Respuesta :
Answer:
The probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.
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 sample size is, n = 553 > 30. Thus, the Central limit theorem is applicable in this case.
Compute the mean and standard deviation as follows:
[tex]\mu_{\hat p}=p=0.04\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.04\times 0.96}{553}}=0.0083[/tex]
Compute the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% as follows:
 [tex]P(\hat p>0.05)=P(\frac{\hat p-\mu_{\hat p}}{\sigm_{\hat p}}>\frac{0.05-0.04}{0.0083})\\\\=P(Z>1.20)\\\\=1-P(Z<1.20)\\\\=1-0.88493\\\\=0.11507\\\\\approx 0.1151[/tex]
Thus, the probability that the proportion of airborne viruses in a sample of 553 viruses would be greater than 5% is 0.1151.
