ma316f03hw6

ma316f03hw6 - 1 3 2 e-( x- ) 2 2 2 ( x- ) and f 00 ( x )...

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MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #6 SOLUTIONS: SECTION 3.4 FALL 2003 3.39: If X N (75 , 100) then μ = 75 and σ = 10 so for part a) we have P ( X < 60) = P ± X - μ σ < 60 - 75 10 = P ( Z < - 1 . 5) = Φ( - 1 . 5) = 1 - Φ(1 . 5) = 0 . 067 using the table in the back of the text. For part b) we have similarly that P (70 < X < 100) = P ± 70 - 75 10 < X - μ σ < 100 - 75 10 = P ( - 0 . 5 < X < 2 . 5) = Φ(2 . 5) - Φ( - 0 . 5) = 0 . 994 - (1 - 0 . 691) = 0 . 685 . 3.51: If X N ( μ,σ 2 ) and P ( X < 89) = 0 . 9 then P ± X - μ σ < 89 - μ σ = P ± Z < 89 - μ σ = 0 . 9. Using the tables in the back of the text we see that since Φ(1 . 282) 0 . 9 then z = 89 - μ σ = 1 . 282. Similarly from P ± X - μ σ < 94 - μ σ = 0 . 95 we have that z = 94 - μ σ = 1 . 645 . Solving the resulting system of two equations in two unknowns for μ and σ yields μ = 71 . 3 and σ = 13 . 7 . 3.54: If f ( x ) = 1 σ 2 π e - ( x - μ ) 2 2 σ 2 is the density function of the normal distribution with parameters μ and σ then f 0 ( x ) = 1 σ 2 π e - ( x - μ ) 2 2 σ 2 ± - 2( x - μ ) 1 2 σ 2 = -
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Unformatted text preview: 1 3 2 e-( x- ) 2 2 2 ( x- ) and f 00 ( x ) =-1 3 2 e-( x- ) 2 2 2 1 + ( x- ) e-( x- ) 2 2 2 -2( x- ) 2 2 . Factoring e-( x- ) 2 2 2 from the expression above and setting the result equal to 0 yields ( x- ) 2 = 2 so that x- = or x = are the inection points. 3.60: If X N (1 , 4) then = 1 and = 2 so that P (1 &lt; X 2 &lt; 9) = P (-3 &lt; X &lt;-1) P (1 &lt; X &lt; 3) = P -3-1 2 &lt; X-1 2 &lt;-1-1 2 + P 1-1 2 &lt; X-1 2 &lt; 3-1 2 = (-1)-(-2) + (1)-(0) = 0 . 477 using the Z distribution tables in the back of the text. 1...
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