Given: 2x + 3y + 4 = 0 and SD of x is 6

To find: SD of y

Approach:

We know that the formula for SD is given by:
SD = √(Σ(xi-μ)²/N)

where,
xi = individual value of the variable
μ = mean of the variable
N = number of observations

We can use the above formula to find the SD of y.

Solution:

1. Finding the mean of x:

2x + 3y + 4 = 0
2x = -3y - 4
x = (-3/2)y - 2

The above equation represents a straight line.
For a straight line, the mean is given by the midpoint of the line segment joining the extremes.

The extremes of the line are (-4/3, 0) and (0, -4/3).
So, the midpoint is ((-4/3 + 0)/2, (0 - 4/3)/2) = (-2/3, -2/3).

Therefore, the mean of x is -2/3.

2. Finding the mean of y:

2x + 3y + 4 = 0
3y = -2x - 4
y = (-2/3)x - (4/3)

The above equation represents a straight line.
For a straight line, the mean is given by the midpoint of the line segment joining the extremes.

The extremes of the line are (-3, 2) and (0, -4/3).
So, the midpoint is ((-3 + 0)/2, (2 - 4/3)/2) = (-3/2, 5/6).

Therefore, the mean of y is 5/6.

3. Finding the SD of y:

We are given that the SD of x is 6.
So, we can use the formula for covariance to find the covariance between x and y:

cov(x, y) = Σ((xi - μx)(yi - μy))/N

where,
xi = individual value of x
μx = mean of x
yi = individual value of y
μy = mean of y
N = number of observations

We can rearrange the given equation 2x + 3y + 4 = 0 to get y in terms of x:

y = (-2/3)x - (4/3)

Using this equation, we can calculate the covariance between x and y:

cov(x, y) = Σ((x - (-2/3))(y - 5/6))/N
= Σ((3x + 2)(-2x/3 - 1/6))/N
= -2/3 Σ(x²) - 1/3 Σ(x) + 5/9 N/2
= -2/3 (var(x) + mean(x)²) - 1/3 (-2/3)N + 5/18 N
= -2/3 (6² + (-2/3)²) + 2/9 N
= -248/9

where,
var(x) = variance of x
mean(x) = mean of x

We know that the formula for the covariance between x and