Chapter 12.1, Problem 8E

Referring to Exercise 7, suppose that the standard deviation of the random deviation e is 350 psi.

1. 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?
2. b. Repeat part (a) with 2500 in place of 2000.
3. 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?
4. 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?

To determine

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 $\left(y\right)$ and accelerated strength $\left(x\right)$ is $y=1800+1.3x$. The unit of measurement for both the variables is (psi) and the standard deviation of the variable 28-day strength is 350 psi.

Calculation:

Linear equation:

The general form of linear equation with one independent random variable is given as,

$y={\beta }_0+{\beta }_{1}x$

Where ${\beta }_0$ and ${\beta }_1$ are constants, x is the independent variable and y is the dependent variable. Here ${\beta }_0$ be the y-intercept of the linear equation and ${\beta }_1$ be the slope of the linear equation.

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 ${\widehat{y}}_E={\beta }_0+{\beta }_1{x}_E$, where $x_E$ be particular value of predictor variable x.

Here, the regression equation is $y=1800+1.3x$. Where y represents the variable 28-day strength and x represents the variable accelerated strength.

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:

$\begin{array}{c}{\mu }_y=1800+1.3x\\ =1800+1.3\times 2000\\ =4400\end{array}$

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,

${\sigma }_y=350\text{psi}$.

Desired probability value:

The probability that the observed value of 28-day strength will exceed 5000 psi is obtained as follows:

$\begin{array}{c}P\left(Y>5000\right)=P\left(\frac{Y-4400}{350}>\frac{5000-4400}{350}\right)\\ =P\left(\frac{Y-{\mu }_y}{{\sigma }_y}>\frac{5000-4400}{350}\right)\\ =P\left(Z>1.71\right)\\ =1-P\left(Z\le 1.71\right)\end{array}$

$=1-\Phi \left(1.71\right)$

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,

$\begin{array}{c}P\left(Y>5000\right)=1-\Phi \left(1.71\right)\\ =1-0.9564\\ =0.0436\end{array}$

Thus, the probability that the observed value of 28-day strength will exceed 5000 psi for 2000 psi accelerated strength is 0.0436.

To determine

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 $y=1800+1.3x$. Where y represents the variable 28-day strength and x represents the variable accelerated strength.

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:

$\begin{array}{c}{\mu }_y=1800+1.3x\\ =1800+1.3\times 2500\\ =5050\end{array}$

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,

${\sigma }_y=350\text{psi}$.

Desired probability value:

The probability that the observed value of 28-day strength will exceed 5000 psi is obtained as follows:

$\begin{array}{c}P\left(Y>5000\right)=P\left(\frac{Y-5050}{350}>\frac{5000-5050}{350}\right)\\ =P\left(\frac{Y-{\mu }_y}{{\sigma }_y}>\frac{5000-5050}{350}\right)\\ =P\left(Z>-0.14\right)\\ =1-P\left(Z\le -0.14\right)\end{array}$

$\begin{array}{l}=1-\Phi \left(-0.14\right)\\ =1-\left(1-\Phi \left(0.14\right)\right)\\ =\Phi \left(0.14\right)\end{array}$

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,

$\begin{array}{c}P\left(Y>5000\right)=\Phi \left(0.14\right)\\ =0.5557\end{array}$

Thus, the probability that the observed value of 28-day strength will exceed 5000 psi for 2500 psi accelerated strength is 0.5557.

To determine

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 $y=1800+1.3x$. Where y represents the variable 28-day strength and x represents the variable accelerated strength.

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 $Y_1$ be the observation of 28-day strength when the accelerated strength is 2000 psi and $Y_2$ be the observation of 28-day strength when the accelerated strength is 2500 psi.

Therefore, $Y_2-Y_1$ be the difference between second observation and first observation.

From part (b), the expected mean of $Y_2$ is ${\mu }_{y_2}=5050$ and from part (a), the expected mean of $Y_1$ is ${\mu }_{y_1}=4400$.

Expected mean of $Y_2-Y_1$:

The expected mean of $Y_2-Y_1$ is obtained as follows:

$\begin{array}{c}{\mu }_{y_2-y_1}={\mu }_{y_2}-{\mu }_{y_1}\\ =5050-4400\\ =650\end{array}$

Thus, the expected mean of $Y_2-Y_1$ is 650.

Here, $Y_1$ and $Y_2$ are independent and the standard deviation of both the variable is 350 psi.

From the properties of covariance, the covariance between two independent variables is “0”.

