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 =______."
Answer: our function is:
f(x) = A*sin(2*pi*270s*x)
Step-by-step explanation:
We have that the frequency of a given engine is 270 hz, it starts at 0, so we may represent this with a sin function (because sin(0) = 0)
Then we could use the function
f(x) = A*sin(c*x)
A is the maximum intensity of the hum, and c depends on te frequency, we know that the period of a sin function is 2pi, then:
f = 270/s
T = 1/270s
c*1/270s = 2pi
c = 2*pi*270s
Then our function is:
f(x) = A*sin(2*pi*270s*x)
15x<130+20, 15x<150 divided both sides by 15 ,the answer is 10
