So 1 hk1 is the probability the lightbulb survives

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Unformatted text preview: X (u) 0 .5 Therefore, Var(X ) = E [X 2 ] − µ2 = E [X 2 ], so we find X −0.5 9 = − 32 . ) ∞ u2 fX (u)du E [X 2 ] = −∞ 1 3 u2 (1 − u2 )du 4 −1 = Thus, Var(X ) = 0.2 and σX = 3.3 1 3 4 = Var(X ) = (u2 − u4 )du = 0.2. −1 √ 0.2 ≈ 0.45. Uniform distribution Let a < b. A random variable X is uniformly distributed over the interval [a, b] if 1 b−a fX (u) = 0 a≤u≤b else. The mean of X is given by 1 b−a E [X ] = b udu = a u2 2(b − a) b = a b2 − a2 (b − a)(b + a) a+b = = . 2(b − a) 2(b − a) 2 Thus, the mean is the midpoint of the interval. The second moment of X is given by b 1 E [X ] = b−a u3 u du = 3(b − a) b 2 2 a = a b3 − a3 (b − a)(b2 + ab + a2 ) a2 + ab + b2 = = . 3(b − a) 3(b − a) 3 So Var(X ) = E [X 2 ] − E [X ]2 = a2 + ab + b2 − 3 a+b 2 2 = (a − b)2 . 12 Note that the variance is proportional to the length of the interval squared. A useful special case is when a = 0 and b = 1, in which case X is uniformly distributed over the unit interval [0, 1]. In that case, for any k...
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