Practice Exam: Statistical Analysis, Financial Management, and AccountingConceptsWe know that for the normal distribution the Mean±3σ(standard deviation ) intervals contains 99.7%of the total observations.So here, (0.07-3*0.01,0.07+3*0.01) = 0.04 and 0.10 are the required numbers within which 99.7% of thestudents lies.Here X = BAC of students who drink five beer ~Normal(0.07,0.01),Z = Standard normal variate =𝑋−0.070.01~Normal(0,1)Required probability,P(X < 0.09) = P(Z <0.09−0.070.01=2) = 0.9773 (from standard normal table)= 97.73%Let that number be k so,P(X>k) = 0.15P(X < k) = 0.85P(Z <𝑘−0.070.01) =0.85 = P(Z <1.04) ;From normal tableComparison gives,𝑘−0.070.01= 1.04K = 0.07+1.04*0.01 =0.0804Thus the required number is 0.0804.Preview Mode
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