Answer:
0.6848
Step-by-step explanation:
Mean of \hat{p} = 0.453
Answer = 0.453
Standard deviation of \hat{p} :
= \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.453(1-0.453)}{100}} = 0.0498
Answer = 0.0498
P(0.0453 - 0.05 < p < 0.0453 + 0.05)
On standardising,
= P(\frac{0.0453-0.05-0.0453}{0.0498} <Z<\frac{0.0453+0.05-0.0453}{0.0498})
= P(-1.0044 < Z < 1.0044) = 0.6848
Answer = 0.6848
Using the Central Limit Theorem, it is found that the mean of the sampling distribution of the sample proportions is 0.453.
The Central Limit Theorem establishes 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 sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem:
By the Central Limit Theorem, the mean is [tex]\mu = p = 0.453[/tex].
A similar problem is given at https://brainly.com/question/15413688
