revised
prat_agl wrote:The members of the newest Recruiting class of a certain military organization are taking their physical conditioning test, and those who score in the bottom 16% will have to retest. If the scores are normally distributed and have an arithmetic mean of 72, what is the score at or below which the recruits will have to retest?
1) There are 500 Recruits in the class.
2) 10 Recruits scored 82 or higher.
normally, normally distributed probability concept isn't tested on GMAT. But since you touched on the concept seemingly addressed by must be non-official source or some "exalted" test prep company, i will try to solve it for the rule of
'2-14-34'. Also look my post at
https://www.beatthegmat.com/mean-and-sta ... tml#402966 for explanations of the rule selected in bold.
Prompt: bottom 16% implies two standard deviations covered from zero or one standard deviation less than average. We are given the average, 72 and need to know standard deviation, as the required score will be 72 less (subtracted) one standard deviation.
st(1) 500 recruits in the class allow us to ascertain only the number of elements in the set but not their dispersion, as we must know exactly the value of each element. Not Sufficient;
st(2) 10 recruits scored 82 or higher again suggests that the value (in score) of 10 recruits is 82+, but we don't know the values of other recruits or how many of them are in the set to figure out their dispersion. Not Sufficient;
combining st(1&2): there are 500 recruits and 10 of them makes 2% or one standard deviation [spoiler](Hurray)[/spoiler]. Since 82>72 we infer that this is 98-100% or 2% of set values distributed normally. So total makes three standard deviations from the mean as 82>72, two standard deviations placed before the interval 98-100% and the score of less than 82 is achieved by 50% (mean) + 34% + 14% of all recruits (98% recruits). (82-72)/2=One standard deviation (S) and S=10/2. 16% will be 72-10/2=67 Sufficient
answer
C, score 67
