Answer:
a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45
![\bar X = 41.538](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%2041.538)
![s = 5.317](https://tex.z-dn.net/?f=s%20%3D%205.317)
b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25
![\bar X = 43.615](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%2043.615)
![s= 5.583](https://tex.z-dn.net/?f=%20s%3D%205.583)
c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75
![\bar X= 3.462](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%203.462)
![s = 0.443](https://tex.z-dn.net/?f=%20s%20%3D%200.443)
d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).
And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.
And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.
Step-by-step explanation:
For this case we can use the following formulas for the mean and standard deviation:
![\bar X = \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
![s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
Part a
Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45
And if we calculate the mean we got:
![\bar X = 41.538](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%2041.538)
![s = 5.317](https://tex.z-dn.net/?f=%20s%20%3D%205.317)
Part b
For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:
44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25
And if we calculate the new mean and deviation we got:
![\bar X = 43.615](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%2043.615)
![s= 5.583](https://tex.z-dn.net/?f=%20s%3D%205.583)
Part c
The new dataset would be each value divided by 12 so we have:
3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75
And the new mean and deviation would be:
![\bar X= 3.462](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%203.462)
![s = 0.443](https://tex.z-dn.net/?f=%20s%20%3D%200.443)
Part d
As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).
And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.
And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.