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 drive to campus. A random 90 students where selected, what is the probability that less than 64 of them drive to campus. Find answer without continuity correction, using the binomial normal approximation.

(round answer to three decimal places)

P(x<64)=

Solution

Here,

mean = np = 90*0.75 = 67.5
standard deviation = sqrt(n p (1-p)) = sqrt(90*0.75*(1-0.75)) = 4.107919181

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    64      
u = mean =    67.5      
          
s = standard deviation =    4.107919181      
          
Thus,          
          
z = (x - u) / s =    -0.852012867      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.852012867   ) =    0.197103475 [ANSWER]