The following relative frequency distribution was constructed from a population of 400. Calculate the population mean, the population varian

Question

The following relative frequency distribution was constructed from a population of 400. Calculate the population mean, the population variance, and the population standard deviation. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Class Relative Frequency −20 up to −10 0.22 0.22 −10 up to 0 0.340.32 0 up to 10 0.380.34 10 up to 20 0.06

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Emery 3 weeks 2021-12-31T00:47:10+00:00 1 Answer 0 views 0

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    2021-12-31T00:48:29+00:00

    Answer:

     E(X) = \sum_{i=1}^n X_i P(X_i)

     E(X) = -15*0.22 + (-5)*0.34 +5*0.38 +15*0.06= -2.2

    For the variance first we need to find the second moment given by:

     E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

    And if we replace we got:

     E(X^2) =(-15)^2*0.22 + (-5)^2*0.34 +(5)^2*0.38 +(15)^2*0.06= 81

    And then the variance is given by:

     Var(X) = E(X^2)- [E(X)]^2 = 81 -[-2.2]^2 = 76.16

    And the deviation is given by:

     Sd(X)= \sqrt{76.16}= 8.727

    Step-by-step explanation:

    Previous concepts

    In statistics and probability analysis, the expected value “is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values”.

    The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

    And the standard deviation of a random variable X is just the square root of the variance.  

    Solution to the problem

    Let X our random variable of interest.

    For this case we can calculate the mean , variance and deviation creating the following table:

    Class          Midpoint(xi)    fi           xi*fi     xi^2 *fi

    -20 to-10         -15           0.22        -3.3      49.5

    -10-0                 -5           0.34        -1.7        8.5

    0-10                   5           0.38         1.9        9.5

    10-20                15           0.06        0.9       13.5

    And for this case we can find the expected value with the following formula:

     E(X) = \sum_{i=1}^n X_i P(X_i)

    And if we replace we got:

     E(X) = -15*0.22 + (-5)*0.34 +5*0.38 +15*0.06= -2.2

    For the variance first we need to find the second moment given by:

     E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

    And if we replace we got:

     E(X^2) =(-15)^2*0.22 + (-5)^2*0.34 +(5)^2*0.38 +(15)^2*0.06= 81

    And then the variance is given by:

     Var(X) = E(X^2)- [E(X)]^2 = 81 -[-2.2]^2 = 76.16

    And the deviation is given by:

     Sd(X)= \sqrt{76.16}= 8.727

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