Unformatted text preview: students in the sample of size n = 100 that live off campus. Then X still has a binomial model with n = 100 and p = 0.70. However the sample size is large enough to use the normal approximation (could you verify this?). So to find the probability of at least 80, we use that X is approximately normal with a mean of 100(0.70) = 70 and a standard deviation of sqrt[100(0.70)(0.30)] = 4.58. 80  70 P( X 80) = P Z = P( Z 2.18) = 1  0.9854 = 0.0146 4.58 4. (ii) is the correct interpretation.
5. a. P(X < 65) = P Z < b. (0.0013)2 = 0.00000169 (very unlikely). c. 6. a. 65  80 = P(Z < 3)= 0.0013 5 x  80 so x = 80  1.28(5) = 73.6 5...
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 Fall '08
 Gunderson
 Normal Distribution, Standard Deviation, 14%, 15 years, 25 years, 16 years

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