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A set of eight numbers has mean 5 and SD 4. The number 21 and another integer n are added to the set. The standard deviation of the set is now 6. Find n and the mean of the set now.

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What have you tried? –  JavaMan Feb 27 '12 at 14:49
    
Finding the sum and the sums of squares, trying to form equations about variance. –  Adam Granger Feb 27 '12 at 15:07
    
I fooled around some and using SD equal to the square root of $(1/8)\sum(x_i-\bar{x})^2$, got an equation that seems not to give integer $n$. But it was done quickly and I am accident-prone. –  André Nicolas Feb 27 '12 at 16:26
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1 Answer

One way to do this is to concoct $8$ numbers which have a mean of $5$ and a standard deviation of $4$. For example, take $$ -1,1,3,5,5,7,9,11, $$ and label these points $x_1 , x_2, \dots , x_8$. Then, one can easily check: $$ \frac{1}{8}\sum_{i=1}^8 x_i = 5 $$ and $$ \sqrt{\sum_{i=1}^{8} \frac{(x_i - 5)^2}{8 - 1}} = 4. $$

Adding $21$ and $n$ gives $$ \mu = \frac{8 \cdot \mu_1 + 21 + n}{10} = \frac{61 + n}{10}. $$ The new standard deviation is $$ 6 = \sigma = \sqrt{\sum_{i=1}^{8} \frac{(x_i - \mu)^2}{10 - 1} + \frac{(21 - \mu)^2}{10 - 1} + \frac{(n - \mu)^2}{10 - 1}}. $$ Expanding gives a quadratic equation in $\mu$:

$$\begin{align} 324 = & (-1 - \mu)^2 + (1 - \mu)^2 + (3 - \mu)^2 + 2(5 - \mu)^2 \\ &+ (7 - \mu)^2 + (9 - \mu)^2 + (11 - \mu)^2 + (21 - \mu)^2 + (9 \mu - 61)^2 \end{align}$$ Note that the last term is $(10 \mu - 61 - \mu)^2$ as $\mu_2 = \frac{61 + n}{10}$. Solving this for $\mu$ yields $\mu = 8$. Hence, $n = 19$.


Edit: For clarification, this does not totally answer the question at hand. My example does not prove that $n = 19$ is the only (integer) solution. However, it is likely that this is the case. The method I used to find the set of integers which worked was by noticing that we needing to find $8$ squares that sum to $112 = 4^2 \cdot (8 -1)$. This gave a natural choice of $$ 112 = 6^2 + 6^2 + 4^2 + 4^2 + 2^2 + 2^2 + 0^2 + 0^2 $$ which implied that $$\begin{array}{c} x_1 = 5 - 6 \\ x_2 = 5 + 6 \\ x_3 = 5 - 4 \\ x_4 = 5+4 \\ x_5 = 5 - 2 \\ x_6 = 5+2 \\ x_7 = 5 - 0 \\ x_8 = 5 + 0 \end{array}$$ This happened to lead to a integral value of $\mu$ and therefore an integral value of $n$.

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This is a proof by "My professor says it must be true", i.e., you do one example and assume it gives the same answer as any other example. –  Graphth Feb 27 '12 at 16:53
    
@Grapth: I never claimed that this solves the complete problem. Your point is valid in the sense that the question has not been fully answered, but this is not a proof, and hence your comment doesn't apply. –  JavaMan Feb 27 '12 at 17:52
    
I wonder why you are dividing by 7, and not by 8 (in your calculation of the standard deviation)? I realize there are circumstances where division by $n-1$ is correct, but I would have thought that in this case $n$ would be the correct denominator. –  Gerry Myerson Feb 28 '12 at 0:25
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