Suppose that a random sample of size 64 is to be selected from a population with mean 50 and standard deviation 5. (a) What are the mean and standard deviation of the sampling distribution of x

Mean of sampling distribution = 50

Standard deviation of sampling distribution, = 0.625

Step-by-step explanation:

Given :

Mean, μ = 50

Standard deviation, σ = 5

Sample size, n = 64

The mean of sampling distribution, μxbar = population mean, μ

μxbar = μ

According to the central limit theorem, the sampling distribution converges to the population mean as the sample size increases, hence, , μxbar = μ = 50

Standard deviation of sampling distribution, σxbar = σ/√n

σxbar = 5/√64 = 5 / 8 = 0.625

6, 10

9, -13

12, -21

18, -23

step-by-step explanation:

you just add 6 to the x-coordinate and subtract 10 from the y-coordinate.