(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Normally distributed probability problem

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