Average Calculator

They are three different ways to find the 'center' of your data. The mean is the sum divided by the count — what most people call the average. The median is the middle value once the numbers are sorted, so half the data sits above it and half below; with an even count, you average the two middle values. The mode is the value that appears most often, and a dataset can have one mode, several, or none. Take the data 1, 2, 2, 3, 10: the sum is 18 over 5 values, so the mean is 3.6; the sorted middle value is 2, so the median is 2; and 2 occurs twice, making it the mode. Here the single high value (10) pulls the mean above the median, which is a clear sign of skew.

Use the median whenever your data has extreme outliers or a heavily skewed distribution, because the median is barely affected by a few unusual values. Income is the classic example. Suppose 9 people earn $50,000 and 1 person earns $1,000,000: the total is $1,450,000 across 10 people, so the mean is $145,000 — a figure no one in the group actually earns. The median, the average of the 5th and 6th sorted values, is $50,000, which describes the typical person far better. This is exactly why official statistics report median household income and median home prices rather than the mean. Reach for the mean instead when the data is roughly symmetric and free of outliers, since it then uses all the information available.

A weighted average lets some values count more than others, instead of treating every value equally. You multiply each value by its weight, add those products together, and divide by the sum of the weights. Grades are the everyday example: if homework averages 90 and counts for 30% of the grade while exams average 80 and count for 70%, the weighted average is (90 × 0.30) + (80 × 0.70) = 27 + 56 = 83. Notice that this is lower than the simple average of 85, because the exams — where the score was lower — carry more weight. Whenever the items you are averaging are not equally important, or represent groups of different sizes, the weighted average gives the correct result and the simple average does not.

Yes, but only by taking a simple average when each percentage represents a group of the same size; otherwise you must weight them. Imagine Class A has 20 students averaging 85% and Class B has 30 students averaging 75%. The naive answer, (85 + 75) ÷ 2 = 80%, is wrong because it ignores that Class B has more students. The correct figure weights each class by its size: (20 × 85 + 30 × 75) ÷ 50 = (1,700 + 2,250) ÷ 50 = 79%. The result leans toward Class B's lower score because that class is larger. The same caution applies to averaging rates, ratios, and speeds — when the underlying groups differ in size, always use a weighted average.

It simply means the sum of your values is below zero, so the negative numbers outweigh the positive ones. There is nothing unusual or invalid about this — a negative mean is perfectly meaningful wherever data can fall below zero. You see it constantly in finance, where months of losses can outweigh gains to produce a negative average return, and in temperature data, where a run of sub-zero readings gives a negative average for the period. For example, the values -5, -3, and 2 sum to -6 over 3 values, giving a mean of -2. The calculation is exactly the same as always: add everything up and divide by the count. The sign of the result just tells you which side of zero the center of your data falls on.