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ankoles [38]
3 years ago
13

Range of scores making up the interquartile ranger

Mathematics
1 answer:
Nostrana [21]3 years ago
4 0

For a standard normally distributed random variable <em>Z</em> (with mean 0 and standard deviation 1), we get a probability of 0.0625 for a <em>z</em>-score of <em>Z</em> ≈ 1.53, since

P(<em>Z</em> ≥ 1.53) ≈ 0.9375

You can transform any normally distributed variable <em>Y</em> to <em>Z</em> using the relation

<em>Z</em> = (<em>Y</em> - <em>µ</em>) / <em>σ</em>

where <em>µ</em> and <em>σ</em> are the mean and standard deviation of <em>Y</em>, respectively.

So if <em>s</em> is the standard deviation of <em>X</em>, then

(250 - 234) / <em>s</em> ≈ 1.53

Solve for <em>s</em> :

16/<em>s</em> ≈ 1.53

<em>s</em> ≈ 10.43

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Answer:

  (4, -2, 3)

Step-by-step explanation:

You want the final augmented coefficient matrix to look like ...

\left[\begin{array}{ccc|c}1&0&0&4\\0&1&0&-2\\0&0&1&3\end{array}\right]

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To get there, you do a series of row operations. The usual Gauss-Jordan elimination algorithm has you start by arranging the rows so the highest leading coefficient is in the first row. Dividing that row by that coefficient immediately generates a bunch of fractions, so gets messy quickly. Instead, we'll start by dividing the given first row by 2 to make its leading coefficient be 1:

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And dividing that row by -3 makes it ...

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Continuing the process of zeroing out the first column, we can subtract the third row from 3 times the first to get ...

  0x +5y +11z = 23

After these operations, our augmented matrix is ...

\left[\begin{array}{ccc|c}1&2&3&9\\0&1&2&4\\0&5&11&23\end{array}\right]

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Conveniently, the second row has a 1 on the diagonal, so we can use that directly to zero the second column of the other rows. Subtracting 2 times the second row from the first, the new first is ...

  {1, 2, 3 | 9} -2{0, 1, 2 | 4} = {1, 0, -1 | 1}

Subtracting 5 times the second row from the 3rd, the new 3rd row is ...

  {0, 5, 11 | 23} -5{0, 1, 2 | 4} = {0, 0, 1 | 3}

After these operations, our augmented matrix is ...

\left[\begin{array}{ccc|c}1&0&-1&1\\0&1&2&4\\0&0&1&3\end{array}\right]

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Conveniently, the third row has 1 on the diagonal, so we can use that directly to zero the third column of the other rows.

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Subtracting twice the third row from the second gives the new second row ...

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So, our final augmented matrix is ...

\left[\begin{array}{ccc|c}1&0&0&4\\0&1&0&-2\\0&0&1&3\end{array}\right]

This tells us the solution is (x, y, z) = (4, -2, 3).

_____

<em>Comment on notation</em>

It is a bit cumbersome to write the equations represented by each row of the matrix, so we switched to a bracket notation that just lists the coefficients in order. It is more convenient and less space-consuming, and illustrates the steps adequately. For your own work, you need to use a notation recognized by your grader, or explain any notation you may adopt as a short form.

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