If a group of 50 students are sampled at random, the probability that their total weight does not exceed 3600 Kg is:
A. 0.9961 B. 0.9977 C. 0.9989 D. 0.9952 E. 0.9931
I know that you must find z-score; (3600-3500)/5
But the official answers say that the standard deviation must be written as so, σS= sqrt(50) × 5 = 35.35
The part where I get confused is when you multiply 5 by sqrt (50) to reflect the new standard deviation, as I think I should instead be writing 5/sqrt 50
Any help would be appreciated!
Let $X_i$ be a random variable with mean $\mu_x$ and variance $\sigma_x^2$ for all $i = 1, 2, ..., n$. Assume $X_i$ are independent and identically distributed.
Let $Z = \sum_i X_i $.
Then, the mean of $Z$, $\mu_z$, is given by
$$ \mu_z = E[Z] $$ $$ = E\left[\sum_i X_i\right] $$ $$ = \sum_i{E\left[X_i\right]} $$ $$ = nE\left[X_i\right] $$ $$ = n\mu_x $$
The variance of $Z$, $\sigma_z^2$, is given by
$$ \sigma_z^2 = V[Z] $$ $$ = V\left[\sum_i X_i\right] $$ $$ = \sum_i{V\left[X_i\right]} $$ $$ = nV\left[X_i\right] $$ $$ = n\sigma^2_x $$
So, the standard deviation of $Z$ is $\sigma_z = \sqrt{n}\sigma_x$.
Now, with $X_i$ normally distributed, we further have that $Z$ is also normally distributed - the sum of normally distributed random variables is itself normally distributed.
In this case, the mean of the sum is
$$ \mu_z = n\mu_x = 50\cdot 70 = 3500 . $$
That's the 3500 you're comparing 3600 against in your $Z$-score.
The standard deviation of the sum is
$$ \sigma_z = \sqrt{n}\sigma_x = \sqrt{50}\cdot 5 . $$
That's what you need to divide by in your $Z$-score.