In: Statistics and Probability
The calculations in excel is given below
salary(x) | (x-xbar)^2 | y=X+5 | (y-ybar)^2 | z=x-2 | (z-zbar)^2 |
55 | 56.829 | 60 | 56.829 | 53 | 56.829 |
38 | 89.52 | 43 | 89.52 | 36 | 89.52 |
53 | 30.675 | 58 | 30.675 | 51 | 30.675 |
58 | 111.06 | 63 | 111.06 | 56 | 111.06 |
41 | 41.751 | 46 | 41.751 | 39 | 41.751 |
41 | 41.751 | 46 | 41.751 | 39 | 41.751 |
55 | 56.829 | 60 | 56.829 | 53 | 56.829 |
38 | 89.52 | 43 | 89.52 | 36 | 89.52 |
53 | 30.675 | 58 | 30.675 | 51 | 30.675 |
29 | 340.827 | 34 | 340.827 | 27 | 340.827 |
58 | 111.06 | 63 | 111.06 | 56 | 111.06 |
55 | 56.829 | 60 | 56.829 | 53 | 56.829 |
43 | 19.905 | 48 | 19.905 | 41 | 19.905 |
xbar | Sx | ybar | Sy | zbar | Sz |
47.4615 | 9.4747 | 52.4615 | 9.4747 | 45.4615 | 9.4747 |
(a) Find the sample mean and the sample standard deviation.
The sample mean is $47.4615 and sample standard deviation is $9.4747
b) Each employee in the sample is given a $5000 raise. Find the sample mean and the sample standard deviation .
The sample mean is $52.1615 K and sample standard deviation is $9.4747
(c) each employee in the sample takes a pay cut of $2000 from the original salary. Find the sample mean and the sample standard deviation for the revised data set.
The sample mean is $45.1615 K and sample standard deviation is $9.4747
D) From resulta a, b and c,we conclude that if the increment/ decrement in the salary implies that sample mean is also get increased/decreased but no change in sample standard deviation.
In short the sample mean is affected by change in origin whereas sample standard deviation does not affected.