121616949-math.230

# 121616949-math.230 - variable X and that the mean μ...

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216 Chapter 9 Applications of Integration center of mass is ¯ x = Z b a xf ( x ) dx Z b a f ( x ) dx . If we extend the beam to infinity, we get ¯ x = Z -∞ xf ( x ) dx Z -∞ f ( x ) dx = Z -∞ xf ( x ) dx, because R -∞ f ( x ) dx = 1. In the center of mass interpretation, this integral is the total mass of the beam, which is always 1 when f is a probability density function. EXAMPLE 9.35 The mean of the standard normal distribution is Z -∞ x e - x 2 / 2 2 π dx. We compute the two halves: Z 0 -∞ x e - x 2 / 2 2 π dx = lim D →-∞ - e - x 2 / 2 2 π fl fl fl fl fl 0 D = - 1 2 π and Z 0 x e - x 2 / 2 2 π dx = lim D →∞ - e - x 2 / 2 2 π fl fl fl fl fl D 0 = 1 2 π . The sum of these is 0, which is the mean. Suppose that f : R R is the probability density function for the continuous random
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Unformatted text preview: variable X , and that the mean μ , exists (and is ﬁnite). We would like to measure how far a “typical” value of X is from μ . One way to measure this distance is ( X-μ ) 2 ; we square the diﬀerence so as to measure all distances as positive. The expected value of this quantity is V ( X ) = Z ∞-∞ ( x-μ ) 2 f ( x ) dx. This quantity is called the variance, and is the expected value of the squared distance to μ . The standard deviation , denoted σ , is the square root of the variance. By taking...
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