# Quartile deviation is 1âˆ’1 =0.2063 (solving F (x)=x3

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```					2
√      R 1.02      ³       2      2
´
e
1. (a) Pr(0.99 < X < 1.02) =                 3· 218
√                exp − 218·3 ·(x−1) dx = 48.33%
2π     0.99                 2
3
Quartile deviation is 1 − 21/3 = 0.2063 (solving F (x) = x = 1 for
1
2   4
x yields lower quartile).
√ √     R        ³            ´
(b) Pr(0.99 < X ¯ < 1.02) = 10· 218 1.02 exp − 218·10·(x−1)2 dx =
√
2π     0.99           2
1
50.45% (since σ 2 =       10
).

2.
¡¢          ¡¢
(a) p = 1 − 2.8e−1.8 = 0. 537163, 4 p2 q2 + 4 p3 q + p4 = 74.11% since
2           3
F (x) = 1 − (1 + x)e−x
R
4! ∞
(b) 21 0 x (1 − (1 + x)e−x ) (1 + x)2 e−2x xe−x dx = 1.438
√      R 0.29      ³       2         2
´
e
3. Pr(X < 0.29) =          2· 301
√                exp − 301·2 ·(x−0.25) dx = 91.74% based
2π     −∞                    2
√
on F (x) = x
9/16−1/16
Quartile deviation is         2
= 1.
4

4.
√       ¡¢         ¡¢         ¡¢
(a) p =      .6, 1 − 8 p5 q 3 − 8 p6 q 2 − 8 p7 q − p8 = 8.2585%
5          6          7

8!
R √ 4
1        √                    R1       √ 3   √
(b)   4!3!
1
x x (1− x)3 2√x dx = 1 , 140 (x− 1 )2 x (1− x)3 dx =
3           3
1
33
0                                            0

5. f (x, y) = n(n−1)(y−x)n−2 for 0 < x < y < 1 ⇒ f (x, r) = n(n−1)rn−2
for 0 < x < 1 and 0 < r < 1 − x (where r = y − x) ⇒ f (r) =
R
1−r
n(n − 1)rn−2      dx = n(n − 1)rn−2 (1 − r) for 0 < r < 1, which is
0
beta(n − 1, 2).
R1                               R1
7!                                7!
When n = 7, E[X(1) ] = 0!×6! 0 x(1−x)6 dx = 1 , E[X(7) ] = 6!×0! 0 x7 dx =
8
7
R1 Ry
8
and E[X(1) · X(7) ] = 7 × 6 0 y 0 x(y − x)5 dxdy = 1 , which implies
9
7
that Cov[X(1) , X(7) ] = 1 − 64 = 0.001736
9
R 1 R 0.1
Finally, Pr[X(1) < 0.1 ∩ X(7) > 0.9] = 7 × 6 0.9 0 (y − x)5 dxdy =
25.31%

```
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 views: 8 posted: 8/31/2010 language: English pages: 1