STAT 400 Lecture AL1 Answers for 3.1 Spring 2015 Dalpiaz 1. Let X be a continuous random variable with the probability density function f ( x ) = k x , 0 x 4, f ( x ) = 0, otherwise. a) What must the value of k be so that f ( x ) is a probability density function? 1) f (x) 0, 2) 1d xxf . 402140ddd1 xxkxx k xxf 316043223kxk . k = 163 = 0.1875. b) Find the cumulative distribution function of X, F X ( x ) = P ( X x ). F X ( x ) = P ( X x ) = x dyyf . x 0 F X ( x ) = 0. 0 x 4 F X ( x ) = 2323081032163163xxyyy x d . x 4 F X ( x ) = 1.c) Find the probability P ( 1 X 2 ). P ( 1 X 2 ) = 21dxxf = 1232163d1632321 xxx 0.22855. OR P ( 1 X 2 ) = F X ( 2 ) – F X ( 1- ) = 2323181281 0.22855. d) Find the median of the distribution of X. That is, find m such that P ( X m ) = P ( X m ) = 1/2. 2323081032163d163d21 m mx xx xxf m m . 234 m . m = 324 2.51984. e) Find the 30th percentile of the distribution of X. That is, find a such that P ( X a ) = 0.30. 2323081032163163300 aax xx xxf a d a d . . 2342 a . . a = 324.2 1.79256.f) Find X = E ( X ). E ( X ) = 402340d163d163d)(Xxxxxxxxfx . = 045216325 x = 2.4. g) Find X = SD ( X ). Var ( X ) = 240252224.2d163Xd)(X xxxxfx = 2274.20472163 x = 1.09714. X = SD ( X ) = 09714.1Var X = 1.04745. 2. Let X be a continuous random variable with the cumulative distribution function F ( x ) = 0, x < 0, F ( x ) = 83 x, 0 x 2, F ( x ) = 1 – 2 1x, x > 2. a) Find the probability density function f ( x ). f ( x ) = F' ( x ).x < 0 f ( x ) = F' ( x ) = 0, 0 x 2 f ( x ) = F' ( x ) = 83, x > 2 f ( x ) = F' ( x ) = 3 2x. b) Find the probability P ( 1 X 4 ). P ( 1 X 4 ) = F ( 4 ) – F ( 1- ) = 83 1615 = 169 = 0.5625. OR P ( 1 X 4 ) = 41xxf d = 42321 2 83xxx d d = 0.5625. c) Find X = E ( X ). E ( X ) = 2320 283)(Xμ xxxxxxxfx d d d . = 212 022832 xx = 1 43 = 47 = 1.75. d) Find X = SD ( X ). Var ( X ) = 222X)(Xμσ d xxfx . However, 23220222 83 )( xxxxxxxfx d d d diverges. Therefore, Var ( X ) and SD ( X ) are not
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