Answer :
Answer:
0.06
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the standard deviation is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex].
In this problem, we have that:
[tex]p = 0.2, n = 50[/tex]
So
[tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]s = \sqrt{\frac{0.2*0.8}{50}}[/tex]
[tex]s = 0.06[/tex]