In: Statistics and Probability
Find the following probabilities for two normal random variablesZ=N(0,1) andX=N(−1,9).
(a)P(Z >−1.48).
(b)P(|X|<2.30).
(c) What is the type and the parameters of the random variableY= 3X+ 5?
(a) P(Z > - 1.48) = 1 - P(Z
- 1.48) = 1 -
(-1.48) = 1 - 0.0694 = 0.9306
[(.)
is the cdf of N(0,1)]
(b) X ~ N(-1, 9) i.e. (X + 1)/3 ~ N(0,1)
P(|X|
2.30) = P-2.30
X
2.30) = P[(-2.30 + 1)/3
(X + 1)/3
(2.30 + 1)/3] = P[-0.4333
(X + 1)/3
1.1] = P[(X + 1)/3
1.1] - P[(X + 1)/3
- 0.4333] =
(1.1) -
(-0.4333) = 0.8643 - 0.3324 = 0.5319
(c) We have, E(Y) = 3 * E(X) + 5 = -3 + 5 = 2
V(Y) =
* V(X) = 9 * V(X) = 9 * 9 = 81
Since, Y is a linear function of X, Y ~ N(2, 81)