My Notes (a) If we have a distribution of x values that is more or less mound-sh

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My Notes (a) If we have a distribution of x values that is more or less mound-shaped and somewhat symmetrical, what is the sample size n needed to claim that the distribution of sample means x from random samples of that size is approximately normal? n 2 (b) If the original distribution of x values is known to be normal, do we need to make any restriction about sample size in order to claim that the distribution of sample means x taken from random samples of a given size is normal? O Yes O No Need Help? My Notes Ask Your Suppose x has a distribution with ? 24 and ?-17. (a) If random samples of size n 16 are selected, can we say anything about the x distribution of sample means? Yes, the x distribution is normal with mean ? O No, the sample size is too small. O Yes, the x distribution is normal with mean ?.. 24 and ? 24 and ? 4.25 = 1.1 Yes, the x distribution is normal with mean, = 24 and = 17 (b) If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16? O No, the sample size is too small. O Yes, the x distribution is normal with mean , 24 and ?? 1.1 Yes, the x distribution is normal with mean = 24 and = 4.25 O Yes, the x distribution is normal with mean ?,= 24 and ? = 17 Find P(20 sx 25). (Round your answer to four decimal places.) Need Help?T to Tuer Submit AnswerSave Pess Pracice Another Version

Explanation / Answer

2)
mean = 24 , s = 17
a) n = 16

Yes, xbar distribution is normal with mean = 24 and std.dev = 4.25

std.dev. = s/sqrt(n) = 17/sqrt(16) = 4.25

b)
Yes, xbar distribution is normal with mean = 24 and std.dev = 4.25

p(20 < = x < =25)

= p((20 - 24)/4.25 <z <(25-24)/4.25)
= P(-0.9412 < z < 0.2353)

P(20 < = x < =25) = P(-0.9412 < z < 0.2353) = 0.4197 by using standard normal table

3)
mean = 10 , s = 5

a) n =40
mean = 10 , std.dev = s/sqrt(n) = 5/sqrt(40) = 0.7906
P(10<=x<=12)
= P((10-10)/0.7906 <z <(12-10)/0.7906)
= P( 0 < 2.5297)

P(10<=x<=12) = P( 0 < 2.5297) = 0.4943 by using standard normal table

b)
n = 75
mean = 10 , std.dev = s/sqrt(n) = 5/sqrt(75) = 0.5774

P(10<=x<=12)
= P((10-10)/0.5774 <z <(12-10)/0.5774)
= P( 0 < 3.4638)

P(10<=x<=12) = P( 0 < 3.4638) = 0.4997 by using standard normal table

c) because sample size is large in part b)
The std.deviation of part b is small because of large sample size

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