Question 1044667
a) z-score = (X - mean) / std dev = (675 - 512) / 113 = 1.4424 approx 1.44
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we consult the table of Z-values for our z-score
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Probability(P) ( X < 675 ) = 0.9251 approx 93%
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b) sample is 1000 and sample mean is the same as population mean = 512
sample std. dev. = std. dev / square root(sample size) = 113 / square root(1000) = 3.5733 approx 3.57
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P ( X > 575 ) = 1 - P ( X < 575 )
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z-score = (575 - 512) / 3.57 = 17.6471 approx 17.65
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P ( X < 575 ) = 100% or 1.00
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We expect no SAT scores > 575 in our sample of 1000
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