Problem Solving

Hi All,

We're told that the variable X takes on integer values between 1 and 7 inclusive as shown above (meaning that out of a total of 15 values, there are three 1s, one 2, three 3s, one 4, etc.). We're asked for the probability that the absolute value of the DIFFERENCE between the MEAN (re: 4) and a randomly chosen value of X will be GREATER than 3/2. While this question looks a bit 'thick', it's based on some basic Probability math and some Arithmetic.

To start, for the DIFFERENCE between 4 and a value in the list to be GREATER than 3/2, the value of X would have to be 1, 2, 6 or 7 (for example, 4 - 1 = 3 and 6 - 4 = 2).

Based on the frequencies, we know that there are three 1s, one 2, one 6 and three 7s. That's a total of 8 values out of the 15 total values.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorLU wrote:Source: GMAT Prep

X frequency
1 3
2 1
3 3
4 1
5 3
6 1
7 3

The variable x takes on integer values between 1 and 7 inclusive as shown above. What is the probability that the absolute value of the difference between the mean of the distribution which is 4 and a randomly chosen value of x will be greater than 3/2?

A. 8/15
B. 4/7
C. 4/5
D. 6/7
E. 8/7

The only values of x that give us |x - 4| > 3/2 are 1, 2, 6, and 7. Since we have a total of 3 + 1 + 3 + 1 + 3 + 1 + 3 = 15 values in the list and a total of 3 + 1 + 1 + 3 = 8 values that can be x, the probability is 8/15.

Answer: A