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

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern