5 it thus has mean x np 500 and standard deviation np1

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Unformatted text preview: variable, and use X − 10 15 − 10 15 − 10 ≥ =Q = Q(1.25) = 1 − Φ(1.25) = 0.1056. 4 4 4 X − 10 5 − 10 5 − 10 P {X ≤ 5} = P ≤ =Φ = Q(1.25) = 0.1056. 4 4 4 X − 10 X − 10 ≥ 2.5 + P ≤ −7.5 P {X 2 ≥ 400} = P {X ≥ 20} + P {X ≤ −20} = P 4 4 = Q(2.5) + Q(7.5) ≈ Q(2.5) = 1 − Φ(2.5) = 0.0062. (Note: Q(7.5) < 10−12 .) P {X ≥ 15} = P Finally, P {X = 2} = 0, because X is a continuous-type random variable. Example 3.6.5 Suppose X is a random variable with mean 10 and variance 3. Find the numerical √ value of P {X < 10 − 3} (or, nearly equivalently, P {X < 8.27}) for the following two choices of distribution type: (a) Assuming X is a Gaussian random variable, (b) Assuming X is a uniform random variable. Solution: (a) If X is N (10, 3), P {X < 10 − √ 3} = P X − 10 √ ≤ −1 3 = Φ(−1) = 1 − Φ(1) = Q(1) ≈ 0.1587. 92 CHAPTER 3. CONTINUOUS-TYPE RANDOM VARIABLES a+b (b − a)2 and variance . Hence, 2 12 a + b = 20 and b − a = 6, giving a = 7, b = 13. That is, X is u...
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