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By the elevator at our math department it is written 'max 1125 kg or 15 persons'. Assume the mean weight of a student is 74 kg. Assume the probability that the total weight of 15 students exceeds 1125 kg is 5%. What is then the standard deviation of the weight of a student?

What I did: $$ Mean = 74 * 15 = 1110 $$ $$ Z = \frac{X - M}{\sigma} $$ $$ .05 = \frac{1125 - 1110}{\sigma} $$ $$ \sigma = 300 $$ $$ \sigma student = 20kg $$

I know the correct answer is 2.354397. What is the proper way to solve this?

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    $\begingroup$ You need to use the distribution somewhere. You are probably supposed to assume a normal approximation. $\endgroup$ – Henry Nov 23 '14 at 15:46
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Hints:

  • Assume a normal approximation (perhaps justified by the Central Limit Theorem)
  • For a standard normal distribution, the point above which there is $5\%$ of the distribution is how many standard deviations above the mean?
  • If $1125-1110=15$ is that many standard deviations above the mean of the sum of $1110$, then what is the standard deviation of the sum?
  • So what is the variance of the sum?
  • Assuming that the students are independent (heavy students do not necessarily have heavy companions), what is the variance of the weights of individual students?
  • So what is standard deviation of the weights of individual students?

Incidentally, R gives me an answer of $2.354607$ so there may be a slight rounding issue somewhere.

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  • $\begingroup$ for a standard normal distribution there are 2 standard deviations above the mean when there is 5% of the distribution. Correct? $\endgroup$ – burb Nov 23 '14 at 18:37
  • $\begingroup$ No. For a normal distribution, about $2.3\%$ is more then $2$ standard deviations above the mean; about $2.5\%$ is more than $1.96$ standard deviations above the mean. So, with symmetry, about $5\%$ is more than $1.96$ standard deviations away from the mean (above or below). About $5\%$ is more than $1.644854$ standard deviations above the mean. $\endgroup$ – Henry Nov 23 '14 at 18:59
  • $\begingroup$ I managed to get the right answer but one thing confuses me. Why do you have turn the standard deviations into variance and then divide by 15? Why doesn't it work to just divide the initial standard deviations by 15? I know it doesn't give the right answer but I don't understand why that is. $\endgroup$ – burb Nov 24 '14 at 19:08
  • $\begingroup$ For independent random variables, the variance of the sum is the sum of the random variables. For perfectly correlated random variables the standard deviation of the sum is the sum of the standard deviations. $\endgroup$ – Henry Nov 24 '14 at 19:35
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Suppose, that X is a random variable, which is $X\sim\mathcal N(74,\sigma_x^2)$

The probability, that one student has a weight, which is smaller or equal to x is then

$P(X \leq x )=\Phi \left( \frac{x- \mu_x}{\sigma} \right)$

$\Phi(z)$ is the function of the standard normal distribution.

Now let $Y=15X$

Thus the equation is $P(Y\leq 15x )=\Phi \left( \frac{15x-15\cdot \mu_x}{ \sigma_x \cdot \sqrt{15}} \right)$

$P(Y\leq 1125 )=\Phi \left( \frac{1125-1110}{\sigma_x \cdot \sqrt{15}} \right)$

Now you have to find $\sigma_x$, if $P(Y\geq 1125 )=0.05$

We know, that $P(Y\geq 1125 )=1-P(Y\leq 1125 )$

Thus $1-P(Y\leq 1125 )=1-\Phi \left( \frac{1125-1110}{\sigma_x \cdot \sqrt{15}} \right)=0.05$

After a Little bit of transforming you get:

$\Phi \left( \frac{1125-1110}{\sigma_x \cdot \sqrt{15}} \right)=0.95$

$\frac{1125-1110}{\sigma_x \cdot \sqrt{15}}=\Phi^{-1} \left( 0.95 \right) $

If you look in a table of standard normal distribution you will see, that $\Phi^{-1} \left( 0.95 \right)=1.645$

$\frac{1125-1110}{\sigma_x \cdot \sqrt{15}}=1.645 $

Now you can solve the equation for $\sigma_x$

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