Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores $x_{\text{new}}=a+bx$ that has mean 100 and standard deviation 20.

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is $x_{\text{new}}=-40+2x$.

Explanation of Solution

Given: Each grade average of 100 and standard deviation is 20.

Calculation: The reading ability has mean 70 and standard deviation 10 provided to third-graders and has mean 80 and standard deviation 11 on the same test and the required new marks are $\left(100,20\right)$.

The new score transformation is given by $x_{\text{new}}=a+bx$.

Now, the mean of the new marks will be:

$\begin{array}{c}{\displaystyle \sum x_{\text{new}}}={\displaystyle \sum \left(a+bx\right)}\\ \frac{{\displaystyle \sum x_{\text{new}}}}{n}=\frac{1}{n}{\displaystyle \sum \left(a+bx\right)}\\ {\overline{x}}_{new}=\frac{1}{n}\left({\displaystyle \sum a}+b{\displaystyle \sum x}\right)\\ =\left(\frac{an}{n}+b\frac{{\displaystyle \sum x}}{n}\right)\end{array}$

That is,

${\overline{x}}_{new}=a+b\overline{x}$…… (1)

And the standard deviation of new score is:

$\text{SD}\left(x_{\text{new}}\right)=\text{SD}\left(a+bx\right)$

$\text{SD}\left(x_{\text{new}}\right)=b\cdot \text{SD}\left(x\right)$…… (2)

For third graders standard deviation is $\text{SD}\left(x\right)=10$ and standard deviation of new score is $\text{SD}\left(x_{\text{new}}\right)=20$.

Substitute the values in the equation (2) and get,

$\begin{array}{c}\text{SD}\left(x_{\text{new}}\right)=b\cdot \text{SD}\left(x\right)\\ 20=b\left(10\right)\\ b=\frac{20}{10}\\ b=2\end{array}$

For third graders mean is $\overline{x}=70$ and mean of new score is $x_{\text{new}}=100$.Substitute the values in the equation (1) and get,

$\begin{array}{c}{\overline{x}}_{new}=a+b\overline{x}\\ 100=a+2\left(70\right)\\ 100=a+140\\ a=-40\end{array}$

Interpretation: The linear transformation that change third-grade scores x into new scores is $x_{\text{new}}=-40+2x$.

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores $x_{\text{new}}=a+bx$ that has mean 100 and standard deviation 20.

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is $x_{\text{new}}=-45.448+1.8181x$.

Explanation of Solution

Given: The reading ability has mean 70 and standard deviation 10 provided to sixth-graders and has mean 80 and standard deviation 11 on the same test.

The third-grade marks are $\left(70,10\right)$ and the sixth-grade marks are $\left(80,11\right)$ and the required new marks are $\left(100,20\right)$.The new score transformation is given by $x_{\text{new}}=a+bx$.Now, the mean of the new marks can be calculated as:

$\begin{array}{c}{\displaystyle \sum x_{\text{new}}}={\displaystyle \sum \left(a+bx\right)}\\ \frac{{\displaystyle \sum x_{\text{new}}}}{n}=\frac{1}{n}{\displaystyle \sum \left(a+bx\right)}\\ {\overline{x}}_{new}=\frac{1}{n}\left({\displaystyle \sum a}+b{\displaystyle \sum x}\right)\\ =\left(\frac{an}{n}+b\frac{{\displaystyle \sum x}}{n}\right)\end{array}$

That is,

${\overline{x}}_{new}=a+b\overline{x}$…… (1)

And the standard deviation of new score is:

$\begin{array}{c}\text{SD}\left(x_{\text{new}}\right)=\text{SD}\left(a+bx\right)\\ \text{SD}\left(x_{\text{new}}\right)=b\cdot \text{SD}\left(x\right)\end{array}$…… (2)

For sixth-graders standard deviation is $\text{SD}\left(x\right)=11$ and standard deviation of new score is $\text{SD}\left(x_{\text{new}}\right)=20$.Substitute the values in the equation (2) and get,

$\begin{array}{c}\text{SD}\left(x_{\text{new}}\right)=b\cdot \text{SD}\left(x\right)\\ 20=b\left(11\right)\\ b=\frac{20}{11}\\ b=1.8181\end{array}$

For sixth-graders mean is $\overline{x}=80$ and mean of new score is $x_{\text{new}}=100$.Substitute the values in the equation (1) and get,

$\begin{array}{c}{\overline{x}}_{new}=a+b\overline{x}\\ 100=a+1.8181\left(80\right)\\ 100=a+145.448\\ a=-45.448\end{array}$

Interpretation: The linear transformation that change sixth-grade scores x into new scores is $x_{\text{new}}=-45.448+1.8181x$.

(c)

To determine

To find: The transformed score of David and Nancy if David is a third-grade student and scores 72 in the test and Nancy is a sixth-grade student and scores 78 in the test and to find who scores more in the test after transformation.

Answer to Problem 176E

Solution: The transformed score of David is 104 and the transformed score of Nancy is 96.3638. David scores more than Nancy after transformation into new scores.

Explanation of Solution

Given: The linear transformation that change third-grade scores x into new scores is

$x_{\text{new}}=-40+2x$ and sixth-grade scores x into new scores is $x_{\text{new}}=-45.448+1.8181x$.

Calculation: David scores 72 in the test with the third-grade. So, his transformed new score can be calculated as:

$\begin{array}{c}{x}_{\text{new}}=-40+2x\\ =-40+2\left(72\right)\\ =-40+144\\ =104\end{array}$

This score is above the mean of 100. And Nancy scores 72 in the test with the third-grade. So, her transformed new score can be calculated as:

$\begin{array}{c}{x}_{\text{new}}=-45.448+1.8181x\\ =-45.448+1.8181\left(78\right)\\ =-45.448+141.8118\\ =96.3638\end{array}$

This score is below the mean of 100.

Interpretation: Therefore, David scores more than Nancy after transformation into new scores.

(d)

To determine

To find: What percent of third-graders and sixth-graders have scored less than 75 if the distribution of scores in both grade is normal and have $N\left(100,20\right)$ distribution.

Answer to Problem 176E

Solution: Around 10.56% of third-graders and sixth-graders have scored less than 75.

Explanation of Solution

Calculation: The Z score is given by the formula,

$Z=\frac{X-\mu }{\sigma }$

Substitute the values in the formula for standardized score of 75 as:

$\begin{array}{c}Z=\frac{X-\mu }{\sigma }\\ =\frac{75-100}{20}\\ =-\frac{25}{20}\\ =-1.25\end{array}$

From the standard normal table $P\left(Z<-1.25\right)=0.1056$. Hence, the required probability is 0.1056.

Interpretation: Therefore, it can be concluded that 10.56% of third-graders and sixth-graders have scored less than 75.