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skad [1K]
4 years ago
8

Suppose that Y1,....Yn are i.i.d random variables with a N (μγ, σ^2y) distribution. How would the probability density of Y chang

e as the sample size n increases?
a. As the sample size increases, the variance of ȳ decreases. So, the distribution of ȳ becomes highly concentrated around μy.
b. As the sample size increases, the variance of ȳ decreases. So, the distribution of ȳ becomes less concentrated around μ μy.
c. As the sample size increases, the variance of ȳ increases. So, the distribution of ȳ becomes highly concentrated around μy.
d. As the sample size increases, the variance of ȳ increases. So, the distribution of ȳ becomes less concentrated around μy
Mathematics
1 answer:
klemol [59]4 years ago
7 0

Answer:

A

Step-by-step explanation:

As the sample size n increases, the sample mean (μy) becomes a more accurate estimate of the parametric mean, so the standard error of the mean becomes smaller. Therefore, the variance of y decreases and the distribution of y becomes highly concentrated around μy.

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Solve \ x^4 - 17x^2 + 16 = 0

To solve this equation , we need to write it in quadratic form

ax^2+bx+c=0

To get the equation in quadratic form we replace x^2 with u

x^2 = u

x^4 - 17x^2 + 16 = 0 can be written as

(x^2)^2 - 17x^2 + 16 = 0, Replace u for x^2

So equation becomes

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