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In a local university, 66% of the students live in the dormitories. A random sample of...

In a local university, 66% of the students live in the dormitories. A random sample of 80 students is selected for a particular study. We know that the standard error of the proportion is 0.0530. Find the probability that the sample proportion (the proportion living in the dormitories) is between 0.65 and 0.68.

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Answer #1

0=0.66 n=80 us= } = 0.66 F = 1340- $) = 50.66 (1-0.66) ✓ 80 -0.0530 we have to find, P(0-65<P< 0.68) = P(0.65-up < Pelộ < 0.6

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