Solution:

Given:
R = -1
Mean of x series = 4.5
Mean of y series = 5.5
Square of X series standard deviation = 5.25
Square of y series standard deviation = 5.25
N = 8
One pair observation (x=9 y=10) omitted to be included and hence to be included

To calculate the correct coefficient of correlation, we need to follow the below steps:

Step 1: Find the sum of x and y variables
- Sum of x variables = Mean of x series * N
= 4.5 * 8
= 36
- Sum of y variables = Mean of y series * N
= 5.5 * 8
= 44

Step 2: Find the sum of squares of x and y variables
- Sum of squares of x variables = (Square of X series standard deviation + Square of mean of x series) * (N-1)
= (5.25 + 4.5^2) * 7
= 212.25
- Sum of squares of y variables = (Square of y series standard deviation + Square of mean of y series) * (N-1)
= (5.25 + 5.5^2) * 7
= 266.25

Step 3: Find the sum of products of x and y variables (before the one pair observation is included)
- Sum of products of x and y variables = R * square root of (Square of X series standard deviation * Square of y series standard deviation) * (N-1)
= -1 * sqrt(5.25 * 5.25) * 7
= -18.375

Step 4: Include the omitted pair observation (x=9 y=10) and find the new sum of x and y variables, sum of squares of x and y variables, and sum of products of x and y variables
- New sum of x variables = 36 + 9
= 45
- New sum of y variables = 44 + 10
= 54
- New sum of squares of x variables = 212.25 + 9^2 + 2*9*(4.5*8) + (4.5^2)*(8-1)
= 356.25
- New sum of squares of y variables = 266.25 + 10^2 + 2*10*(5.5*8) + (5.5^2)*(8-1)
= 490.25
- New sum of products of x and y variables = -18.375 + 9*10
= 71.625

Step 5: Calculate the corrected coefficient of correlation (r)
- r = (N*sum of products of x and y variables - sum of x variables * sum of y variables) / sqrt((N*sum of squares of x variables - (sum of x variables)^2) * (N*sum of squares of y variables - (sum of y variables)^2))
= (8*71.625 - 45*54) / sqrt((8*356.25 - 45^2) * (8*490.25 - 54^2