Question

Given a random sample of four (x, y) pairs of data points:
(3.1,9) (3.5,13) (5,15) (4.6,11)
What is the covariance?
Please give your answer to four decimal places.
1 point

Answer 1

Answer:

don't know

Explanation:

sorry................

Answer 2

To find the covariance of the following data:

(3.1,9) (3.5,13) (5,15) (4.6,11)

Step-by-Step Explanation:

Formula of covariance =    ∑(x-u)(y-v)

                                            ----------------

                                                  N-1

The given number of data (N) = 4

X = 3.1, 3.5, 5, 4.6

Y = 9, 13, 15, 11

MEAN OF X is u

u= ∑x / n

 =3.1 + 3.5 + 5 + 4.6 = 16.2/4 = 4.05

Mean of Y is v

v = ∑y / n

  = 9 + 13 +15 +11 = 48/4 = 12

(x-u)(y-v) =  (3.1-4.05)(9-12) + (3.5-4.05)(13-12) + (5-4.05)(15-12) +

                   (4.6-4.05) (11-12)

               = (2.85) + (-0.55) + (2.85) + (0.55) = 5.7

∑(x-u)(y-v)                  5.7            5.7  

--------------   =         -----------  =   -------

    N-1                      (4-1)              3

                       

                           =  1.9

The covariance is 1.9