Consider the following set of data.

(18, 12), (29, 48), (65, 29), (82, 24), (115, 56), (124, 13)

(a) Calculate the covariance of the set of data. (Give your answer correct to two decimal places.)
.

(b) Calculate the standard deviation of the six x-values and the standard deviation of the six y-values. (Give your answers correct to three decimal places.)
sx =
sy =

(c) Calculate r, the coefficient of linear correlation, for the data in part (a). (Give your answer correct to two decimal places.)

To calculate the covariance of a set of data, you can use the following formula:

cov(X, Y) = Σ((x - μx)(y - μy)) / (n - 1)

Where:
- X and Y are the variables (x and y values in this case).
- Σ denotes the sum of the values.
- x and y are the individual values.
- μx and μy are the means (averages) of the x and y values.
- n is the total number of data points.

Therefore, to calculate the covariance of the given data:

Step 1: Calculate the means (μx and μy):
To calculate the mean of x values (μx):
μx = (18 + 29 + 65 + 82 + 115 + 124) / 6 = 53.83 (rounded to two decimal places)

To calculate the mean of y values (μy):
μy = (12 + 48 + 29 + 24 + 56 + 13) / 6 = 31.33 (rounded to two decimal places)

Step 2: Calculate the covariance using the formula:
cov(X, Y) = Σ((x - μx)(y - μy)) / (n - 1)
= ((18 - 53.83)(12 - 31.33) + (29 - 53.83)(48 - 31.33) + (65 - 53.83)(29 - 31.33)
+ (82 - 53.83)(24 - 31.33) + (115 - 53.83)(56 - 31.33) + (124 - 53.83)(13 - 31.33)) / (6 - 1)

= (-35.83 * -19.33 + -24.83 * 16.67 + 11.17 * -2.33 + 28.17 * -7.33 + 61.17 * 24.67 + 70.17 * -18.33) / 5

= 77.30 (rounded to two decimal places)

Therefore, the covariance of the given set of data is 77.30.

To calculate the standard deviation of the x-values and y-values, you can use the following formula:

Standard deviation (s) = √(Σ(x - μ)² / (n - 1))

Where:
- Σ denotes the sum of the values.
- x is the individual value.
- μ is the mean (average) of the values.
- n is the total number of data points.

Step 1: Calculate the standard deviation of the x-values (sx):
sx = √(Σ(x - μx)² / (n - 1))
= √(((18 - 53.83)² + (29 - 53.83)² + (65 - 53.83)² + (82 - 53.83)² + (115 - 53.83)² + (124 - 53.83)²) / (6 - 1))

= √(((-35.83)² + (-24.83)² + (11.17)² + (28.17)² + (61.17)² + (70.17)²) / 5)

= 44.431 (rounded to three decimal places)

Step 2: Calculate the standard deviation of the y-values (sy):
sy = √(Σ(y - μy)² / (n - 1))
= √(((12 - 31.33)² + (48 - 31.33)² + (29 - 31.33)² + (24 - 31.33)² + (56 - 31.33)² + (13 - 31.33)²) / (6 - 1))

= √(((-19.33)² + (16.67)² + (-2.33)² + (-7.33)² + (24.67)² + (-18.33)²) / 5)

= 16.340 (rounded to three decimal places)

Therefore, the standard deviation of the x-values (sx) is 44.431 and the standard deviation of the y-values (sy) is 16.340.

To calculate the coefficient of linear correlation (r), you can use the following formula:

r = cov(X, Y) / (sx * sy)

Where:
- cov(X, Y) is the covariance of the x and y values.
- sx is the standard deviation of the x-values.
- sy is the standard deviation of the y-values.

Step 1: Calculate r using the provided values:
r = 77.30 / (44.431 * 16.340) = 0.681 (rounded to two decimal places)

Therefore, the coefficient of linear correlation (r) for the given data is 0.681.