Suppose the mean cost of a coffee drink at a local coffee establishment is $4.35

Question

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

Explanation / Answer

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]

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.