Standard deviation:

The variance of $Y_2-Y_1$ is,

$\begin{array}{c}V\left(Y_2-Y_1\right)=V\left(Y_2\right)+V\left(Y_1\right)-2\times \mathrm{cov}\left(Y_2,Y_1\right)\\ ={\left(350\right)}^2+{\left(350\right)}^2-0\\ =245,0000\end{array}$

The general formula for standard deviation is,

$\begin{array}{c}\text{S}\text{.D}\left(Y_2-Y_1\right)=\sqrt{V\left(Y_2-Y_1\right)}\\ =\sqrt{245,000}\\ =494.97\end{array}$

Thus, the standard deviation of $Y_2-Y_1$ is 494.97.

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:

$\begin{array}{c}P\left(Y_2-Y_1>1000\right)=P\left(\frac{\left(Y_2-Y_1\right)-650}{494.97}>\frac{1000-650}{494.97}\right)\\ =P\left(\frac{\left(Y_2-Y_1\right)-{\mu }_{\left(y_2-y_1\right)}}{{\sigma }_{\left(y_2-y_1\right)}}>\frac{1000-650}{494.97}\right)\\ =P\left(Z>0.71\right)\\ =1-P\left(Z\le 0.71\right)\end{array}$

$=1-\Phi \left(0.71\right)$

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,

$\begin{array}{c}P\left(Y_2-Y_1>1000\right)=1-\Phi \left(0.71\right)\\ =1-0.7611\\ =0.2389\end{array}$

Thus, the probability that the second observation of 28-day strength will exceed the first observation by more than 1000 psi is 0.2389.

To determine

Find the value of $\left(x_2-x_1\right)$ in case of $P\left(Y_2>Y_1\right)=0.95$

Answer to Problem 8E

In case of $P\left(Y_2>Y_1\right)=0.95$ the value of $\left(x_2-x_1\right)$ is 626.33.

Explanation of Solution

Calculation:

Here, the regression equation is $y=1800+1.3x$. Where y represents the variable 28-day strength and x represents the variable accelerated strength.

Let $Y_1$ be the observation of 28-day strength when the accelerated strength is $x=x_1$ and $Y_2$ be the observation of 28-day strength when the accelerated strength is $x=x_2$.

Therefore, $Y_2-Y_1$ be the difference between second observation and first observation.

Expected mean of $Y_2-Y_1$:

The expected mean of $Y_2-Y_1$ is obtained as follows:

$\begin{array}{c}{\mu }_{y_2-y_1}={\mu }_{y_2}-{\mu }_{y_1}\\ =1800+1.3x_2-\left(1800+1.3x_1\right)\\ =1.3\left(x_2-x_1\right)\end{array}$

Thus, the expected mean of $Y_2-Y_1$ is $1.3\left(x_2-x_1\right)$.

Here, $Y_1$ and $Y_2$ are independent and the standard deviation of both the variable is 350 psi.

From part (c), the standard deviation of $Y_2-Y_1$ is 494.97.

Desired probability value:

The probability that the second observation of 28-day strength will exceed the first observation is,

$\begin{array}{c}P\left(Y_2-Y_1>0\right)=P\left(\frac{\left(Y_2-Y_1\right)-1.3\left(x_2-x_1\right)}{494.97}>\frac{1000-1.3\left(x_2-x_1\right)}{494.97}\right)\\ =P\left(\frac{\left(Y_2-Y_1\right)-{\mu }_{\left(y_2-y_1\right)}}{{\sigma }_{\left(y_2-y_1\right)}}>\frac{1000-1.3\left(x_2-x_1\right)}{494.97}\right)\\ =P\left(Z>\frac{1000-1.3\left(x_2-x_1\right)}{494.97}\right)\end{array}$

The probability that the second observation exceeds the first observation is 0

.95.

The value of $\left(x_2-x_1\right)$ is obtained as follows:

$P\left(Y_2-Y_1>0\right)=P\left(Z>\frac{1000-1.3\left(x_2-x_1\right)}{494.97}\right)=0.95$

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 $\left(x_2-x_1\right)$ value is,

$\begin{array}{c}\frac{1000-1.3\left(x_2-x_1\right)}{494.97}=-1.645\\ 1000-1.3\left(x_2-x_1\right)=-814.2257\\ -1.3\left(x_2-x_1\right)=-1814.2257\\ \left(x_2-x_1\right)=626.33\end{array}$

Thus, the $\left(x_2-x_1\right)$ is 626.33 when $P\left(Y_2>Y_1\right)=0.95$.