stats 363 HW 7-5

stats 363 HW 7-5 - Section 7.5 Exercise 7.19 The mean and...

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Unformatted text preview: Section 7.5 Exercise 7.19 . The mean and standard deviation of the sampling distrib- ution of the sample mean x are, respectively, given by E ( x ) = & and SE ( x ) = ¡ p n : ( a ) If n = 36, & = 10, and ¡ 2 = 9, then E ( x ) = 10 and SE ( x ) = 3 p 36 = 1 2 = 0 : 5 : ( b ) If n = 100, & = 4, and ¡ 2 = 4, then E ( x ) = 4 and SE ( x ) = 2 p 100 = 1 5 = 0 : 2 : ( c ) If n = 8, & = 120, and ¡ 2 = 1, then E ( x ) = 120 and SE ( x ) = 1 p 8 = p 2 4 = 0 : 3536 : Exercise 7.24 . A random sample of n = 25 observations is selected from a normal population with mean & = 106 and standard deviation ¡ = 12. ( a ) E ( x ) = 106 and SE ( x ) = 12 p 25 = 12 5 = 2 : 4 : ( b ) By the Central Limit Theorem , it follows that P ( x > 110) & P ( z > 1 : 67) = 1 ¡ P ( z ¢ 1 : 67) = 1 ¡ : 9525 = 0 : 0475 ; where the standardized value of 100 is 110 ¡ 106 2 : 4 & 1 : 67 : ( c ) By the Central Limit Theorem , it follows that P ( j x ¡ & j < 4) = P ( ¡ 4 < x ¡ & < 4) & P ( ¡ 1 : 67 < z < 1 : 67) = P...
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stats 363 HW 7-5 - Section 7.5 Exercise 7.19 The mean and...

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