# SOLUTION: Normal Distribution The scores for the intelligence quotient (IQ) test are normally distributed with mean 100 and standard deviation 16. (a) What is the probability that at leas

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: Normal Distribution The scores for the intelligence quotient (IQ) test are normally distributed with mean 100 and standard deviation 16. (a) What is the probability that at leas      Log On

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 Question 258722: Normal Distribution The scores for the intelligence quotient (IQ) test are normally distributed with mean 100 and standard deviation 16. (a) What is the probability that at least two of five randomly selected people have IQ test score higher than 120? (b) A person is randomly selected from all people whose IQ are higher than 110. What is the probability that this person has IQ higher than 120? (c) One thousand people are randomly selected. Approximately what is the probability that fewer than 90 of them have IQ score higher than 120?Answer by stanbon(57407)   (Show Source): You can put this solution on YOUR website!Normal Distribution The scores for the intelligence quotient (IQ) test are normally distributed with mean 100 and standard deviation 16. (a) What is the probability that at least two of five randomly selected people have IQ test score higher than 120? --- Find the z-score of 120: z(120) = (120-100)/16 = 1.25 P(z>1.25) = normalcdf(1.25,100) = 0.1056 Find P(2<= x <=5 with n = 5 and p = 0.1056) = 1-binomcdf(5,0.1056,1) = 0.0899.. ------------------------------------------------------------- (b) A person is randomly selected from all people whose IQ are higher than 110. What is the probability that this person has IQ higher than 120? z(110) = 10/16 = 0.625 P(x>110) = P(z> 0.625) = 0.26598.. P(x>120 | x>110) = P(x > 120)/P(x>110) = 0.1056/0.26598 = 0.3970.. ======================================================================== (c) One thousand people are randomly selected. Approximately what is the probability that fewer than 90 of them have IQ score higher than 120? P(x>120) = 0.1056 P(k< 90) = binomcdf(1000,0.1056,89) = 0.04624..... Cheers, Stan H.