military taking physical conditioning test PR bin4 #3

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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

My Question
How can you determine the standard deviation using this information. The explanation states that the fact that we now know the top 2% have an 82 or above and the scores are normally distributed, the top 2% represent the 3rd standard deviation above the mean. How do we know this? Why can't 82 represent the fourth or 2nd standard deviation above the mean? Thanks

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by Ian Stewart » Thu Aug 23, 2012 5:57 pm
This is a very bad practice question, and you should simply ignore it. It's bad for several reasons:

* mathematically, a finite set simply can't be "normally distributed". Only infinite sets can be normally distributed. At best a finite set can be 'approximately normal'. So the question makes no mathematical sense to begin with;

* in any case, you do not need to know a single thing about "normally distributed" sets on the GMAT. Since this question relies entirely on that knowledge, it's not at all realistic;

* to answer these types of questions properly, you need to consult a statistics table, and obviously you don't have books full of stats tables at hand when taking the GMAT. I've seen the original solution to this question, and to other similar questions from the same source, and they all have incorrect solutions because they've rounded off the numbers you'd find in those stats tables.

You won't learn anything from this question that will help you on the GMAT, so just ignore it and work on something else.
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by pemdas » Thu Aug 23, 2012 6:42 pm
wow, Ian
the finite set can be normally distributed, and it's not that unusual for such sets to be symmetric around their means (bell-shaped). The z-score and other stats are quite handy for navigating the distributed values. Yet it says here 82+ scores fall in the interval 98-100%. If we follow you then math is not useful (?), as the whole science of math is about making assumptions and basing the premises on such assumptions. The infinite sets can be tested for normally distributed properties - we need to move into calculus and apply differentials, derivatives to find delta of increasing/decreasing function(s) -> +- infinity

here it provides normal distribution property and suggests the set is dispersed within 6 st.dev-s. By using st(2) we know that 10/500=2% which corresponds to one st.dev. Hence, we find 1 st.dev from the other 2 st.dev (82-72)/2=5. The question says below 16% will retest and this means 2%+14% will retest or 76-5-5=66 score (students earning less than 66 score will have to retest). The 2nd statement is Sufficient and the 1st is Not.

However, I agree with Ian, GMAT won't ask normal distribution and other probability distributions because the questions involving stat parameters (t,f,z-score, chi square) may appear very ambiguous. I have already seen that GMAT only asks questions which are 100% accurate and devoid of ambiguity. This one is very difficult to prove as we are dealing with discrete values (the number of students) and continuous values (scores). The values corresponding to students and scores should be precisely estimated within their data sets not to look insensible.
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by Ian Stewart » Thu Aug 23, 2012 7:22 pm
pemdas wrote:wow, Ian
the finite set can be normally distributed, and it's not that unusual for such sets to be symmetric around their means (bell-shaped). The z-score and other stats are quite handy for navigating the distributed values. Yet it says here 82+ scores fall in the interval 98-100%. If we follow you then math is not useful (?)
Finite sets cannot be normally distributed. Just quoting wikipedia, "In probability theory, the normal (or Gaussian) distribution is a continuous probability distribution". Continuous sets must be infinite sets. Finite sets with similar properties to normal distributions are normally called "binomial distributions" or "approximately normal distributions". A finite set cannot be "normally distributed" by definition. A "normal distribution" is not simply a distribution of data shaped like a "bell curve". It has precise statistical properties, properties a finite set simply cannot have.

And that's one of the problems with the question in the original post. In a truly normal distribution, 2.25% of data is more than 2 standard deviations above the mean. So if Statement 2 is intended to refer precisely to the range of test takers at least 2 standard deviations above average, and the population consists of 500 people, then it ought to be 11 recruits who scored 82 or higher, and not 10 recruits. When I say it's not a good question, it's not only because it's irrelevant for GMAT test takers to study; the math in the question is also simply wrong.

And it's not clear to me how, from my post, you arrived at the conclusion that I think "math is not useful". I do think math is quite useful, provided the math is correct.


pemdas wrote: here it provides normal distribution property and suggests the set is dispersed within 6 st.dev-s. By using st(2) we know that 10/500=2% which corresponds to one st.dev. Hence, we find 1 st.dev from the other 2 st.dev (82-72)/2=5. The question says below 16% will retest and this means 2%+14% will retest or 76-5-5=66 score and less earning students will have to retest. The 2nd statement is Sufficient and the 1st is Not.
As for the above, when you analyze statement 2 you seem to be using the fact that there are 500 people in total, a fact you only know if you use statement 1.
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by pemdas » Fri Aug 24, 2012 12:49 pm
yes, 500 is provided with st(1) and the answer requires both statements.
Ian, thanks for clarifying the math side.
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by pemdas » Fri Aug 24, 2012 6:39 pm
Ian, as you noted good approximation of discrete value i.e binomial probability to normal one is achieved after adjustment. Binomial param-s of k=10, and p=.02 (rather 0.98 since p for k>25 isn't found in the tables :roll: ) and normal parm-s of z-score (0.98/2 is placed between 0 and z for which 0,4901 is z=2.33) ==> 1-0,4901=0,5099 and our approximation falls beyond 3 z-scores. So we have no good adjustment then? As 3 z-scores describe 0,998 confidence interval, the question is statistically deficient. That's what you meant?
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