Respuesta :
The given data are the scores turned in by the amateur golfer, from
which the sample standard deviation can be calculated.
Responses:
The mean and sample standard deviation for the 2017 season are;
- Mean = 74
- Sample standard deviation ≈ 2.07
The mean and sample standard deviation for the 2018 season are;
- Mean = 74
- Sample standard deviation ≈ 5.26
What are the equations for the mean and sample standard deviation?
The mean is given by the formula;
[tex]\overline x= \mathbf{\dfrac{\sum \left x_i }{N}}}[/tex]
N = 8
The mean for the 2017 season is therefore;
[tex]\overline x[/tex] = (72 + 76 + 77 + 75 + 73 + 71 + 73 + 75) ÷ 8 = 74
The sample standard deviation is given by the formula;
[tex]s =\mathbf{\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{N -1}}}[/tex]
[tex]\sum \left (x_i- \overline x \right )^2[/tex] = 30
N = 8
The sample standard deviation for 2017 season is therefore;
[tex]\sigma =\sqrt{\dfrac{30}{8-1}} \approx \mathbf{2.07}[/tex]
The mean and sample standard deviation for the 2017 season are;
- Mean = 74
- Sample standard deviation ≈ 2.07
The mean for the 2018 season is found as follows;
[tex]\overline x[/tex] = (62 + 68 + 73 + 75 + 83 + 78 + 83 + 78) ÷ 8 = 74
The sample standard deviation is given by the formula;
[tex]s=\mathbf{\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{N - 1}}}[/tex]
[tex]\sum \left (x_i- \overline x \right )^2[/tex] = 194
N = 8
The sample standard deviation for 2018 season is therefore;
[tex]Standard \ deviation \ for \ the \ 2018 \ seaon, \sigma =\sqrt{\dfrac{194}{8 - 1}} \approx \mathbf{5.26}[/tex]
The mean and sample standard deviation for the 2018 season are;
- Mean = 74
- Standard deviation ≈ 5.26
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