Hard problem

[hiebiz]

Pls solve this problem with clear explanation.

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 percent 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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

[KillGmat]

To me answer is E.
Total/No. of students = 72 ->given data
bottom 16% will have to take retest.
option 1) we know No of students. we get the total. Yet with all the data ,we wont be able to find the bottom 16% os the class.
2)10 recruits scored 82 or higher doesnt give much information either

Same, when we consider both the options.
So, it has to be E.

[hiebiz]

THE ANSWER IS C.
Can anyone give a clear explanation for this problem?

[krusta80]

Originally Posted by hiebiz

Pls solve this problem with clear explanation.

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 percent 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.

(1) There are 500 recruits in the class.

This tells us nothing of the distribution. INSUFF

(2) 10 recruits scored 82 or higher.

This tells us nothing of the PERCENTAGE of recruits. INSUFF


(1) and (2)

We know that 10/500 = 2 percent of recruits scored 82 or better. So...

P(Z >= (82-72)/std.dev) = .02

Assuming that we have access to a z chart (or if there are certain z values that GMAT takers need to memorize), we can solve for the standard deviation and then solve the following:

P(Z <= (x-72)/std.dev) = .16

Using a Z-chart, I see that the 98th percentile is about 2 standard deviations, and the 84th percentile is about 1 standard deviation. So, I'm guessing that we're supposed to know these estimates as test takers.

Therefore, from the first equation we have std. dev = 5, which means that 67 is the "magic score" we're looking for.

C


*** I just realized that we don't need to generate the number, since it's a DS question. Where did you get this question?

[DWarrior]

To define a normal curve, we need to know the mean and standard deviation.

St 1 is insufficient because we have mean and population, but we don't have any info on standard deviation.

St 2 is insufficient because we have 10 recruits scoring 82, but we don't know what percentage of the population that 10 represents.

St 1+2 is sufficient because we know 10 recruits represents 2% of the population, and that the top 2% scored 10 points above the mean. From that we can derive the standard deviation:

Z*=(x*-x)/std

Z* we get out of a textbook (you don't actually have to solve it), x* is 82, x is 72, now to get standard deviation:
std=(82-72)/Z*, where Z* is known. Now we have the mean (72), and standard deviation, and thus the normal curve is defined, and we can find out what bottom 16% represents.

[Girasol]

Another way to short cut this answer is to remember that in any normal distribution follows the pattern: 2%, 14%, 34%, mean, 34%, 14%, 2%. Hence the lower 2% of the class were two standard deviations from the mean. We know the mean, and we know the standard deviation since we know that 82 is at two standard deviations from the mean. The recruits would have to retest at or below 62. Hence 1 and 2 together are suffcient, C.