Question 1110831
mean(mu) = 9.3 mg/L
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(a)the 95% confidence interval = 9.3 + or - 2.6
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margin of error(me) = cv * standard deviation
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the me is 2.6, so we have
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cv * standard deviation = 2.6
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the cv for a 95% confidence interval is the z-score associated with a critical probability(p*) of 0.975 which is 1.96
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Note p* is calculated this way, p* = 1 - (a/2), alpha(a) = 1 - (95/100)
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standard deviation = 2.6 / 1.96 = 1.3265 
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(b)  z-score = (8 - 9.3) / 1.3265 = -0.9800 
Probability (P) ( X < 8 ) = 0.1635
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(c) P( X > 12 ) = 1 - P ( X < 12 )
z-score = (12 - 9.3) / 1.3265 = 2.0354
P ( X < 12 ) = 0.9791
P ( X > 12 ) = 1 - 0.9791 = 0.0209
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