The median of a data set with an even number of elements is the mean of the middle two numbers, assuming the elements are arranged in order; since the data set has eight elements, the median will be the mean of the fourth-lowest element and the fourth-highest element.

If  and , the data set becomes ; if  and , the data set becomes . In both data sets, the middle two elements are 25 and 35, making both medians equal to . The quantities are equal.

Note: You don't need to do the last step and actually find the median. As long as you know that the two medians are the same, you've answered the question!

The first step is to reorder the set, , in order from smallest to largest. 

The mode (which occurs most often) is 3.3, so . 

The median is the middle number in the set (which is 6.8, having 3 numbers to its right and 3 to its left in the ordered list), so . 

The mean (or average) is , so 

Therefore, . 

To find her average grade for the class, we need to multiply Mary's test scores by their corresponding weights and then add them up.

The five quizzes were each worth 10%, or 0.1, of her grade, and the midterm and final were both worth 25%, or 0.25.

average = (0.1 * 89) + (0.1 * 74) + (0.1 * 84) + (0.1 * 92) + (0.1 * 90) + (0.25 * 92) + (0.25 * 91) = 88.95 = 89. 

Looking at the answer choices, they are all spaced 2 percentage points apart, so clearly the closest answer choice to 88.95 is 89.

The median of the set is the fifth-highest value, which is 70; this is also the mode, being the most commonly occurring element.

The mean is the sum of the elements divided by the number of them. This is

The midrange is the mean of the least and greatest elements, This is 

The midrange is the greatest of the four.

The mean of the data set is the sum of the elements divided by 5:

The median of the data set is the middle value when the values are arranged in ascending order, which here is the third-highest value. This is 79.

The mean exceeds the median by 

Mark's grade is a weighted mean in which his hourly tests have weight 1 and his final has weight 2. If we call  his final, then his term average will be 

,

which simplifies to 

.

Since Mark wants his score to be 80 or better, we solve this inequality:

Mark must score 77 or better on his final.

To calculate whether Sally or Sue has the higher average, it is only necessary to add, for each contestant, all of their scores except for their highest and lowest. Since both sums are divided by 5, the higher sum will result in the higher mean score.

(a) For Sally, the highest and lowest scores are 9.7 and 9.1. The sum of the other five scores is:

(b) For Sue, the highest and lowest scores are 10.0 and 9.1. The sum of the other five scores is:

Sue's total - and, subsequently, her score - is higher than Sally's, so (b) is the greater quantity.

The highest and lowest scores of the seven are 9.9 and 9.3, so Kathy's score is the mean of the other five:

This makes (a) greater.

For Chuck to achieve an average of , his scores on the five tests must total . At current, his scores total , so he needs  points to achieve his average. This makes (a) greater.

The sum of the elements in the data set  is

Divide by 5 to get the mean: