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

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Practice Exam: Statistical Analysis, Financial Management, and Accounting Concepts We know that for the normal distribution the Mean ± 3 σ (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 the students lies. Here X = BAC of students who drink five beer ~Normal(0.07,0.01), Z = Standard normal variate = 𝑋 − 0 . 07 0 . 01 ~Normal(0,1) Required probability, P(X < 0.09) = P(Z < 0 . 09 − 0 . 07 0 . 01 = 2 ) = 0.9773 (from standard normal table)= 97.73% Let that number be k so, P(X>k) = 0.15  P(X < k) = 0.85  P(Z < 𝑘 − 0 . 07 0 . 01 ) =0.85 = P(Z < 1.0 4) ;From normal table Comparison gives, 𝑘 − 0 . 07 0 . 01 = 1.04  K = 0.07+1.04*0.01 = 0.0804 Thus 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 it is random. That is one game does not affect another game. So each hands are independent thus the correct option is b.
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Probability means the chance of the occurrence of that event, now according to the classical definition of probability is nothing but the number of favorable cases out of a large number of cases. So here the correct 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 = 0.02 P(X >2) = 1 - P(X =2) = 1 - 0.42 = 0.58 P(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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