In a large midwestern university 30 if the students live in
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)
Solution
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]
