Solution:
Given, coefficient of correlation between x and y is 0.46

We need to find the coefficient of correlation between x and y/2

To solve this problem, we can use the formula for correlation coefficient:

r = (nΣxy - ΣxΣy) / sqrt[(nΣx^2 - (Σx)^2) (nΣy^2 - (Σy)^2)]

where n is the number of data points, Σxy is the sum of the products of x and y, Σx and Σy are the sums of x and y respectively, and Σx^2 and Σy^2 are the sums of the squares of x and y respectively.

Let us assume that we have n data points for x and y. Then, we can write:

- Σx = sum of all x values
- Σy = sum of all y values
- Σxy = sum of all (x * y) values
- Σx^2 = sum of all (x^2) values
- Σy^2 = sum of all (y^2) values

We know that y/2 is half of y, so we can write:

y/2 = 0.5 * y

Therefore, we can replace y with 0.5y in the formula for correlation coefficient:

r' = (nΣx(0.5y) - ΣxΣ(0.5y)) / sqrt[(nΣx^2 - (Σx)^2) (nΣ(0.5y)^2 - (Σ(0.5y))^2)]

Simplifying this expression, we get:

r' = (0.5nΣxy - 0.5ΣxΣy) / sqrt[(nΣx^2 - (Σx)^2) (n(0.25)Σy^2 - (0.5Σy)^2)]

r' = (0.5nΣxy - 0.5ΣxΣy) / sqrt[(nΣx^2 - (Σx)^2) (0.25nΣy^2 - 0.25(Σy)^2)]

r' = (0.5nΣxy - 0.5ΣxΣy) / sqrt[(nΣx^2 - (Σx)^2) (0.25nΣy^2 - (0.25Σy)^2)]

r' = 0.5(nΣxy - ΣxΣy/2) / sqrt[(nΣx^2 - (Σx)^2) (0.25nΣy^2 - (0.25Σy)^2)]

r' = 0.5(2r) (since coefficient of correlation between x and y is 0.46)

r' = 0.92

Therefore, the coefficient of correlation between x and y/2 is 0.92.

Conclusion:
The coefficient of correlation between x and y/2 is 0.92. This means that there is a strong positive correlation between x and y/2.

0.46