Respuesta :
Answer:
Part a: When the value of 2 years is subtracted from the individual values of participant's ages, the mean will be reduced by 2 years from 19.2 to 17.2 whereas the standard deviation will remain the same.
Part b: When the individual values of participant's weights are converted to pounds by multiplication with 2.2, the new mean will be a given as 143.7 (2.2*65.3 kg) whereas the standard deviation is 17.2 (2.2*7.83).
Explanation:
Part a:
As a specific value is subtracted from all the values thus the impact is as given as follows. The mean is given as follows
[tex]\mu=\frac{\sum x}{n}\\[/tex]
For the specific value k is subtracted from the each value x, this is given as
[tex]\mu'=\frac{\sum (x-k)}{n}\\[/tex]
Thus the new mean is given as
[tex]\mu'=\frac{\sum (x-k)}{n}\\\mu'=\frac{\sum (x)}{n}-\frac{\sum (k)}{n}\\\\\mu'=\mu-k\\[/tex]
This indicates that the new mean is simply equal to the difference of old mean and the value subtracted.
The standard deviation is given as
[tex]\sigma=\sqrt{\frac{\sum{(x-\mu)}^2}{n-1}[/tex]
The the specific value k is subtracted from the each value of x, the new value of standard deviation is given as
[tex]\sigma'=\sqrt{\frac{\sum{((x-k)-\mu')}^2}{n-1}[/tex]
The new standard deviation is given as
[tex]\sigma'=\sqrt{\frac{\sum{(x-k-\mu+k)}^2}{n-1}}\\\sigma'=\sqrt{\frac{\sum{(x-\mu)}^2}{n-1}}\\\sigma'=\sigma[/tex]
This indicates that the new standard deviation is simply equal to the original standard deviation.
So when the value of 2 years is subtracted from the individual values of participant's ages, the mean will be reduced by 2 years from 19.2 to 17.2 whereas the standard deviation will remain the same.
Part b
As a specific value is multiplied with all the values thus the impact is as given as follows. The mean is given as follows
[tex]\mu=\frac{\sum x}{n}\\[/tex]
For the specific value k is multiplied with each value x, this is given as
[tex]\mu'=\frac{\sum (kx)}{n}\\[/tex]
Thus the new mean is given as
[tex]\mu'=\frac{\sum (kx)}{n}\\\mu'=k\frac{\sum (x)}{n}\\\mu'=k\mu\\[/tex]
This indicates that the new mean is simply equal to the multiple of old mean and the value multiplied with each value.
The standard deviation is given as
[tex]\sigma=\sqrt{\frac{\sum{(x-\mu)}^2}{n-1}[/tex]
Then the specific value k is multiplied with each value of x, the new value of standard deviation is given as
[tex]\sigma'=\sqrt{\frac{\sum{(kx-\mu')}^2}{n-1}[/tex]
The new standard deviation is given as
[tex]\sigma'=\sqrt{\frac{\sum{(kx-k\mu)}^2}{n-1}}\\\sigma'=\sqrt{\frac{\sum{k^2(x-\mu)}^2}{n-1}}\\\sigma'=\sqrt{\frac{k^2\sum{(x-\mu)}^2}{n-1}}\\\sigma'=k\sqrt{\frac{\sum{(x-\mu)}^2}{n-1}}\\\sigma'=k\sigma[/tex]
This indicates that the new standard deviation is simply equal to the multiple of the original standard deviation with k.
So when the individual values of participant's weights are converted to pounds by multiplication with 2.2, the new mean will be a given as 143.7 (2.2*65.3 kg) whereas the standard deviation is 17.2 (2.2*7.83).
