Calculating standard deviation and variance using the definitional formula Consider a data set containing the following values: 92 84 85 93 95 89 86 91 The mean of the preceding values is 89.375. The deviations from the mean have been calculated as follows: 2.625 -5.375 -4.375 3.625 5.625 -0.375 -3.375 1.625 If this is sample data, the sample variance is __ and the sample standard deviation is _.If this is population data, the population variance is and the population standard deviation is _.Suppose the smallest value of 84 in the data was misrecorded as 85, If you were to recalculate the variance and standard deviation with the 85 instead of the 84, your new values for the variance and standard deviation would be______.

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20-02-2020
Mathematics

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Calculating standard deviation and variance using the definitional formula Consider a data set containing the following values: 92 84 85 93 95 89 86 91 The mean of the preceding values is 89.375. The deviations from the mean have been calculated as follows: 2.625 -5.375 -4.375 3.625 5.625 -0.375 -3.375 1.625 If this is sample data, the sample variance is __ and the sample standard deviation is _.If this is population data, the population variance is and the population standard deviation is _.Suppose the smallest value of 84 in the data was misrecorded as 85, If you were to recalculate the variance and standard deviation with the 85 instead of the 84, your new values for the variance and standard deviation would be______.

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oeerivona oeerivona · Answer 1 · 2022-03-24T19:42:40+01:00

From the given data, mean, and deviations from the mean, the

completed statement is presented as follows;

If this is sample data, the sample variance is 16.268 and the sample

standard deviation is 4.033. If this is population data, the population

variance is 14.234, and the population standard deviation is 3.773.

Suppose the smallest value of 84 in the data was misrecorded as 85, if

you were to recalculate the variance and standard deviation with 85

instead of the 84, your new values for the variance and standard

deviation would be; for the sample, 14.875, 3.857, for the population,

13.016, 3.608, respectively.

How can the variance and the standard deviation be found?

The definitional formula for the variance is presented as follows;

[tex]Sample \ variance, \, s^2 = \mathbf{\dfrac{\sum \left(x - \overline x \right)^2}{N - 1}}[/tex]

[tex]Population \ variance, \, \sigma^2 =\mathbf{\dfrac{\sum \left(x - \overline x \right)^2}{N }}[/tex]

Where;

s = The standard deviation of the sample

σ = The standard deviation of the population

N = The sample size

From the given deviations from the mean, (x - [tex]\overline x[/tex])², we have;

∑(x - [tex]\overline x[/tex])² = (2.625)² + (-5.375)² + (-4.375)² + (3.625)² + (5.625)² + (-0.375)² + (-3.375)² + (1.625)² = 113.875

[tex]s^2 = \dfrac{113.875}{8 - 1} \approx \mathbf{16.268}[/tex]

The sample variance is approximately 16.268

The sample standard deviation is therefore; s ≈ √(16.268) ≈ 4.033

The population variance, σ², is found as follows;

[tex]\sigma ^2 = \dfrac{113.875}{8 } \approx \mathbf{14.234}[/tex]

The population variance is approximately 14.234

The population standard deviation is, σ = √(14.234) ≈ 3.773

Given that 84 is replaced with 85, we have;

The lowest deviation from the mean, -5.375, will be -4.375, which gives;

∑(x - [tex]\overline x[/tex])² = (2.625)² + (-4.375)² + (-4.375)² + (3.625)² + (5.625)² + (-0.375)² + (-3.375)² + (1.625)² = 104.125

The new values are as follows;

[tex]\mathbf{s_{new}^2} = \dfrac{104.125}{8 - 1} \approx \underline{14.875}[/tex]

[tex]\mathbf{s_{new}}= \sqrt{14.875} \approx \underline{3.857}\alpha[/tex]

[tex]\mathbf{\sigma_{new}^2} = \dfrac{104.125}{8 } \approx \underline{ 13.016}[/tex]

[tex]\mathbf{\sigma_{new}} = \sqrt{13.016} \approx \underline{3.608}[/tex]

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