The linear transformation that change third-grade scores x into new scores x new = a + b x that has mean 100 and standard deviation 20.

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−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.

(c)

Expert Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

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(100,20) distribution.

(d)

Expert Solution

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=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−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.

(c)

Expert Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

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(100,20) distribution.

(d)

Expert Solution

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=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

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Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−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.

(c)

Expert Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

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(100,20) distribution.

(d)

Expert Solution

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=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

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Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−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.

(c)

Expert Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

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(100,20) distribution.

(d)

Expert Solution

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=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−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.

(c)

Expert Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

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(100,20) distribution.

(d)

Expert Solution

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=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

Answer 6

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

Answer 7

Chapter 1, Problem 176E

(a)

To determine

The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

Answer 8

(a)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 (100,20).

The new score transformation is given by xnew=a+bx.

Now, the mean of the new marks will be:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=b⋅SD(x)…… (2)

For third graders standard deviation is SD(x)=10 and standard deviation of new score is SD(xnew)=20.

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

SD(xnew)=b⋅SD(x)20=b(10)b=2010b=2

For third graders mean is x¯=70 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+2(70)100=a+140a=−40

Interpretation: The linear transformation that change third-grade scores x into new scores is xnew=−40+2x.

Answer 13

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

Answer 14

(b)

To determine

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

Answer 15

(b)

Expert Solution

Answer to Problem 176E

Solution: The linear transformation that change sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 (70,10) and the sixth-grade marks are (80,11) and the required new marks are (100,20).The new score transformation is given by xnew=a+bx.Now, the mean of the new marks can be calculated as:

∑xnew=∑(a+bx)∑xnewn=1n∑(a+bx)x¯new=1n(∑a+b∑x)=(ann+b∑xn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=b⋅SD(x)…… (2)

For sixth-graders standard deviation is SD(x)=11 and standard deviation of new score is SD(xnew)=20.Substitute the values in the equation (2) and get,

SD(xnew)=b⋅SD(x)20=b(11)b=2011b=1.8181

For sixth-graders mean is x¯=80 and mean of new score is xnew=100.Substitute the values in the equation (1) and get,

x¯new=a+bx¯100=a+1.8181(80)100=a+145.448a=−45.448

Interpretation: The linear transformation that change sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

Answer 20

(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.

(c)

Expert Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

This score is below the mean of 100.

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

Answer 21

(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 22

(c)

Expert Solution

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.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

xnew=−40+2x and sixth-grade scores x into new scores is xnew=−45.448+1.8181x.

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

xnew=−40+2x=−40+2(72)=−40+144=104

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:

xnew=−45.448+1.8181x=−45.448+1.8181(78)=−45.448+141.8118=96.3638

This score is below the mean of 100.

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

Answer 27

(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(100,20) distribution.

(d)

Expert Solution

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=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

Answer 28

(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(100,20) distribution.

Answer 29

(d)

Expert Solution

Answer to Problem 176E

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

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation: The Z score is given by the formula,

Z=X−μσ

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

Z=X−μσ=75−10020=−2520=−1.25

From the standard normal table P(Z<−1.25)=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.

Answer 34

Ch. 1.1 - Prob. 1UYK Ch. 1.1 - Prob. 2UYK Ch. 1.1 - Prob. 3UYK Ch. 1.1 - Prob. 5UYK Ch. 1.1 - Prob. 4UYK Ch. 1.1 - Prob. 6UYK Ch. 1.1 - Prob. 7UYK Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

The linear transformation that change third-grade scores x into new scores x new = a + b x that has mean 100 and standard deviation 20.

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The linear transformation that change third-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

Answer to Problem 176E

Explanation of Solution

The linear transformation that change sixth-grade scores x into new scores xnew=a+bx that has mean 100 and standard deviation 20.

Answer to Problem 176E

Explanation of Solution

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

Explanation of Solution

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(100,20) distribution.

Answer to Problem 176E

Explanation of Solution

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