Respuesta :
Answer:
The probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.
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:
n = 596
p = 0.09
As the sample size is quite large, i.e. n = 596 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by a Normal distribution.
The mean and standard deviation are:
[tex]\mu_{\hat p}=p=0.09\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.09(1-0.09)}{596}}=0.012[/tex]
Compute the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% as follows:
[tex]P(\hat p-p>0.03)=P(\hat p>0.12)[/tex]
[tex]=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.12-0.09}{0.012})\\\\=P(Z>2.5)\\\\=1-P(Z<2.5)\\\\=1-0.99379\\\\=0.00621\\\\\approx 0.006[/tex]
Thus, the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.
