LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)
LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)
8th Edition
ISBN: 9781464133404
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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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
Check Mark

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+bx)=(ann+bxn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)

SD(xnew)=bSD(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)=bSD(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
Check Mark

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+bx)=(ann+bxn)

That is,

x¯new=a+bx¯…… (1)

And the standard deviation of new score is:

SD(xnew)=SD(a+bx)SD(xnew)=bSD(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)=bSD(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
Check Mark

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
Check Mark

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=Xμσ

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

Z=Xμσ=7510020=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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Chapter 1 Solutions

LaunchPad for Moore's Introduction to the Practice of Statistics (12 month access)

Ch. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.2 - Prob. 16UYKCh. 1.2 - Prob. 17UYKCh. 1.2 - Prob. 18UYKCh. 1.2 - Prob. 19UYKCh. 1.2 - Prob. 20UYKCh. 1.2 - Prob. 21UYKCh. 1.2 - Prob. 22UYKCh. 1.2 - Prob. 23UYKCh. 1.2 - Prob. 24UYKCh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.3 - Prob. 47UYKCh. 1.3 - Prob. 48UYKCh. 1.3 - Prob. 49UYKCh. 1.3 - Prob. 50UYKCh. 1.3 - Prob. 51UYKCh. 1.3 - Prob. 52UYKCh. 1.3 - Prob. 53UYKCh. 1.3 - Prob. 54UYKCh. 1.3 - Prob. 55UYKCh. 1.3 - Prob. 56UYKCh. 1.3 - Prob. 57UYKCh. 1.3 - Prob. 58UYKCh. 1.3 - Prob. 59UYKCh. 1.3 - Prob. 60UYKCh. 1.3 - Prob. 67ECh. 1.3 - Prob. 69ECh. 1.3 - Prob. 61ECh. 1.3 - Prob. 62ECh. 1.3 - Prob. 63ECh. 1.3 - Prob. 64ECh. 1.3 - Prob. 65ECh. 1.3 - Prob. 66ECh. 1.3 - Prob. 74ECh. 1.3 - Prob. 75ECh. 1.3 - Prob. 76ECh. 1.3 - Prob. 71ECh. 1.3 - Prob. 68ECh. 1.3 - Prob. 70ECh. 1.3 - Prob. 77ECh. 1.3 - Prob. 78ECh. 1.3 - Prob. 79ECh. 1.3 - Prob. 80ECh. 1.3 - Prob. 81ECh. 1.3 - Prob. 82ECh. 1.3 - Prob. 83ECh. 1.3 - Prob. 84ECh. 1.3 - Prob. 85ECh. 1.3 - Prob. 86ECh. 1.3 - Prob. 87ECh. 1.3 - Prob. 88ECh. 1.3 - Prob. 89ECh. 1.3 - Prob. 90ECh. 1.3 - Prob. 91ECh. 1.3 - Prob. 92ECh. 1.3 - Prob. 93ECh. 1.3 - Prob. 94ECh. 1.3 - Prob. 95ECh. 1.3 - Prob. 96ECh. 1.3 - Prob. 72ECh. 1.3 - Prob. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 73ECh. 1.4 - Prob. 101UYKCh. 1.4 - Prob. 102UYKCh. 1.4 - Prob. 103UYKCh. 1.4 - Prob. 104UYKCh. 1.4 - Prob. 105UYKCh. 1.4 - Prob. 106UYKCh. 1.4 - Prob. 107UYKCh. 1.4 - Prob. 108UYKCh. 1.4 - Prob. 109ECh. 1.4 - Prob. 110ECh. 1.4 - Prob. 111ECh. 1.4 - Prob. 112ECh. 1.4 - Prob. 113ECh. 1.4 - Prob. 114ECh. 1.4 - Prob. 115ECh. 1.4 - Prob. 116ECh. 1.4 - Prob. 117ECh. 1.4 - Prob. 118ECh. 1.4 - Prob. 119ECh. 1.4 - Prob. 120ECh. 1.4 - Prob. 121ECh. 1.4 - Prob. 122ECh. 1.4 - Prob. 123ECh. 1.4 - Prob. 124ECh. 1.4 - Prob. 125ECh. 1.4 - Prob. 126ECh. 1.4 - Prob. 127ECh. 1.4 - Prob. 128ECh. 1.4 - Prob. 129ECh. 1.4 - Prob. 130ECh. 1.4 - Prob. 131ECh. 1.4 - Prob. 132ECh. 1.4 - Prob. 133ECh. 1.4 - Prob. 134ECh. 1.4 - Prob. 135ECh. 1.4 - Prob. 136ECh. 1.4 - Prob. 137ECh. 1.4 - Prob. 138ECh. 1.4 - Prob. 139ECh. 1.4 - Prob. 140ECh. 1.4 - Prob. 141ECh. 1.4 - Prob. 142ECh. 1.4 - Prob. 143ECh. 1.4 - Prob. 144ECh. 1.4 - Prob. 145ECh. 1.4 - Prob. 146ECh. 1.4 - Prob. 147ECh. 1.4 - Prob. 148ECh. 1.4 - Prob. 149ECh. 1.4 - Prob. 150ECh. 1.4 - Prob. 151ECh. 1.4 - Prob. 152ECh. 1.4 - Prob. 153ECh. 1.4 - Prob. 154ECh. 1.4 - Prob. 155ECh. 1 - Prob. 156ECh. 1 - Prob. 157ECh. 1 - Prob. 158ECh. 1 - Prob. 159ECh. 1 - Prob. 160ECh. 1 - Prob. 161ECh. 1 - Prob. 162ECh. 1 - Prob. 163ECh. 1 - Prob. 164ECh. 1 - Prob. 165ECh. 1 - Prob. 166ECh. 1 - Prob. 167ECh. 1 - Prob. 168ECh. 1 - Prob. 169ECh. 1 - Prob. 170ECh. 1 - Prob. 171ECh. 1 - Prob. 172ECh. 1 - Prob. 173ECh. 1 - Prob. 174ECh. 1 - Prob. 175ECh. 1 - Prob. 176ECh. 1 - Prob. 177E
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