{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Answerkey2

1 3 x 2 1 3 924 p x 1 x 2 1 9734 1 p x 1 x 2 1 9734 1

Info iconThis preview shows pages 5–6. Sign up to view the full content.

View Full Document Right Arrow Icon
1 3 + X 2 1 3 ± ± 0 : 924 ± = P ( X 1 + X 2 ± 1 : 9734) = 1 P ( X 1 + X 2 > 1 : 9734) = 1 P ( X 1 = 1 ;X 2 = 1) = 8 9 Note that we use P ( A ) = 1 P ² A C ³ for some set A and its complement A C in the fourth equality. Furthermore, there is only one pair of ( X 1 ;X 2 ) whose sum is greater than 1.9734, i.e. only when X 1 = X 2 = 1 : Lastly, since X 1 and X 2 are independent, P ( X 1 = 1 ;X 2 = 1) = P ( X 1 = 1) ² P ( X 2 = 1) = 1 3 ² 1 3 = 1 9 : (c) Similarily, when n = 3 ; P ( Z 3 ± 0 : 924) = P 1 p 3 3 P i =1 ( X i ) ± 0 : 924 ± = P 1 p 3 X 1 1 3 + X 2 1 3 + X 3 1 3 ± ± 0 : 924 ± = P ( X 1 + X 2 + X 3 ± 2 : 6004) = 1 P ( X 1 + X 2 + X 3 > 2 : 6004) = 1 P ( X 1 = 1 ;X 2 = 1 ;X 3 = 1) = 26 27 ³ 0 : 963 (d) So far, we have calculated the probability of P ( Z n ± 0 : 924) for n = 2 in (b) and n = 3 in (c). Now, we calculate the probability of P ( Z n ± 0 : 924) for large n using the central limit 5
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
theorem. As n ! 1 ; the central limit theorem states, p n X n ± ± ! d N (0 ; 1) where X n = 1 n P n i =1 X i is the sample average of independent random variables with ± = E ( X i ) and 2 = V ar ( X i ) for i = 1 ; 2 ;:::;n: Note that " ! d " denotes the "covergence in distribution", i.e. the distribution of p n X n ± ± becomes arbitrarily well approximated by the standard normal distribution N (0 ; 1) as n ! 1 : In order to apply the central limit theorem for the given random variable Z n ; dividing Z n by the standard deviation yields, Z n = 1 1 p n ² n P i =1 ( X i ± ) ³ = 1 1 p n ² n P i =1 X i n ± ± ³ = p n ² 1 n n P i =1 X i ± ³ = p n X n ± ± ! d N (0 ; 1) by the central limit theorem Z n ! d Z with Z ~ N (0 ; 1) : Given this, we can calculate P ( Z n ² 0 : 924) as follows, P ( Z n ² 0 : 924) = P ´ Z n ² 0 : 924 µ ³ P 0 @ Z n ² 0 : 924 q 2 9 1 A ³ P ( Z ² 1 : 96) ³ 0 : 975 where P ( Z ² 1 : 96) ³ 0 : 975 is found from the table 1 in the appendix (Given P ( Z 1 : 96) = 0 : 025 P ( Z ² 1 : 96) = 1 P ( Z ´ 1 : 96) = 1 & 0 : 025 = 0 : 975 where P ( Z ´ 1 : 96) = P ( Z 1 : 96) = 0 : 025 due to the symmetry of the standard normal distribution. 6
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page5 / 6

1 3 X 2 1 3 924 P X 1 X 2 1 9734 1 P X 1 X 2 1 9734 1 P X 1...

This preview shows document pages 5 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online