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3 years ago

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A set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. Find the value that is –2 standard devi

ations away from the mean.

2 answers:

dybincka [34]3 years ago

6 0

45+8.3 = 53.3

<span>45 - (2(8.3)) = 28.4

The value is </span> 28.4 <span>that is –2 standard deviations away from the mean.</span><span>

</span>

Kamila [148]3 years ago

4 0

Answer: 28.4

Step-by-step explanation:

Given : A set of data with a mean of $\mu= 45$ and a standard deviation of $\sigma=8.3$ is normally distributed.

Now, the value that is –2 standard deviations away from the mean is given by :-

$x=\mu-2\sigma\\\\\Rightrrrow\ x=45-2(8.3)\\\\\Rightarrow\ x=45-16.6\\\\\Rightrrrow\ x=28.4$

Hence, the value that is –2 standard deviations away from the mean. = 28.4

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and the constant matrix is

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We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so

$\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:

multiply the 1st row by 1/2

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right]$

add -3 times the 1st row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right]$

multiply the 2nd row by -2/7

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right]$

add 7/2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right]$

multiply the 3rd row by 1/3

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add 12/7 times the 3rd row to the 2nd row

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$x=3\\y=4\\z=1$

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.

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