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contexamp - More Examples Continuous Random Variables...

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More Examples- Continuous Random Variables Example 1 Suppose the reaction temperature X (in ± C) in a certain chemical process has a uniform distribution with A = ¡ 5 and B = 5 . a. Compute P ( X < 0) , P ( ¡ 2 < X < 2) , and P ( ¡ 2 · X · 3) . Solution The pdf for X is: f ( x ) = ( 1 10 ¡ 5 · x · 5 0 otherwise ) : Therefore: P ( X < 0) = Z 0 ¡ 5 1 10 dx = x 10 j 0 ¡ 5 = 0 10 ¡ ¡ 5 10 = 5 10 = 0 : 5 P ( ¡ 2 < X < 2) = Z 2 ¡ 2 1 10 dx = x 10 j 2 ¡ 2 = 2 10 ¡ ¡ 2 10 = 4 10 = 0 : 4 P ( ¡ 2 · X · 3) = Z 3 ¡ 2 1 10 dx = x 10 j 3 ¡ 2 = 3 10 ¡ ¡ 2 10 = 5 10 = 0 : 5 : b. For k satisfying ¡ 5 < k < k + 4 < 5 , compute P ( k < X < k + 4) . Solution P ( k < X < k + 4) = R k +4 k 1 10 dx = x 10 j k +4 k = 1 10 [( k + 4) ¡ k ] = 4 10 = 0 : 4 : Example 2 The actual tracking weight of a stereo cartridge that is set to track at three grams on a particular changer can be regarded as a continuous random variable with pdf: f ( x ) = ( k [1 ¡ ( x ¡ 3) 2 ] 2 · x · 4 0 otherwise ) : a. Find the value of k . Solution f ( x ) to be a valid pdf, R 4 2 f ( x ) dx must equal 1. Therefore: Z 4 2 k [1 ¡ ( x ¡ 3) 2 ] dx = k Z 1 ¡ 1 [1 ¡ u 2 ] du = k " u ¡ u 3 3 # 1 ¡ 1 = 4 3 k: Hence k = 3 4 . b. Sketch the graph of f ( x ) . Solution See Maple Supplement.
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contexamp - More Examples Continuous Random Variables...

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