Respuesta :
The question is incomplete. The complete question is :
Suppose X and Y have a bivariate normal distribution with [tex]$\sigma_X = 0.04, \sigma_Y = 0.08, \mu_X = 3.00, \mu_Y = 7.70$[/tex] and ρ = 0.
Determine the following. Round your answers to three decimal places (e.g. 98.765).
(a). [tex]$P(2.90 < X < 3.10) =$[/tex]
(b). [tex]$P(7.60 < X < 7.80) =$[/tex]
(c). P(2.90 < X < 3.10, 9.60 < Y < 7.80) =
Solution :
We have
[tex]$X \sim N (\mu_X=3, \sigma_X^2=0.04^2)$[/tex] and
[tex]$Y \sim N (\mu_Y=7.7, \sigma_Y^2=0.08^2)$[/tex]
a). The required probability is
[tex]$P(2.90 < X < 3.10) =$[/tex] [tex]$P\left(\frac{2.9-3}{0.04}< Z < \frac{3.1-3}{0.04}\right)$[/tex]
= P(-2.5 < Z < 2.5) = P(Z < 2.5) - P(Z < -2.5)
= P(Z < 2.5) - [1-P(Z < 2.5)]
= 2P(Z < 2.5) - 1
= 2(0.99379) - 1
= 0.98758
≈ 0.988
b). The required probability is
[tex]$P(7.6 < X < 7.8) =$[/tex] [tex]$P\left(\frac{7.6-7.7}{0.08}< Z < \frac{7.8-7.7}{0.08}\right)$[/tex]
= P(-1.25 < Z < 1.25) = P(Z < 1.25) - P(Z < -1.25)
= P(Z < 1.25) - [1-P(Z < 1.25)]
= 2P(Z < 1.25) - 1
= 2(0.89435) - 1
= 0.7887
≈ 0.789
c). Since, ρ = 0, so the X and Y are independent. Thus the required probability is :
P(2.90 < X < 3.10, 9.60 < Y < 7.80) = P(2.9 < X < 3.1) x P(7.6 < Y < 7.8)
= (0.98758)(0.7887)
= 0.77890
≈ 0.779
