Solution:

Given,
cov(x,y) = 8/20 = 0.4
variance of x, σ²(x) = 16
sum of squares of deviation of y from its mean, ∑(y-ȳ)² = 250

We know that,
cov(x,y) = E[(x-μ(x))(y-μ(y))] = E[xy] - μ(x)μ(y)

where,
E[xy] = covariance of x and y
μ(x) = mean of x
μ(y) = mean of y

Calculating the mean of x
μ(x) = E[x] = 0 [since, we don't have any information about the mean of x]

Calculating the mean of y
cov(x,y) = E[xy] - μ(x)μ(y)
0.4 = E[xy] - 0*μ(y)
E[xy] = 0.4

Also, we know that,
variance of x, σ²(x) = E[(x-μ(x))²]

Substituting the value of μ(x) and σ²(x), we get
16 = E[x²] - 0²
E[x²] = 16

Calculating the number of observations
∑(y-ȳ)² = ∑y² - n(ȳ)²

where,
∑y² = sum of squares of y
n = number of observations
ȳ = mean of y = E[y]

Substituting the given values, we get
250 = ∑y² - n(ȳ)²

We don't have any information about ∑y² and ȳ.
So, we need one more equation to solve for n.

Using the formula,
variance of y, σ²(y) = E[(y-μ(y))²]

σ²(y) = ∑[(y-ȳ)²]/n [definition of variance]

Substituting the given values, we get
σ²(y) = ∑[(y-ȳ)²]/n = cov(x,y)/σ²(x) = 0.4/16 = 0.025

Calculating the number of observations (continued)
∑[(y-ȳ)²]/n = 0.025
∑[(y-ȳ)²] = 0.025n

Substituting this value in the previous equation, we get
250 = ∑y² - n(ȳ)²
250 = ∑y² - (0.025n)ȳ²

We still need one more equation to solve for n.
Let's use the formula,
covariance of x and y, cov(x,y) = E[xy] - μ(x)μ(y)

cov(x,y) = 0.4
μ(x) = 0
μ(y) = E[y]

Substituting the given values, we get
0.4 = E[xy]

We know that,
E[xy] = ∑xy/n

Let's solve for ∑xy using the above equation,
∑xy = E[xy] * n = 0.4 * n

Using this value in the formula,
∑y² = ∑[(y