In a survey on campus it is reported that about 75 of studen

In a survey on campus it is reported that about 75% of students graduate in time. Suppose 90 students are considered and let r represent the number of students who will graduate in time. Use the normal approximation to the binomial without correction for continuity to find the probability that at least 64 students graduate in time. (give z-score to two decimal places, and probability to three decimal places)

Find the Z-score

Find the probability

Please make the final answer in bold so I can easily find it

Solution

We first get the z score for the critical value:          
          
x = critical value =    64      
u = mean = np =    67.5      
          
s = standard deviation = sqrt(np(1-p)) =    4.107919181      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -0.852012867   [ANSWER, Z SCORE]  
          
Thus, the right tailed area is          
          
P(z >   -0.852012867   ) =    0.802896525 [ANSWER, PROBABILITY]