Respuesta :
Answer:
The probability that in a random sample of 100 CSU graduates the error is within 5% of the population proportion of 60% is 0.6923.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
The information provided is:
p = 0.60
n = 100
As n = 100 > 30, the central limit theorem can be applied to approximate the sampling distribution of sample proportions.
The distribution of sample proportion is [tex]\hat p\sim N(0.60, 0.049^{2})[/tex].
Compute the probability that in a random sample of 100 CSU graduates the error is within 5% of the population proportion of 60% as follows:
[tex]P(|\hat p-\mu_{\hat p}|=0.05)=P(-0.05<\hat p-\mu_{\hat p}<0.05)[/tex]
[tex]=P(\frac{-0.05}{0.049}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.05}{0.049})\\\\=P(-1.02<Z<1.02)\\\\=P(Z<1.02)-P(Z<-1.02)\\\\=0.84614-0.15386\\\\=0.69228\\\\\approx 0.6923[/tex]
Thus, the probability that in a random sample of 100 CSU graduates the error is within 5% of the population proportion of 60% is 0.6923.
