Probability Law and Variance Calculation

Extract

[...] Given that 16 > the variations remain unchanged. [...]

[...] Let = x and = (x+7)² = x²+14x+49 So = 1 and = 2x + 14 So we have = So it = Study of the sign of for all x of > 0 so the sign of is that of -x²+49 Or, -x² + 49 = 0 is equivalent to x² = 49 is equivalent to x = 7 or x = -7. We keep the positive solution because we are on R+. For x > we have -x² +49 0 We deduce the following variation table: X = 7/196 f Question 5 The function f is similar to the expectation of the variable to a factor of 16. [...]

[...] Determine the probability law of X1. X1 corresponds therefore to the variable for the following configuration: - 3 red sectors - 4 white sectors - 1 green sector - On the first spin, the wheel has 3 chances out of 8 to stop on the red sector, corresponding to a gain of 16 So P(X = 16) = - On the first spin, the wheel has 4 chances out of 8 to stop on the white sector, corresponding to a loss of 12 So P(X = = = - On the first spin, the wheel has 1 chance out of 8 to stop on the green sector, in which case a second spin takes place: on the second spin, the wheel has 3 chances out of 8 to stop on the red out of 8 on the white 2 and 1 out of 8 on the green So P(X = = x = Therefore, we deduce the following probability law: [...]

[...] Two out of the three probabilities are unknown, which does not allow us to find the variance immediately. = 0 is equivalent to 0 = P(X = x + P(X = x 0 + P(X = x 1 is equivalent to 0 = P(X = x + is equivalent to 2 x P(X = = is equivalent to P(X = = This value then allows us to calculate the variance of the variable, defined by the following formula: Knowing that = we deduce that P(X = x - E(X))² = 0 And = x - 0)² + x - 0)² Question 2 The probability law of X is determined by knowing all the probabilities of the values it can take. [...]