The lifetimes of all the batteries produced

[sushanta_debnath]

The lifetimes of all the batteries produced by a certain company in a year have a distribution that is symmetric about the mean m. If the distribution has a standard deviation of d, what percent of the distribution is greater than m+d?

(1) 68 percent of the distribution lies in the interval from m-d to m+d, inclusive.
(2) 16 percent of the distribution is less than m-d.

please help me with this one.
The ans. is D

[nitin_prakash_khanna]

The question stem tells you that Distribution is symmetrical around mean m so 50% above m and 50% below.

St-1 68% is between m-d and m+d, this tells you that on the side which is higher 34% is between m and m+d so, remaining 16% has to be above m+d. SUFFICIENT

St-2 tells you that 16% is below m-d, so on the other side 16% will be above m+d. SUFFICIENT as well.

This is actually a normal distribution, where 68% is between 1 Standard deviation (SD), 96% between 2 SD and rest within 3 SD.

I am sure instructors will have more to add to it.
hope it helps

npk

[anoo.anand]

I think there is some rule for this:

which says that 1 group would be within 68%
2nd group within 16%

[RonPurewal]

as usual, in problems that give a numerical value for standard deviation, you don't actually have to know what standard deviation is. in this problem, you can just treat d as a totally arbitrary variable, and solve the problem as is.

this is a VERY consistent pattern in past gmat problems, so, unless the people who write the test do a complete about-face, it's unlikely to change anytime soon.

see here:
post17297.html#p17297
and
post31646.html#p31646

--

imagine that you have the dsitribution on a number line:


------REGION 1-----(m - d)-------REGION 2-------(m)-------REGION 3------(m + d)-------REGION 4-------

we want region 4.

since the distribution is SYMMETRIC, you must have the same percentage in region 1 as in region 4, and the same percentage in region 2 as in region 3.

statement 1:
regions 2 and 3 total 68%. therefore, they are 34% apiece.
two ways to go from here, both of which prove "sufficient":
(a) each HALF of the distribution is 50%, so region 4 = 50% - 34%, or 16%.
(b) regions 1 and 4 together are 100% - 68% = 32%, so each of regions 1 and 4 is half of that (= 16%).
sufficient.

statement 2:
region 1 is 16%.
since regions 1 and 4 are the same, because of symmetry, it follows that region 4 is also 16%.
sufficient.

ans = D

note that, as promised above, this problem requires zero actual knowledge about how standard deviations work.