Mean or Median

Hypothetically speaking, given a choice between a question on median and one on mean, which would you choose? (Not that we are fortunate enough to have a choice on test day, but no harm in dreaming!) It would be wiser to pick the question testing on median, and here is why: Median is the value at a point – to be precise, the point which divides the increasing data set into two equal halves. You don’t care what is on the left and what is on the right of this point, so an outlier will do nothing to the median. The mean, however depends on every value in the set. If you increase one element of data, the mean of the set changes – outliers can drastically change the value of the mean. Hence, every element has to be kept in mind! With the median, there is a lot less to worry about. Let’s illustrate this with an example data sufficiency question: Question on Median: At a bakery, cakes are sold every day for a certain number of days. If 6 or more cakes were sold for 20% of the total number of days, is the median number of cakes sold less than 4? Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4. Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more. The following is the number of cakes sold on any of the days mentioned in the question:         Say there were 100 days (since all figures are in terms of percentages, we can assume a number to simplify our understanding). The question stem tells us that 6 or more cakes were sold for 20% of the days, so for 20 days, 6 or more cakes were sold. Then for 80 days, 1/2/3/4/5 cakes were sold. With this information in mind, is the median number of cakes sold in one day less than 4? We know how to get the median. When we arrange all figures in increasing order, the median will be the average of the 50th and the 51st terms. We need to know if the average of the 50th and 51st term is less than 4. Let’s tackle the statements one at a time: Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4. The number of days that less than 6 cakes were sold = 80. 75% of these 80 days will be 60 days. In 60 days, less than 4 cakes were sold. So the 50th and 51st terms will be less than 4 and so will their average. Hence, the median will be less than 4. This statement alone is sufficient. Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more. In 20 days, 6 or more cakes were sold. This constitutes 50% of the days during which 4 or more cakes were sold, so in another 20 days, 4 or 5 cakes were sold. Hence, during the leftover 60 days, less than 4 cakes were sold. The 50th and 51st terms will be less than 4 and so will their average. Hence, the median will be less than 4. This statement alone is also sufficient, so our answer is D. All we needed to worry about here were the 50th and 51st terms, however the whole problem changes when we talk about mean instead of median. Same Question on Mean: At a bakery, cakes are sold every day for a certain number of days. If 6 or more cakes were sold for 20% of the total number of days, is the average number of cakes sold less than 4? Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4. Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more. Again, the question stem tells us that 6 or more cakes were sold for 20% of the days, so for 20 days, 6 or more cakes were sold. Then for 80 days, 1/2/3/4/5 cakes were sold. We now need to ask ourselves is the average number of cakes sold in one day less than 4? This question asks us about the average. – that is far more complicated than the median. Every value matters when we talk about the average. We need to know the number of cakes sold on each of these 100 days to get the average. 6 or more cakes were sold in 20 days. Note that the number of cakes sold during these 20 days could be any number greater than 6, such as 20 or 50 or 120, etc. The minimum number of cakes sold on these 20 days would be 6*20 = 120. There is no limit to the maximum number of cakes sold. With this in mind, let’s examine the statements: Statement 1: On 75% of the days that less than 6 cakes were sold, the number of cakes sold each day was less than 4. In 80 days, less than 6 cakes were sold. Of this number, 75% is 60 days. In 60 days, less than 4 cakes were sold. So in 60 days, you have a minimum of 1*60 = 60 cakes sold and a maximum of 3*60 = 180 cakes sold. During the leftover 20 days 4 or 5 cakes were sold, so you have a minimum of 4*20 = 80 cakes and a maximum of 5*20 = 100 cakes. The minimum value of the average is (120 + 60 + 80)/ 100 = 2.6 cakes, but the maximum average could be anything. Therefore, this statement alone is not sufficient. Statement 2: On 50% of the days that 4 or more cakes were sold, the number of cakes sold each day was 6 or more. The 20 days when 6 or more cakes were sold make up 50% of the days when 4 or more cakes were sold. So for another 20 days, 4 or 5 cakes were sold. This gives us a minimum of 4*20 = 80 cakes and a maximum of 5*20 = 100 cakes. For 60 days, 1/2/3 cakes were sold. So in 60 days, you have minimum of 1*60 = 60 cakes sold and a maximum of 3*60 = 180 cakes sold. The minimum value of the average is (120 + 60 + 80)/ 100 = 2.6 cakes, but again, the maximum average could be anything. This statement alone is also not sufficient. Note that both statements give you the same information, so if they are not sufficient independently, they are not sufficient together. The answer of this modified question would be E. Here, we had to assume the minimum and maximum value for each data point to get the range of the average – we couldn’t just rely on one or two data points. Finding the mean during a CAT question requires much more information than finding the median!