At a large local university, 40% of the students live in the dormitories. A random sample of 80 students is selected for a parti
cular study. The probability that the sample proportion (the proportion living in the dormitories) is between 0.30 and 0.50 is what?
1 answer:
Answer:
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
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does this make sense to you? Do you have any questions about raising numbers to a negative value?