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skad [1K]
3 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]3 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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What percent of the June budget was for babysitting and movies (to the nearest tenth)?
Aleonysh [2.5K]
The monthly expenses are found below:


Mortgage - $752 Tim's Chevron - $23Johnny's Allowance - $8Electric - $176Jenny-babysitting - $12Movie - $14Dry Cleaning - $41Tithe - $85,Credit Card - $150Food - $101Phone - $45Water - $16 Car - $272 


Adding all the expenses, will give us a total of $1695.
 
To get the percent which is allocated for movie and the baby sitting:


Add the amount for the two then divide that to the total then multiply it by 100%
= 12 + 14 / 1695 x 100%
= 26 / 1695 x 100%
= 0.015339233 x 100%

=1.5%
The answer is 1.5%



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