BUS 308 QUIZ 2 [ ***BUS 308 QUIZ 2 ***] (BUS 308 QUIZ 2)

1. P(A∩B) = 0 represents. (Points : 1)

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        Independent events

        Mutually exclusive events

        Conditional events

        Dependent events

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2. When two events are independent and we are calculating conditional probability P(A | B) then it follows that (Points : 1)

        P(A) = P(B)

        P(A | B) = P(A)

        P(A ∩ B)=2

        P(A U B)=0

 

3. The _______________ describes the number of occurrences of an event over a specified interval of time or space. (Points : 1)

        Binomial random variable

        Poisson random variable

        Discrete random variable

        Continuous random variable

 

4. In a statistical study, the random variable X = 1, if the house is colonial, and X = 0 if the house is not colonial, then it can be stated that the random variable is continuous. (Points : 1)

        True

        False

 

5. Which two distributions are useful in analyzing queues? (Points : 1)

        Binomial and normal

        Normal and exponential

        Poisson and normal

        Poisson and exponential

 

6. If a random variable x has a uniform distribution with a mean of 10 and the lowest value of x is 5 what is the largest value of x that can exist? (Points : 1)

        5

        15

        10

        20

 

7. The height of a continuous probability curve over a given point is (Points : 1)

        Equal to the mean

        Equal to the standard deviation

        Equal to zero

        Equal to Z

 

8. Events that have no sample space outcomes in common, and, therefore, cannot occur simultaneously are (Points : 1)

        independent.

        mutually exclusive.

        intersections.

        unions.

 

9. A standard normal distribution has a mean of ____and standard deviation of ____ (Points : 1)

        zero, zero.

        zero, one.

        one, one.

        one, zero.

  

10. The rule of complements is represented by (Points : 1)

        P(A | B) = P(A ∩ B)

        P(A U B) = P(A) + P(B) – P(A ∩ B)

        P(Ā) = 1 – P(A)

        P(A ∩ B) = P(A) * P(B)

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