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7 boys have a mean height of 1.80m. Peter joins the boys and now the mean height is 1.75m. Find Peter's height.

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Do you really mean $18$ metres?! Are you sure that it isn’t $1.80$ m, or $180$ cm? – Brian M. Scott Oct 2 '12 at 9:05
I think you are having some problems with your units. I have never met someone who is $18$ metres tall. – Michael Albanese Oct 2 '12 at 9:05

I’m going to work this with more plausible figures: Robert Wadlow, the tallest person in history for whom there is irrefutable evidence, was ‘only’ $2.72$ m tall.

If the original $7$ boys have a mean height of $1.80$ m, their total height when laid end to end is $7\cdot1.80=12.60$ m. Now Peter comes along, and their mean height is now only $1.75$ m. This means that the total height of all $8$ boys is $8\cdot 1.75=14.00$ m. How much did Peter add to the total height by joining the group?

An important lesson taught by this exercise is that if you know how many numbers are involved, knowing their mean and knowing their total are pretty much the same thing. If the mean of $n$ numbers is $\bar x$, their total is $n\bar x$; if their total is $y$, their mean is $\dfrac{y}n$.

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The (arithmetic) mean of $n$ numbers $x_1, \dots, x_n$ is $$\frac{1}{n}(x_1 + \dots + x_n).$$

Let $x_1, \dots, x_7$ be the heights of the original $7$ boys. Then we know that $$\frac{1}{7}(x_1 + \dots x_7) = 180.$$

Let $P$ be Peter's height. Then we know that $$\frac{1}{8}(x_1 + \dots + x_7 + P) = 175.$$

Rearranging the first equation, we see that $x_1 + \dots + x_7 = 1260$. Substituting that into the second equation, you can then determine $P$, Peter's height.

Note: At no point do we need to know any of the individual heights of the other seven boys, only the sum of their heights (or equivalently, their mean height).

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