At a large local university, 40% of the students live in the dormitories. a random sample of 80 students is selected for a particular study. the probability that the sample proportion (the proportion living in the dormitories) is between 0.30 and 0.50 is what?

0.9325

Step-by-step explanation:

Given  

n = sample size = 80

p = probability = 0.4

q = 1 – p = 0..6

standard deviation for the proportion = √ (p * q) /n = √(0.4*0.6)/80 = 0.0547

for the proportion mean is 0.4

now we can find z and the probability  

P (0.3<mean<0.5) = P((0.3– 0.4)/0.0547 < z < (0.5– 0.4)/0.0547)

P (0.3<mean<0.5) = P(-1.828< z < 1.828)

Using a z table

P (0.3<mean<0.5) = 0.9325

answer: plz put a pic in so i can answer it

step-by-step explanation: