# Help with 2 Probability Problems, should be simple for most

1. Jun 19, 2004

### maccaman

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?

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

2. Jun 19, 2004

### AKG

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.

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.

3. Jun 19, 2004

### maccaman

we have answers to these btw, just not the worked solutions. On the sheet it says the pass mark is 45%, i.e. what you got. Although ive never seen it done the way youve done it, thanks. Yeh we also have the table of values in our text book, so we must use this http://www.statsoftinc.com/textbook/sttable.html

Last edited: Jun 19, 2004
4. Jun 19, 2004

### AKG

I guessed.
Yeah, with that you'll be fine.

5. Jun 19, 2004

### maccaman

ive solved it
mean = 52, s.d. = 14

Using normal approximation, the area below the z score is 0.3085. so to find the z score:

0.5 - 0.3085 = 0.1915

by looking at the table, this is equal to a z score of 0.5. Because it is in the lower half, it is -0.5.

To find the x score

z = (x - mean) / s.d.
-0.5 = (x - 52) / 14
-7 = x - 52
x = 45