Practice Exam: Statistical Analysis, Financial Management, and Accounting Concepts

This practice exam covers statistical analysis and financial management concepts, beneficial for exam preparation.

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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 =𝑋0.070.01~Normal(0,1)Required probability,P(X < 0.09) = P(Z <0.090.070.01=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 <𝑘0.070.01) =0.85 = P(Z <1.04) ;From normal tableComparison gives,𝑘0.070.01= 1.04K = 0.07+1.04*0.01 =0.0804Thus the required number is 0.0804.

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In poker the cards are drawn and dealt randomly and after each game the cards are mixed up so that itis random. That is one game does not affect another game. So each hands areindependent thus thecorrect option is b.

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Probability means the chance ofthe occurrence of that event, now according to the classical definitionof probability is nothing but the number of favorable cases out of a large number of cases. So here thecorrect option is B.Assuming X = number of people in a family.We know that the total probability is 1. So,P( X = 7 or more) = 1(P(X=2) +P(X=3) +P(X=4) +P(X=5) +P(X=6))= 1-0.98 = 0.02P(X >2) = 1-P(X =2) = 1-0.42 = 0.58P(X = 3 or 4) = P(X = 3) + P(X =4) ;because if a family has 3 members it cant have 4 members thus.= 0.23+ 0.31 = 0.44
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