Averages and Statistics: Practice Questions & Explanations

Average (mean) = sum of values ÷ number of values. Sum: 70 + 75 + 80 + 85 + 90 = 400. Count: 5 values. Average: 400 ÷ 5 = 80. Three types of averages: Mean — what most people call 'average', the sum divided by count. Median — the middle value when sorted (or mean of two middle values if even count); in this case 80 is also the median. Mode — most frequent value. For evenly spaced values, the mean equals the median. The ASVAB often tests problems where you need to find a missing value. Example: 'Three test scores average 80; two of them are 75 and 85. What is the third?' Set up: (75 + 85 + x)/3 = 80; solve: 160 + x = 240; x = 80. To find an average when given one set of data and adding more, use the weighted approach: combined sum ÷ combined count.

To average 85 over four tests, the total of all four scores must be 4 × 85 = 340. Current total: 80 + 75 + 90 = 245. Required fourth score: 340 - 245 = 95. Verification: (80 + 75 + 90 + 95) ÷ 4 = 340/4 = 85 ✓. This 'find the missing value to achieve an average' problem appears regularly on the ASVAB. Strategy: (1) Calculate the required total = desired average × number of values; (2) Sum the known values; (3) Subtract to find the missing value. Variations include: 'How many points above/below the average is this score?' (compute score - average for each value). 'If a score is dropped from the average, what is the new average?' (subtract the dropped score from the sum, divide by the new count). Always check: does the missing value make sense in context? (Some problems require values within a possible range, like 0-100 for test scores.)

Average speed over entire trip = total distance / total time. Total distance: 320 + 360 = 680 miles. Total time: 4 + 6 = 10 hours. Average speed: 680/10 = 68 mph. Important: this is NOT the average of the two individual speeds! Speed 1: 320/4 = 80 mph; Speed 2: 360/6 = 60 mph. Simple average of speeds: (80 + 60)/2 = 70 mph — this is WRONG because the driver spent different amounts of time at each speed. Average speed must be calculated from total distance divided by total time, weighting each speed by its duration. This is a common ASVAB trick question. The principle applies to any 'weighted average' problem: a class average where different sections have different student counts, an investor's return over different periods with different amounts invested, etc. Always think: what does 'average' actually mean for this quantity? Distance ÷ time is the actual definition of average speed; it cannot be averaged any other way.

Sum = average × count = 30 × 5 = 150. This is the basic rearrangement of the average formula: average = sum/count, therefore sum = average × count. This relationship is the key to many average problems. Variations: (1) 'Average of 4 numbers is 25; if one number is removed and the new average is 23, what was the removed number?' Original sum: 25 × 4 = 100; new sum: 23 × 3 = 69; removed number: 100 - 69 = 31. (2) 'A class of 30 has an average grade of 85; another class of 20 has an average of 75. What is the combined average?' Combined sum: (30 × 85) + (20 × 75) = 2,550 + 1,500 = 4,050; combined count: 50; combined average: 4,050/50 = 81. (3) Common ASVAB variations of these problems test understanding that you cannot simply average the averages when the groups have different sizes — you must use weighted averages or compute from sums.

Sum = 72 + 85 + 91 + 80 = 328. Mean = 328 ÷ 4 = 82. The mean is the total divided by the number of scores. A quick sanity check: the mean must sit between the lowest (72) and highest (91) score, and 82 is close to the middle of the pack. The distractors 81, 83 and 84 are all plausible-looking neighbours, which is why adding carefully (328, not 324 or 332) matters more than the final division.

Arrange in order: 4, 5, 6, 7, 8. The median (middle value) is 6 miles. Median means the middle value once the numbers are sorted — not the middle of the original list. Here the mean also happens to be 6 (30 ÷ 5), but on the test the two usually differ, so always sort first. With an even count of values, the median is the average of the two middle numbers.

To average 90 over four tests, the total of all four scores must be 4 × 90 = 360. The three known scores add to 82 + 90 + 95 = 267. The needed fourth score is 360 − 267 = 93. Average problems where you must find a missing value are solved by working from the required total: multiply the desired average by the number of items to get the total needed, then subtract the scores you already have. The student needs a 93 on the fourth test to reach a 90 average.

If the average of five numbers is 20, their total is 5 × 20 = 100. The four known numbers add to 18 + 22 + 25 + 15 = 80. The fifth number is 100 − 80 = 20. The method for a missing value in an average is always the same: multiply the average by the count to get the required total, then subtract the known values. The fifth number is 20. The distractor 100 is the total of all five numbers, not the missing one, so be sure to finish the subtraction.