure of center that is occasionally (rarely?) mentioned in introductory statistics is the geometric mean. It is found by multiplying the n numbers together and then taking the nth root. For example, if the numbers are 1, 1 and 27, we multiply them together to get 27. Taking the cube root gives us 3, so 3 is the geometric mean of {1, 1, 27}. What is the geometric mean of these numbers? {1, -1, 2, 4, 0, 6, 5, -4}

Anything multiplied by zero is zero, so the geometric mean must be zero.

9. 1. A group of five children has these weights: 80 lbs, 100 lbs, 120 lbs, 90 lbs, and 110 lbs.

a) What is their median weight?
b) What is their average weight?

Median 100; average 100.

10. The mean is the preferred measure of the center because it has nice mathematical properties. One of those properties is that it minimizes the sum of errors squared, that is, if we compute the differences from the mean, square them, and then sum the squares, the number we get will be the smallest possible number. If we replace the mean with any other number, the sum of errors squared will be larger. If this is too abstract, consider the following example with the numbers 1, 1, 7. The mean of these numbers is 9/3 = 3 and the median is 1. The errors for the mean are -2, -2, and 4, and the errors for the median are 0, 0, 6. If we square these errors and sum them, for the mean we get 4 + 4 + 16 = 24, while for the median we get 0 + 0 + 36 = 36.

What is the sum of errors squared for both the mean and median for these sets of numbers?

a) 4, 8, 4, 5, 9
b) -6, 19, 0, 8, 1, 2, 4

a) mean is 6, median is 5, sum of errors squared for the mean is 4 + 4 + 4 + 1 + 9 = 22, for the median it is 1 + 5 + 1 + 0 + 16 = 27

b) mean is 4, median is 2, sum or errors squared for the mean is 100 + 225 + 16 +16 +9 + 4 + 0 = 370; for the median 64 + 289 + 4 + 36 +1 + 0 + 4 = 398

Back to Problems

 Home .. Text Next