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I don't know if I have enough data to calculate (or even estimate) this because I don't understand statistics, distributions, etc.

550 people took a test to try to gain entry to a school. There were 2 tests: one in English and one in math. The outcome was as follows, in raw numbers (i.e., not percentages):

Mean score in English = 100
Mean score in math = 94

The scores in both subjects were added together to give the candidate's overall score.

Of the 550 people, 221 scored more than the pass mark of 196. The range of scores from all 550 candidates, for each test was 69 to 141.

Although 221 passed the test, only 150 places are available in the school, so 71 people are going to be out of luck.

Assuming a normal distribution of scores, and the top 150 people are offered a place, is it possible to determine whether a particular candidate is likely to be in the top 150 if his score is known? For example, a score of 220.

I don't have values for variance or SD, so I'm not expecting an answer unless some math genius out there can work it out from the limited data I have given.

daOnlyBG
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Phil Rogers
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  • For the English test, the mean is $100$, and the standard deviation should be set so that you would expect one in $550$ to fall outside the $69$-to-$141$ range. That's what my gut is telling me anyhow. – Arthur Oct 27 '14 at 23:45
  • It is not possible to be certain, especially since you do not have a good estimate of standard deviation or variance, though as Arthur says, it might be possible to make an uncertain estimate of these given the range of scores. A further major problem is that you do not know the correlation between English and Maths scores. – Henry Oct 27 '14 at 23:48
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    Another remark: given your data, the mean score of the test is 194. Since 40% scored more than 196 (which is very close to 194), the distribution of scores seems to be skewed, and thus non-normal. – Ian Oct 27 '14 at 23:53

1 Answers1

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Not allowed to comment, so...

from my ecn 221 class notes (business statistics):

"Use a conservative estimate of the population variance such as the range divided by 4."

Range = 141 - 69 = 72; variance = $\dfrac{72}{4}$ = 18, standard deviation = 4.2426

hope this helps

Mlika
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