The sampling distribution of x has a mean μₓ = <u> μ </u> and standard deviation σₓ = <u> σ/√n </u>.
In the question, we are given that a random sample of size n is drawn from a large population with mean μ and standard deviation σ.
We are asked to find the mean and the standard deviation for the sampling distribution of the variable x for this sample.
The sample mean is regularly distributed, with a mean μₓ = μ and standard deviation σₓ = σ/√n, where n is the sample size, for samples of any size taken from populations that have a normal distribution.
Thus, the sampling distribution of x has a mean μₓ= <u> μ </u> and standard deviation σₓ= <u> σ/√n </u>.
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The provided question is incomplete. The complete question is:
"Fill in the blanks to correctly complete the sentence below.
Suppose a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ.
The sampling distribution of x has mean μₓ =______ and standard deviation σₓ =______."
8-13= -5 //hope this helped :)
Answer:
164
Step-by-step explanation:
(-7)^2 + 5 • 25 - 10
49 + 5 • 25 - 10
49 + 125 -10
164
Answer:
Around 67
x=67
Step-by-step explanation:
90 over 100 equals 60 percent over x
60*100/90=
6000/90=
66.66666666666666 (more sixes continue on)
