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Ch 4.1 Discrete Random Variable

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  • Ch 4.1 Discrete Random Variable

    Random Variable

    Random Variable: is a variable (X) that has a single numerical value, determined by chance, for each outcome of a procedure.


    Probability distribution: is a table, formula or graph that gives the probability of each value of the random variable.


    A Discrete Random Variable has a collection of values that is finite or countable similar to discrete data.

    A Continuous Random Variable as infinitely many values and the collection of values is not countable.


    Ex : X = the number of times “four” shows up after tossing a die 10 times is a discrete random variable.

            X = weight of a student randomly selected from a class. X is a continuous random variable

            X = the method a friend contacts you online. X is not a random variable. ( X is not numerical)


    A Probability Distribution (PDF) for a Discrete Random Variable is a table, graph or formula that gives Probability of each value of X.

     A Probability Distribution Function (PDF or PD) satisfies the following requirements:

    1. The value X is numerical, not categorical and each P(x) is associated with the corresponding probability.

    2  ΣP(X) = 1 .  A ΣP(X) = 0.999 or 1.01 is acceptable as a result of rounding.

    3.  0  ≤P(x) ≤ 1 for all P(x) in the PDF.


    Ex1:  PDF for number of heads in a two-coin toss are given as a table and a graph.

    Table                  graph

     Probability distribution   probability distribution in histogram

    Both are valid PDF  because Σ P(X) = 1

     and each value of P(X) is between 0 and 1.

    Ex2:  The number of medical tests a patient receives after entering a hospital is given by the PDF below


    a) Is the table a valid PDF?

    The table is not a probability distribution because  Σ P(x) = 0.02+0.18+0.3+0.4 = 0.9  is not 1

    b) Define the random variable x.

     x = no. of medical tests a patient receives after entering a hospital.

    c) Explain why the x = 0 is not in the PDF?

         A patient always receives at least one medical test in the hospital.


    Parameters of a Probability distribution:

    Mean μ for a probability distribution:   

        \( E(x) = \mu  = \sum{x\cdot P(x)} \)

    Variance σ2 for a probability distribution:

     \( \sigma^2 = \sum{(x-\mu)^2}\cdot P(x) \) 

    Standard deviation for a probability distribution:

    \( \sigma = \sqrt{\sum{(x-\mu)^2}\cdot P(x)} \)

    To calculate Mean, variance and standard deviation of a probability distribution by technology:

    Use Libretext statistics calculator:

    Enter Number of outcomes, each X and P(X), calculate.

    round off rule: one more decimal place than for E(x)

    Two decimal places for σ and σ2.


    Expected value = the long-term outcome of average of x when the procedure is repeated infinitely many times. Round to one decimal place.


    Non-significant values of X.

    1. The range of X from  \( \mu - 2\cdot\sigma \text{ to }\mu +2\cdot\sigma \) are non-significant. (Range of rule of Thumb)

    2. X that are outside of  \( \mu - 2\cdot\sigma \text{ to } \mu +2\cdot\sigma \)    are significant that is unlikely to occur.


    Ex1: X = no. of year a new hire will stay with the company. P(x) = Prob. that a new hire with stay for x year.

    probability distributiona) Find mean, variance, st. deviation and determine the Expected number of years a new hire will stay.




    Use statistics discrete random variable calculator,

    Enter number of outcomes = 7.    Mean = 2.4, σ2 = 2.73, σ =1.65

    The Expected no. of year a new hire will stay is 2.4 years.

    b) Find probability that a new hire will stay for 4 years or more.  

    Add P(4), P(5) and P(6) = P( 4 or more) = 0.1 + 0.1 + 0.05 = 0.25    

    c) Find probability that a new hire will stay for between 3 or 5 years inclusive.

    P( 3 to 5 inclusive) = 0.15 + 0.1+0.1  = 0.35

    d) Find the probability that a new hire will stay for 2 years or fewer.  

    P(2 or fewer) = 0.12 + 0.18 + 0.30 = 0.6  

    e) Find the range of non-significant year of stay.

    2.4 -2(1.65) to 2.4 + 2(1.65) is -0.9 to 5.7


    Ex2: Given x = of number of textbooks a student buy per semester. What is the expected number of textbooks?

    probability distributiona) Find mean, variance and standard deviation.

    Use statistics discrete random variable calculator, Enter number of outcomes = 6

    E(x) = μ = 3.5,    σ2 = 0.61,  σ = 0.78

    Expected number of textbook is 3.5 books.

    b) Find Probability that a student buys at least 5 textbook.  

    P( at least 5)  = P(5 or more) = 0.03 + 0.02  = 0.05,


    c) Find probability that x is at most 2.  

    P(at most 2 ) = P( 2 or fewer) = 0.02 + 0.03 = 0.05


    c) Find the range non-significant.  

    Range of non-significant is 3.5 – 2(0.78) to 2.5 + 2(0.78) is 1.94 to 5.06.  


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