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10 November, 10:32

A sample of 12 measurements has a mean of 38 and a sample standard deviation of 4.25. Suppose that the sample is enlarged to 14 measurements, by including two additional measurements having a common value of 38 each. The sample standard deviation of the 14 measurements is:

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  1. 10 November, 10:57
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    The sample standard deviation of the 14 measurements is 3.93

    Step-by-step explanation:

    The standard deviation = √ (variance)

    The standard deviation is the square root of variance. And variance is an average of the squared deviations from the mean.

    Mathematically,

    Standard deviation = σ = √[Σ (x - xbar) ²/N]

    x = each variable

    xbar = mean

    N = number of variables

    For 12 variables,

    N = 12

    σ = 4.25

    xbar = 38

    Σ (x - xbar) ² = sum of the square of all deviations; let it be equal to D

    4.25 = √[Σ (x - xbar) ²/12]

    4.25 = √ (D/12)

    4.25² = D/12

    D = 216.75.

    For 14 measurements,

    N = 14

    The mean is going to still be 38, because the two new measurements are each 38.

    xbar = 38

    And the new additions to the sum of deviations, will be (38 - 38) ² twice, that is, 0.

    Standard deviation for 14 measurements

    Standard deviation = σ = √[Σ (x - xbar) ²/N]

    σ = √ (216.75/14) = 3.93
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