Standard deviation shows the is difference between the mean and the values,more the values closer to mean,lesser will be standard deviation.

here the adding two values will make the sd lower mean ,since mean is 6 deviation of values will be lesser if the list have more number of values equal to mean(variance 0)

ans option E

my approach:

sd= rt ((avg-x)^2...)/n)

to decrease the sd, we need to decrease the numerator and increase the denominator which is increased to 102
to decrease the numerator, we need to decrease the the number squared, so the closer the number is to the avg, the lower the square
that occurs with choice e

liferocks wrote:Standard deviation shows the is difference between the mean and the values,more the values closer to mean,lesser will be standard deviation.

here the adding two values will make the sd lower mean ,since mean is 6 deviation of values will be lesser if the list have more number of values equal to mean(variance 0)

ans option E

Agree with this answer. This is a question that tests subject knowledge instead of computational skill. You must know that standard deviation is the average distance between any given point in the set and the mean of the set.

If you want a smaller standard deviation, you need numbers that are closer to the mean, on average, than the rest of the set.

Think about it like this; the only way you can increase the data set and decrease the standard deviation (the amount the data deviates from average) would be to add two numbers that equal the average, or are very close to the average. Thus, E is correct.

I agree with the solutions advanced. NO math is Standard deviation roughly measures the average distance between the mean and every value. To minimize standard deviation, the values added must be as close to the mean as possible.

In this case the mean is 6, so the new values that would minimize standard deviation are 6 and 6. The answer is E.

Look for topic="Statistics" and difficulty="500-600" to find similar questions in the drill generator. This is GMATPrep question 1155

Good luck,
-Patrick