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 continuoustype 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. CONTINUOUSTYPE 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...
View
Full
Document
This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
 Spring '08
 Zahrn
 The Land

Click to edit the document details