Question

Grade point averages of students on a large campus follow a normal distribution with a mean of 2.6 and a standard deviation of 0.5. d) A random sample of 400 students is chosen from this campus. what is the probability that at least 80 of these students have grade point averages higher than 3.0? e) Two students are chosen at random from this campus. what is the probability that at least one of them has a grade point average higher than 3.0?

Answer 1

P ( X > 3   ) = P( (X-µ)/σ ≥ (3-2.6) / 0.5)          
= P(Z ≥   0.800   ) = P( Z <   -0.800   ) =    0.2119

d)

Sample size , n =    400                      
Probability of an event of interest, p =   0.2119   
  
                          
Mean = np =    84.742   
std dev ,σ=√np(1-p)=   8.1725                      
                          
P(X ≥   80   ) = P(Xnormal ≥   79.5   )          
                          
Z=(Xnormal - µ ) / σ = (   79.5   -   84.74215943   ) /   8.1725   =   -0.641
                          
=P(Z ≥   -0.641   ) =    0.7394             

e)

P(at least one) = 1 - P(X=0) = 1 - (1-0.2119)² = 0.3788