Selected Answers to Problem Set 7

1. A better way (and the way that we would be looking for, especially given the mean and standard deviation) would use a formulation similar to weighted averages. Using the standard deviation, we can see how far the student's test scores were from the mean, and then divide that by the standard deviation and take the average of the three scores.

(85 - 75.3) / 14.6 + (76 - 55.9) / 12.8 + (77 - 67.6) / 18.5
(0.6644 + 1.5703 + 0.5081) / 3 = 1.0836

To find the lowest grade we simply drop the lowest corrected test score. In this case, the third test was the lowest corrected score. Notice a quick check to make sure that this is the correct way; the new score, after removing the lowest corrected score, is higher than the previous score.

(0.6644 + 1.5703) / 2 = 1.1176

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2. The expected value is the sum of all the possible values divided by the number of values.

(1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5
The standard deviation is the square root of the average of the squared differences.
SD² = ((1-3.5)² + (2-3.5)² + (3-3.5)² + (4-3.5)² + (5-3.5)² + (6-3.5)²) / 6 = 35 / 12
SD = sqrt(35 / 12) = 1.7078

Rolling 100 dice follows the same pattern. But because these are independent trials, the variance will be multiplied by 100. The standard deviation, the square root of the variance, will therefore be multiplied by the square root of 100.

SD² = 100 * 35 / 12
SD = (sqrt(100 * 35 / 12)) = 17.078

Now, instead of multiplying the variance by 100 (for 100 independent trials) we will divide by 100 (for the average of 100 ttrials). The variance of a variable c times X is equal to c^2 times the variance of X.

SD² = (35 / 12) / 100
SD = 0.17078

Look to make sure that these answers make sense. When we roll the dice 100 times and find the total, the standard deviation should be higher than with one die. And when we roll the dice 100 times and find the average, the standard deviation should be lower thatn with one die. Understand why this is true.

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4. A matrix that rotates the plane +60 degrees (Pi/3) will place the vector <1,0> to and the vector <0,1> to <-sin(Pi/3), cos(Pi/3)>. The matrix then takes the form:

| cos(Pi/3) -sin(Pi/3) |
| sin(Pi/3)  cos(Pi/3) |

A matrix that rotates the plane -60 degrees (-Pi/3) will rotate the vector <1,0> to (cos(-Pi/3), sin(-Pi/3)> and the vector <0,1> to <-sin(-Pi/3), cos(-Pi/3)>. Remembering your even-odd functions, you know that sin(-x) = -sin(x) and cos(-x) = cos(x). So the new matrix then takes the form:

|  cos(Pi/3) sin(Pi/3) |
| -sin(Pi/3) cos(Pi/3) |

A matrix that keep x1 (x-axis) fixed will not change the vector. That is, <1,0,0> will move to <1,0,0>. However, the X2(y-axis) and x3(z-axis) will rotate by 60 degrees (Pi/3) on the x2-x3 plane. Therefore, <0,1,0> will map to <0,cos(Pi/3),sin(Pi/3)> and <0,0,1> will map to <0,-sin(Pi/3), cos(Pi/3)>. The matrix will take the form:

| 1      0          0    |
| 0 cos(Pi/3) -sin(Pi/3) |
| 0 sin(Pi/3)  cos(Pi/3) |

6. Function (a) does not manipulate the object; it simply translates it. The object moves horizontally by four and vertically by two.

Function (b) translates horizontally by four and vertically by negative two, but it also stretches e2 by a factor of two. Function (c) flips the object across the e2-axis, stretches it vertically by a factor of two, then translates it. The translation is done after the manipulation; had it been done before the translation would have been forward one and up one because the flip would negate the horizontal translation. Function (d) skews the object so that it's new x-coordinate is the old x-coordinate plus the y-coordinate. The function also translates the new skewed object +1 in both directions.