Suppose the mean cost of a coffee drink at a local coffee es
Suppose the mean cost of a coffee drink at a local coffee establishment is $4.35 with a standard deviation of $0.83. If the manager decides to lower every drink price by 25 cents, what will be the new mean and standard deviation of the distribution of new prices?
[Control]
A. Mean = $4.10; Standard deviation = $0.83
[Control]
B. Mean = $4.10; Standard deviation = $0.58
[Control]
C. Mean = $4.35; Standard deviation = $0.83
[Control]
D. Mean = $4.35; Standard deviation = $0.58
[Control]
E. Mean = $4.35; Standard deviation = $1.08
| [Control] A. Mean = $4.10; Standard deviation = $0.83 | |
| [Control] B. Mean = $4.10; Standard deviation = $0.58 | |
| [Control] C. Mean = $4.35; Standard deviation = $0.83 | |
| [Control] D. Mean = $4.35; Standard deviation = $0.58 | |
| [Control] E. Mean = $4.35; Standard deviation = $1.08 | 
Solution
Subtracting a constant value for each observation mean subtracting that same value to the mean.
However, it doesn\'t change the standard deviation, that is
u(x-c) = u(x) - c
 s(x-c) = s(X)
Thus,
Mean = 4.35 - 0.25 = 4.10
 s = 0.83 (still)
Thus, the answer is OPTION A. Mean = $4.10; Standard deviation = $0.83 [ANSWER, A]

