Concept explainers
Referring to Exercise 7, suppose that the standard deviation of the random deviation e is 350 psi.
- a. What is the
probability that the observed value of 28-day strength will exceed 5000 psi when the value of accelerated strength is 2000? - b. Repeat part (a) with 2500 in place of 2000.
- c. Consider making two independent observations on 28-day strength, the first for an accelerated strength of 2000 and the second for x = 2500. What is the probability that the second observation will exceed the first by more than 1000 psi?
- d. Let Y1, and Y2 denote observations on 28-day strength when x = x1 and .x = x2, respectively. By how much would x2 have to exceed x1, in order that P(Y2 > Y1) = .95?
a.
Find the probability that the observed value of 28-day strength will exceed 5000 psi for 2000 psi accelerated strength.
Answer to Problem 8E
The probability that the observed value of 28-day strength will exceed 5000 psi for 2000 psi accelerated strength is 0.0436.
Explanation of Solution
Given info:
The regression line corresponding to the variables 28-day strength
Calculation:
Linear equation:
The general form of linear equation with one independent random variable is given as,
Where
Assume x be the predictor variable and y be the response variable, of a regression analysis. Then, the predicted or expected value of the response variable is
Here, the regression equation is
To obtain the probability that the observed value of 28-day strength will exceed 5000 psi when the accelerated strength is 2000 psi, the expected mean and standard deviation of the variable 28-day strength is necessary to apply normal approximation for the random variable 28-day strength.
Expected mean:
The expected mean of 28-day strength for 2000 accelerated strength is obtained as follows:
Thus, the expected mean of 28-day strength for 2000 accelerated strength is 4400.
Standard deviation:
The standard deviation of the variable 28-day strength is,
Desired probability value:
The probability that the observed value of 28-day strength will exceed 5000 psi is obtained as follows:
From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to 1.71 area to the right is 0.9564.
Therefore, the probability value is,
Thus, the probability that the observed value of 28-day strength will exceed 5000 psi for 2000 psi accelerated strength is 0.0436.
b.
Find the probability that the observed value of 28-day strength will exceed 5000 psi for 2500 psi accelerated strength.
Answer to Problem 8E
The probability that the observed value of 28-day strength will exceed 5000 psi for 2500 psi accelerated strength is 0.5557.
Explanation of Solution
Calculation:
Here, the regression equation is
To obtain the probability that the observed value of 28-day strength will exceed 5000 psi when the accelerated strength is 2500 psi, the expected mean and standard deviation of the variable 28-day strength is necessary to apply normal approximation for the random variable 28-day strength.
Expected mean:
The expected mean of 28-day strength for 2500 accelerated strength is obtained as follows:
Thus, the expected mean of 28-day strength for 2000 accelerated strength is 5050.
Standard deviation:
The standard deviation of the variable 28-day strength is,
Desired probability value:
The probability that the observed value of 28-day strength will exceed 5000 psi is obtained as follows:
From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to 0.14 area to the right is 0.5557.
Therefore, the probability value is,
Thus, the probability that the observed value of 28-day strength will exceed 5000 psi for 2500 psi accelerated strength is 0.5557.
c.
Find the probability that the second observation of 28-day strength will exceed the first observation by more than 1000 psi.
Answer to Problem 8E
The probability that the second observation of 28-day strength will exceed the first observation by more than 1000 psi is 0.2389.
Explanation of Solution
Given info:
Two independent observations are made on the variable 28-day strength. The first observation of 28-day strength is with 2000 psi accelerated strength and the second observation of 28-day strength is with 2500 psi accelerated strength.
Calculation:
Here, the regression equation is
To obtain the probability that, the second observation of 28-day strength will exceed the first observation by more than 1000 psi, the expected mean and standard deviation of the of difference between the observations of the variable 28-day strength is necessary to apply normal approximation for the random difference of two observations of the variable 28-day strength.
Let
Therefore,
From part (b), the expected mean of
Expected mean of
The expected mean of
Thus, the expected mean of
Here,
From the properties of covariance, the covariance between two independent variables is “0”.
Standard deviation:
The variance of
The general formula for standard deviation is,
Thus, the standard deviation of
Desired probability value:
The probability that the second observation of 28-day strength will exceed the first observation by more than 1000 psi is obtained as follows:
From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to 0.71 area to the right is 0.7611.
Therefore, the probability value is,
Thus, the probability that the second observation of 28-day strength will exceed the first observation by more than 1000 psi is 0.2389.
d.
Find the value of
Answer to Problem 8E
In case of
Explanation of Solution
Calculation:
Here, the regression equation is
Let
Therefore,
Expected mean of
The expected mean of
Thus, the expected mean of
Here,
From part (c), the standard deviation of
Desired probability value:
The probability that the second observation of 28-day strength will exceed the first observation is,
The probability that the second observation exceeds the first observation is 0
.95.
The value of
From Table A.3 of the standard normal distribution in Appendix , the critical value corresponding to left side of probability value 0.95 to the right is -1.645.
Therefore, the
Thus, the
Want to see more full solutions like this?
Chapter 12 Solutions
Probability and Statistics for Engineering and the Sciences
- Calculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,