Finding standard deviation of a student's weight 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?
 A: 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 $\sigma_x$
A: 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.
