Question

A large bakery makes cakes in two shifts: shift A and shift B. Suppose that, on average, cakes from shift A weigh

7.4 kg with a standard deviation of 0.2 kg. For shift B, the mean and standard deviation are 7.0 kg and 0.1 kg,
respectively.
Every day, the bakery takes an SRS of 25 cakes from each shift. They calculate the mean weight for each sample,
then look at the difference (A - B) between the sample means.
What are the mean and standard deviation (in kilograms) of the sampling distribution of CA - TB?

Answer 1

The mean of the sampling distribution is 0.4 kg and the standard deviation is 0.051 kg.

Step-by-step explanation:

The mean of the sampling distribution of CA - TB is the difference in means between shift A and shift B, which is 7.4 kg - 7.0 kg = 0.4 kg. The standard deviation of the sampling distribution is calculated using the formula:

σCA-TB = √(σA2/nA + σB2/nB)

Substituting the given values, we get:

σCA-TB = √((0.22/25) + (0.12/25)) = 0.051 kg

Answer 2

D

Step-by-step explanation:

Khan Academy