Normally distributed probability problem

1. Feb 24, 2009

subopolois

1. The problem statement, all variables and given/known data
an exam is normally distributed with a mean of 75 and a variance of 64. students with a mark of 90 or above recieve an A+
a) what percentage of students recieve an A+?
b) if 12 students write the exam what is the probability 3 get A+?

2. Relevant equations
z-score- z=mark-mean/standard deviation
my teacher gave us a chart that gives probabilities as area under a curve for normal distributions

3. The attempt at a solution

a) since the standard deviation is the square root of the variance:
SD= 8
z-score= 90-75/8= 1.88
now using the area under the curve chart, for a z-score of 1.88 it reads 0.9699. from this i subtract 1 because i want the amount above 90 and this value is below 90, so i get:
= 1-0.9699
= 0.0301 or 3.01%

b) if 12 students write and 3 get an A+ then 9 students must get something else:
p(x=3)= 3(0.0301)/9(0.9699)= 0.0103
since the probability of getting over A+ is 0.0301 and anything else is one subtract 0.0301.

do my answers seen correct? if not can someone guide me in the right direction?

2. Feb 24, 2009

vikkisut88

The answer to part a) is correct, however, I don't really understand what calculation you've done for part b).

Personally I would just use the Choose function i.e. 12C3 * 0.0301^3 * 0.9699^9

3. Feb 24, 2009

subopolois

Ok, would I also put in 12C9 when multlying0.9699?. Also why areyou putting 0.0301 with an exponent also for the other0.9699?

4. Feb 24, 2009

vikkisut88

nope you just have one 12C3 - it's how you use the binomial theorem:

P(X=r) = nCr * p^r * q^(n-r) where q = 1-p :)

5. Feb 24, 2009

vikkisut88

oh and n = total so in this case 12

6. Feb 24, 2009

subopolois

Alright! Thank you

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