maccaman said:
The probability questions are pretty much the same, and i have trouble with these ones
Question 1
The marks in a maths exam are normally distributed with a mean of 52 and a s.d. of 14. The number of marks alotted to each candidate is integral. If 30.85% of the candidates fail, what is the pass mark?
The pass mark is the mark below which 30.85% of the people scored. Now, with a normally distributed sample of data, regardless of the actual values for mean and s.d., there will always be 68% of the data within 1 standard deviaition from the mean. So, in this example, we know that 68% of students had marks between 52-14 and 52+14, or 38 and 66. That's for 1 standard deviation. If you chose 1.234 standard deviations, then there would some other fixed percentage of data (greater than 68%) within that range, the range being from (52 - 14*1.234) and (52 + 14*1.234).
But if we guess 1 standard deviation, with 68% of people within 1 s.d. of the mean, that means that 32% of people are outisde that range. This means that 16% are below that range, and 16% are above that range. However, we want it so that 30.85% of people are below that range. This also means that 30.85% are above that range, and in total, 61.70% of people are outside that range. So we need to find x, where where 38.3% of people lie within x standard deviations of the mean. Find this, and the answer to your question would be the lower boudn of the range (the pass/fail mark) which would be 52 - 14x. Now, how do you find x? That's very difficult. Oh, of course, if you do find x, and 52-14x turns out not to be an integer, then you'll have to round up to find the pass mark. As far as I know, to find x, you'd plug "0.3830" into "CI" in the following equation:
x = \sqrt{2} \mathop{\rm erf}\nolimits ^{-1} (CI)
Where erf^(-1) is the Inverse erf function (erf stands for error function I think) which is a complicated function. Truth is, I don't know how you'd get a nice answer, I'd *guess* x = 1/2., and so the pass mark is 45, but that's just a guess.
Question 2
In a certain exam, the class average was 55% and the s.d. was 19. If the teacher wanted to award the top 1% a grading of A, find the percentage the students had to score to be given an A.
Any help would be greatly appreciated, thanks
Same idea, find the x so that 98% of the data (or CI = 0.98), then your answer is 55 + 19x. Again, I would *guess* x is 3, that is, 98% of the data lies within 3 s.d. of the mean, so 2% lie outside. This means 1% lies below, and 1% lies above, and that's what we want (only 1% with marks high enough to get this grade). So 55 + 19(3) = 112. Clearly, a bad guess because there's no way a student will get 112%, so I don't know how to help you. It would be nice if you had a table of values that correspond CI's to x's.
I guessed.
