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ExampleLet X and Y have a bivariate normal distribution with means px = 5 and py = 6. standard deviations Gx = 3 ad Oy = 2, and covariaiice eV(xy) P= OxY = 2) Let $...

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ExampleLet X and Y have a bivariate normal distribution with means px = 5 and py = 6. standard deviations Gx = 3 ad Oy = 2, and covariaiice eV(xy) P= OxY = 2) Let $ denote the cumulative distribution function of a normal random variable with mean 0 and variance 1. What is P(2 < X ~ Y < 5) in terms of $ ? 0(2) Pc2<2) a=| b-I Covxy 2 (2 <X-Y< 5) '~0 Cewly) Cov= |2 2= 5 4= 6

Example Let X and Y have a bivariate normal distribution with means px = 5 and py = 6. standard deviations Gx = 3 ad Oy = 2, and covariaiice eV(xy) P= OxY = 2) Let $ denote the cumulative distribution function of a normal random variable with mean 0 and variance 1. What is P(2 < X ~ Y < 5) in terms of $ ? 0(2) Pc2<2) a=| b-I Covxy 2 (2 <X-Y< 5) '~0 Cewly) Cov= |2 2= 5 4= 6



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Find the mean, variance, and standard deviation of the random variable $x$ associated with the probability density function over the indicated interval. $$f(x)=\frac{5}{2} x^{3 / 2} ;[0,1]$$

29. In a normal distribution, X equals negative five and Z is equal to negative 3.14 This tells you that X equals five is 3.14 Standard deviations to the left of the mean We know it's to the left of the mean because the Z score is negative.

In a normal distribution. X equals negative too, and see equal six. This tells you that negative to ISS six standard deviations to the right of the mean We know it's to the right of the mean because Z score is positive.

Mm And this problem, random variable X is normally distributed with Mimi equals five and standard deviation sigma equals four. We wish to use a normal distribution table to find the following probabilities A through E that is we forced to find the relevant acts that satisfy these probabilities. So remember that for a standard normal distribution Aziz table maps Z scoring on the probabilities, that's probably easy. Greater than zero equals peanut implies that the area peanut purple is the area to the right, under the normal curve for score, is he? Not As an example, we have probably the greater than 0.5 because zero is the meat of the standard normal distribution, thus half of the total area is right. This example relates excellently to the to points we have here that we'll use to solve this problem, the symmetry of the normal curve and a total area under normal curve is one. So, to compute these probabilities, we need to compute them as the scores and then convert the Z score to our x variable value for x equals z, sigma plus meal. So the probability Z greater than 0.5 give zero equals zero. That's not easy. Not signal plus me equals five. Next, probably exported 6.95 gives equal 1.65. Thus X is 11.6 per. See we have a Z, not uh 0.36 or the problem is, you know, is less than zero, less than one. Where we converted our x equals nine to z. Score of one. This gives a 6.44 for D v solid follow a similar path. See we have TMZ score approaching infinity that his ex is not exist. The reason for this is that our x value of three correspondents t score negative 30.5. However there is no Xenon for which the probability that Z between negative 0.5 and 0.95 that is because 0.95 is around here notice how much area there is another standard normal that's leftover. Thus there can be no doubt that satisfies D finally any we look for negative X and X that satisfied from u minus five. Probably that's not the disease between negative zero and 9.99. This gives you an A plus or minus 2.56 So excess plus minus 2.56 times sigma equals plus or minus 10. 24. Notice how in party we did not add our meal because we subtracted mu originally


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