In a large midwestern university 30% if the students live in apartments. if 200

Question

In a large midwestern university 30% if the students live in apartments. if 200 students are randomly selected find the probability that the number of them living in apartments will be between 55 and 70 inclusive (use normal approximation to binomial) In a large midwestern university 30% if the students live in apartments. if 200 students are randomly selected find the probability that the number of them living in apartments will be between 55 and 70 inclusive (use normal approximation to binomial) In a large midwestern university 30% if the students live in apartments. if 200 students are randomly selected find the probability that the number of them living in apartments will be between 55 and 70 inclusive (use normal approximation to binomial)

Explanation / Answer

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    54.5      
x2 = upper bound =    70.5      
u = mean = np =    60      
          
s = standard deviation = sqrt(np(1-p)) =    6.480740698      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.848668425      
z2 = upper z score = (x2 - u) / s =    1.620185175      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.19803291      
P(z < z2) =    0.947403747      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.749370837   [ANSWER]  

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