MS-Sol-EE-C16

# MS-Sol-EE-C16 - CHAPTER 16 THE NORMAL DISTRIBUTION AND ITS...

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Unformatted text preview: CHAPTER 16 THE NORMAL DISTRIBUTION AND ITS APPLICATIONS EXERCISE 16.1 Section 16.1 Continuous probability distributions (page 401) 1. (a) From the given, f ( x ) ≥ 0 for all x and ∫ ∞ ∞- ) ( dx x f = ∫ 2 2 dx x = 2 2 4 x = 1 ∴ f ( x ) is a probability density function of X . (b) (i) P( X < 1) = 4 1 = 4 = 2 1 1 2 ∫ x dx x (ii) P(0.5 < X < 1.5) = 2 1 = 4 = 2 5 . 1 5 . 5 . 1 5 . 2 ∫ x dx x (iii) P( X < 3) = ∫ 2 2 dx x = 1 2. (a) From the given, f ( t ) 0 for all t and ∞ ∞ ∞- ∞ ∫ ∫ - 1 1 2 1 = 1 = ) ( t dt t dt t f = 1 ∴ f ( t ) is a probability density function of T . (b) P(2 < T < 5) = 10 3 = 1 = 1 5 2 5 2 2 ∫ - t dt t 3. (a) (b) ∫ 30 ) ( dy y f = 1 ∴ 30 k = 1 k = 1 30 (c) P(10 < Y < 28) = (28 - 10) k = 3 5 240 C HAPTER 16 T HE N ORMAL D ISTRIBUTION AND I TS A PPLICATIONS (d) E( Y ) = ∫ ∞ ∞- ) ( dy y f y = ∫ 30 30 1 dy y = 30 2 2 30 1 y = 15 Var ( Y ) = ∫ ∞ ∞-- ) ( ) ( 2 dy y f y μ = ∫ - 30 2 30 1 15) ( dy y = 30 3 3 15) ( 30 1 - y = 75 4. (a) ∫ ∞ ∞- ) ( dx x f = 1 i.e. Area under AB and above the x-axis = 1 ∴ 1 2 (0.2 + k ) (5 - 1) = 1 k = 0.3 (b) Equation of AB is y- 0.2 = 1) ( 1 5 0.2 3 .--- x i.e. y = 1 40 x + 7 40 ∴ The probability density function of X is otherwise , 5 1 for , 40 7 + 40 1 = ) ( ≤ ≤ x x x f (c) P(2 < X < 4) = ∫ 4 2 40 7 + 40 1 dx x = 4 2 2 7 + 2 40 1 x x = 2 1 (d) P( X 4.5) = ∫ ∞ 5 . 4 ) ( dx x f = ∫ 5 5 . 4 40 7 + 40 1 dx x = 5 5 . 4 2 7 + 2 40 1 x x = 47 320 (or 0.146 9) 241 S ECTION 16.1 C ONTINUOUS P ROBABILITY D ISTRIBUTIONS 5. (a) As ∫ ∞ ∞- ) ( dx x f = 1 ∴ ∫- 2 2 2 1) + ( dx x k = 1 2 2 3 + 3- x x k = 1 k = 3 28 (b) P(- 2 ≤ X < 1) = ∫- 1 2 2 1) + ( 28 3 dx x = 1 2 3 + 3 28 3- x x = 9 14 (c) (d) mean = μ = ∫- ⋅ 2 2 2 1) + ( 28 3 dx x x = 2 2 2 4 2 + 4 28 3- x x = Since σ 2 = E( X 2 ) - μ 2 , variance = σ 2 = ∫-- ⋅ 2 2 2 2 2 1) + ( 28 3 μ dx x x = 3 + 5 28 3 2 2 3 5- - x x = 35 68 (or 1.942 9) 6. (a) From ∫ ∞ ∞- 1, = ) ( dx x f we have ∫ 3 3 dx kx = 1 3 4 4 x k = 1 81 4 k = 1 ∴ k = 4 81 242 C HAPTER 16 T HE N ORMAL D ISTRIBUTION AND I TS A PPLICATIONS (b) P( X ≤ m ) = 1 2 ∴ ∫ m dx x 3 81 4 = 1 2 m x 4 4 81 4 = 1 2 m 4 81 = 1 2 m = 81 2 4 = 2.522 7 7. (a) ∫- 4 ) (4 dy y ky = 1 4 3 2 3 2 - y y k = 1 k = 3 32 (b) μ = ∫- ⋅ 4 ) (4 32 3 dy y y y = 4 4 3 4 3 4 32 3 - y y = 2 σ 2 = ∫-- ⋅ 4 2 2 ) (4 32 3 μ dy y y...
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## This note was uploaded on 10/17/2011 for the course IELM 3010 taught by Professor Fugee during the Winter '11 term at HKUST.

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MS-Sol-EE-C16 - CHAPTER 16 THE NORMAL DISTRIBUTION AND ITS...

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