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## Homework Statement

__Problem:__

In an acid-base titration, a base or acid is gradually added to the other until they have completely neutralized each other.

Let X and Y denote the milliliters of acid and base needed for equivalence, respectively.

Assume X and Y have a bivariate normal distribution with ##σ_X## = 5mL, ##σ_Y## = 2mL , ##μ_X## = 120mL , ##μ_Y## = 100mL , and ρ = 0.6.

Determine the following.:

a) Covariance of X and Y

b) Marginal probability distribution of X

c) P(X < 116)

d) Conditional probability distribution of X given that Y = 102

e) P(X < 116 | Y = 102)

__Solutions given by solutions manual:__

a) ##ρ = cov(X,Y)/[σ_x σ_y] =0.6 cov(X,Y)= 0.6*2*5=6##

b) The marginal probability distribution of X is normal with mean ##μ_x##, ##σ_x##.

c) ##P(X < 116) =P(X-120 < -4)=P((X_120)/5 < -0.8)=P(Z < -0.8) = 0.21##

d) The conditional probability distribution of X given Y = 102 is a bivariate normal distribution with mean and variance

μ_{X | Y = 102} = 120 - 100*0.6*(5/2) + (5/2)*0.6*102 = 123

σ^2_{X | Y = 102} = 25(1 - 0.36) = 16

e) ##P(X < 116|Y=102)=P(Z < (116-123)/4)=0.040##

## Homework Equations

##μ_{X | Y = y} = μ_x + ρ σ_X/σ_Y (y – μ_y)##

##σ^2_{Y | X = x} = σ_Y^2 (1 – ρ^2)##

Bivariate normal distribution:

##f_{XY}(x,y; σ_X, σ_Y, μ_x, μ_y, ρ) = 1/[2πσ_Xσ_Y√(1 – ρ^2)] × exp( –1/[2(1 – ρ^2)] × ((x – μ_X)/σ_X^2 – 2ρ(x – μ_X)(y - μ_Y)/[σ_Xσ_Y] + (y – μ_Y)^2/σ_Y^2) )## for –∞ < x < ∞ and –∞ < x < ∞, with parameters σ_X > 0, σ_Y > 0, –∞ < μ_X < ∞, –∞ < μ_Y < ∞ and –1 < ρ < 1.

Univariate normal distribution:

##f(x) = 1/√(2πσ) e^{–(x – μ)^2 / [2σ^2]}## for –∞ < x < ∞

## The Attempt at a Solution

There are a few things I want to address with this post. Here they are.:

1) Would the marginal probability distribution of part b) be given by ##∫^∞_∞ 1/√(2π

**σ_X**) e^{–(x –

**μ_X**)^2 / [2

**σ_X**^2]} dx## ?

2) Would this be calculable using the integral from my question “1)” above?

3) Would the conditional probability distribution of X given that Y = 102 be given by ##∫^∞_∞ 1/√(2π

**σ_{X | Y = 102}**) e^{–(x –

**μ_{X | Y = 102}**)^2 / [2

**σ_{X | Y = 102}**^2]} dx## ?

4) Would this be calculable using the integral from my question “3)” above?

Any input would be GREATLY appreciated!