Gmat_mission wrote: ↑Fri Jan 08, 2021 1:47 am
A certain characteristic in a large population has a distribution that is symmetric about the mean \(m.\) If \(68\) percent of the distribution lies within one standard deviation \(d\) of the mean, what percent of the distribution is less than \(m + d?\)
(A) \(16\%\)
(B) \(32\%\)
(C) \(48\%\)
(D) \(84\%\)
(E) \(92\%\)
Answer:
D
Source: Official Guide
Let's first sketch a distribution that is symmetric about the mean m.
Notice that m+d represents 1 unit of standard deviation ABOVE the mean
Likewise, m-d represents 1 unit of standard deviation BELOW the mean
And m+2d represents 2 units of standard deviation ABOVE the mean, etc.
ASIDE: There are infinitely many distributions that are symmetric about the mean m. The above distribution is just one.
Our goal is to determine what portion of the population is LESS THAN than m+d
First recognize that, since the distribution is symmetric about the mean m, 50% of the population is BELOW the mean, and 50% is ABOVE the mean.
Next, we're told that 68% of the distribution lies within one standard deviation d of the mean
In other words, 68% the population is BETWEEN m-d and m+d
Since the distribution is symmetric about the mean m, this 68%, is divided into two equal populations.
When we COMBINE our two findings, we see that the percentage of the population that's below m+d =
50% + 34% = 84%
Answer: D
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
