Answer:
The standard deviation for the sample sum distribution is 212.13
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
For sums
Distribution is approximately normal with mean [tex]n*\mu[/tex] and standard deviation [tex]\sqrt{n}*\sigma[/tex]
Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?
We have n = 50, [tex]\sigma = 30[/tex].
So
[tex]\sqrt{50}*30 = 212.13[/tex]
The standard deviation for the sample sum distribution is 212.13
