Answer:
Y = 0.009042 + 0.0002457X
Y = 0.1073 seconds
Explanation:
In the given problem we have two variables: the transmission time and average size of file.
Y = transmission time
X = average file size
The linear regression model is given by
Y = a + bX
The slope b is given by
b = correlation coefficient*(SDy/SDx)
Where SDy is the standard deviation of average transmittance time and SDx is the standard deviation of average file size.
b = 0.86(0.01/35)
b = 0.0002457
The y-intercept a is given by
a = y - bx
a = 0.04 - (0.0002457)126
a = 0.04 - 0.030958
a = 0.009042
Therefore, the linear regression model is
Y = 0.009042 + 0.0002457X
Predict the time it will take to transmit a 400 Kbyte file.
Substitute X = 400 in the regression model
Y = 0.009042 + 0.0002457(400)
Y = 0.009042 + 0.09828
Y = 0.1073 seconds
Therefore, the predicted time to transmit a 400 Kbyte file is 0.1073 seconds.