Answer:
Y = 0.009042 + 0.0002457X
Y = 0.1073 seconds
Step-by-step 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.
Answer:
0.34134
Step-by-step explanation:
In other to solve for this question, we would be using the z score formula
z = (x - μ) / σ
x = raw score
μ = mean
σ = Standard deviation
We are told in the question to find the probability that a worker selected at random makes between $350 and $400
let x1 = 350 and x2= 400 with the mean μ = 400 and standard deviation σ = $50.
z1 = (x1 - μ) / σ = (350-400) / 50 = -1
z2 = (x2 - μ) / σ = (400 - 400) / 50 = (0/50) = 0
From tables, P(z <= -1) = 0.15866
P(z <= 0) = 0.5
Then, the probability would give us, P(-1 ≤ z ≤ 0) =0.5 - 0.15866 =
0.34134
Hence, The probability that a worker selected at random makes between $350 and $400 = 0.34134
Answer:
Step-by-step explanation:
The best option is for the consultant to remove these data points because they are outliers. Unusual data points which are located far from rest of the data points are known as outliers.
Answer:
a) 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.
( 0.0761 , 0.3239)
b) Margin of error = 0.1264.
Step-by-step explanation:
<u>Explanation</u>:-
Given '8' persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents.
sample proportion of success 
q = 1=p
q = 1-0.2 = 0.8
a)
95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.


(0.2 - 0.1239,0.2+0.1239)
( 0.0761 , 0.3239)
b) the margin of error for a 95% confidence interval for the population proportion.
For the 95% confidence interval ∝= 0.05 and zₐ = 1.96≅2.


Margin of error for a 95% confidence interval for the population proportion.
Margin of error = 0.1264.
Answer:
Question 13: For age groups y=1 and y=1.3 response is 8 microseconds.
Question 14: The club was making a loss between 11.28 and 4.88 years.
Step-by-step explanation:
Question 13:
The age group y for which the response rate R is 8 microseconds is given by the solution of the equation

We graph this equation and find the solutions to be

Since only positive solutions for y are valid in the real world we take only those.
Thus only for age groups y=1 and y=1.3 the response is 8 microseconds.
Question 14:
The footbal club is making a loss when 
Or

We graph this inequality and find the solutions to be
and 
Since in the real world only positive values for t are valid, we take the the second solution to be true.
Thus the club was making a loss in years 