MA 214 Probability and Statistics for Business
Dr. Ed Donley
Test 1
Name ____________________________
February 15, 1995

Formulas:
     
P(A B) = P(A) + P(B) - P(A B)
P(A B) = P(A|B) P(B)

  1. (6 points) Classify each of the following random variables as discrete or continuous.
    a) The weight of a bunch of bananas.
    Continuous
    b) The number of students in a statistics class.
    Discrete

  2. (11 points) Ford Motor Company's total sales revenues for 1986 are shown in the table below. Construct a relative frequency bar chart.
    RegionRevenue ($billion)
    U.S.50
    Canada10
    Europe13
    Latin Am.3
    All Other2

    Answer:

    Relative Frequencies
    US	50/78 = 0.64
    Can	10/78 = 0.13
    Euro	13/78 = 0.17
    L.A.	 3/78 = 0.04
    Other	 2/78 = 0.03
    

  3. (7 points) Telephone calls put on hold have mean length 55 seconds and standard deviation 15 seconds. Find the z-score for a call that was put on hold for 82 seconds. If the distribution of holding times is mound-shaped, is an 82 second wait unusual? Why?

    z = (x - )/ = (82-55)/15 = 27/15 = 1.8

    This is 1.8 standard deviations above the mean. This is not unusual. 95% lie within 2 standard deviations.

  4. (20 points) The salaries of a sample of 6 CEOs, in millions of dollars are:

    2, 1, 5, 8, 5, 3.

    Compute the following.

    a) mean

    = (2 + 1 + 5 + 8 + 5 + 3)/6 = 24/6 = 4

    b) median

    1, 2, 3, 5, 5, 8 median = (3 + 5)/2 = 4

    c) mode

    5

    d) range

    8 - 1 = 7

    e) first quartile

    i = 25/100 x 6 = 1.5 Round up to 2. Q1 = x2 = 2.

    f) third quartile

    i = 75/100 x 6 = 4.5 Round up to 5. Q3 = x5 = 5.

    g) interquartile range

    5 - 2 = 3

    h) variance

    s2 = [ (2-4)2 + (1-4)2 + (5-4)2 + (8-4)2 + (5-4)2 + (3-4)2 ]/(6-1) = 32/5 = 6.4

    i) standard deviation

  5. (10 points) A boxplot is shown below for the number of farms in each state in the U.S. Estimate the values of the Five Number Summary from the boxplot.

          -----------------
       ---I       |       I------------                    *
          -----------------
       +---------+---------+---------+---------+---------+-----     
         0        35        70       105       140       175

    Minimum: 0
    Q1: 10
    Median: 37
    Q3: 65
    Maximum: 185

  6. (9 points) A consultant has presented his client with three alternative research studies methods. If the client chooses alternative A, the consultant will have to hire 1 additional employee. Two employees will have to be hired if the client chooses B, and 5 if the client chooses C. The probability that the client will select alternative A is 0.1, with P(B) = 0.5 and P(C) = 0.4. Find the expected number of employees that will have to be hired.

        X         P(X)                X P(X)            
        1         0.10                 0.10             
        2         0.50                 1.00             
        5         0.40                 2.00             
                            E(X) =  X P(X) =  3.10                        
    

  7. (9 points) A certain computer has two power supplies. The computer can operate on either one. On any given day, the first is 99% reliable and the second is 99.5% reliable. What is the probability that on any given day the computer will lose its power?

    P(lose power) = P(first fails and second fails)
    = P(first fails second fails)
    = P(first fails) P(second fails)
    = (0.01) (0.005)
    = 0.00005

  8. (9 points) The results of an experiment studying children's understanding of TV commercials are shown in the table.

                                   Age               
                            5-7    8-10      11-13   
    Don't understand        55     65         30     
    Understand              45     35         70     
    

    1. What is the probability that a 9 year old will understand TV commercials?

      P(understanding | 9 years old) = 35/(65 + 35) = 0.35

    2. Is understanding independent of age? Use probabilities to demonstrate your answer.

      P(understanding) = (45+35+70)/300 = 0.50

      P(understanding | 8-10 years old) = 35/(65+35) = 0.35

      These are not equal, so understanding and age are dependent.

  9. (15 points) A production line produces bolts. Machine X produces 60% of the bolts and Machine Y produces the remainder. 0.1% of the bolts produced on Machine X are defective and 0.2% of the bolts produced on Machine Y are defective.

    1. Construct a probability tree for this situation.

    2. What is the probability of a bolt being defective?

      P(D) = P(X and D or Y and D) = P(X D) + P(Y D)
      = 0.0006 + 0.0008 = 0.0014