MyLab and N Take a Test MATH 211 JAN 2018 9002(LEE) Test: Final Exam Part II Thi

Question

MyLab and N Take a Test MATH 211 JAN 2018 9002(LEE) Test: Final Exam Part II This Question: 7 pts Submit Test This Test: 100 pts with 14 minutes and . 2 minutes. Complete parts (a) through (d). a. If you select a random sample of 50 sessions, what is the probability that the sample mean is between 13.8 and 14.2 minutes? (Round to three decimal places as needed.) 14 minutes? (Round to three decimal places as needed.) c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 13.8 and 14.2 minutes? (Round to three decimal places as needed.) d. Explain the difference in the results of (a) and (c). Choose the correct answer below The sample size in (c) is greater than the sample size in (a), so the standard error of tho mean (or he the standard deviation the sample size increases. | han in (a). As when standard deviation ofthe samping distributor)in (c)is | 1. values become | I concentrated around the mean. Therefore, the probability of a region that includes the mean will always Enter your answer in each of the answer boxes

Explanation / Answer

mean = 14 , s= 2

a) n = 50
P(13.8 < x < 14.2)
= P((13.8 - 14 ) / ( 2/sqrt(50) < z < (14.2 - 14 ) / ( 2/sqrt(50))
= P( -0.7071 < z < 0.7071)

P(13.8 < x < 14.2) = P( -0.7071 < z < 0.7071 ) = 0.5206

b)

n = 50
P(13.5 < x < 14)
= P((13.5 - 14 ) / ( 2/sqrt(50) < z < (14 - 14 ) / ( 2/sqrt(50))
= P(-1.7677 < z < 0 )

P(13.5 < x < 14) = P(-1.7677 < z < 0) = 0.4615

c)

n = 100
P(13.8 < x < 14.2)
= P((13.8 - 14 ) / ( 2/sqrt(100) < z < (14.2 - 14 ) / ( 2/sqrt(100))
= P(-1 < z <1 )

P(13.8 < x < 14.2) = P( -1 < z < 1) = 0.6827

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