(Solved): The Elo Chess Rating Systern is a method for predicting scores from a chess match (in this case un ...

The Elo Chess Rating Systern is a method for predicting scores from a chess match (in this case under a certain scenario). The probability of getting a certain score (for any real number \( x \) ) is \[ f(x)=\frac{\mathrm{e}^{-(x-1400), d 00}}{1+\mathrm{e}^{-(x-1400) y 900}} \] The cumulative distribution function (for any real number \( x \) ) is \[ F(x)=\frac{1}{1+e^{-(x-14 D 0), 900}} \] What is the probability of getting a score of 879 ? Use 4 decimal places What is the probability of getting a score less than 879 ? Use 4 decimal places What is the probability of getting a score greater than \( 1468 ? \) Use 4 decimal planes What is the probability of getting a score between 879 and \( 1468 ? \) Use 4 decimal planes Which equation calculates the variance of the scores? \( \int_{-\infty}^{\infty} x^{2} \frac{\mathrm{e}^{-(x-1400) / 900}}{1+\mathrm{e}^{-(x-1400), 300}} \mathrm{dx}-\left(\int_{-\mathrm{e}}^{\infty} x \frac{\mathrm{e}^{-(x-1400) / 900}}{1+\mathrm{e}^{-(x-1400), 900}} d x\right)^{2} \) \( \int_{-\infty}^{=} x^{2} \frac{e^{-(x-1400) / 900}}{1+e^{-i x-1400) / 900}} d x \) \( \left(\int_{-\infty}^{\infty} x \frac{e^{-(x-1400)} d 000}{1+e^{-(x-1400)} / 900} d x\right)^{2} \) \( \int_{-\infty}^{\infty} x \frac{e^{-(x-1400) / 900}}{1+e^{-(x-1400) / 900}} d x-\left(\int_{-\infty}^{\infty} x^{2} \frac{e^{-(x-1400)} / 900}{1+e^{-(x-1400) / 900}} d x\right)^{2} \) \( \left(\int_{-\infty}^{\infty} \frac{e^{-(x-1400) / 900}}{1+e^{-(x-1400) ; 900}} d x\right)^{2}-\int_{-\infty}^{\infty} \frac{e^{-i x-1400) / 900}}{1+e^{-(x-1400) ; 900}} d x \) \( \left(\int_{-\infty}^{\infty} x^{2} \frac{e^{-(x-1400) / 300}}{1+e^{-i x-1400 y) / 900}} d x\right)^{2}-\int_{-=}^{\infty} x \frac{e^{-(x-1400 ; / 900}}{1+e^{-(x-1400) / 900}} d x \) \( \left(\int_{-\pi}^{\infty} x \frac{\mathrm{e}^{-(x-1400) / 9000}}{1+e^{-(x-1400) / 900}} d x\right)^{2}-\int_{-\infty}^{\infty} x^{2} \frac{\mathrm{e}^{-(x-1400) j 900}}{1+\mathrm{e}^{-(x-1400) / 900}} d x \) \( \int_{-\infty}^{\infty} x \frac{1}{1+e^{-(x-1400) / 900}} d x-\left(\int_{-\infty}^{\pi} x^{2} \frac{1}{1+e^{-(x-1400), 900}} d x\right)^{2} \)