Answer:
The correct option is (a) 0.9780.
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]
As the sample selected is quite large, i.e. n = 110 > 30, the central limit theorem can be applied to approximate the sampling distribution of sample proportion by a Normal distribution.
The mean and standard deviation are:
[tex]\mu_{\hat p}=p=0.70\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.70(1-0.70)}{110}}=0.044[/tex]
Compute the probability that the sample proportion of students living in the dormitories falls in between 0.60 and 0.80 as follows:
[tex]P(0.60<\hat p<0.80)=P(\frac{0.60-0.70}{0.044}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.80-0.70}{0.044})[/tex]
[tex]=P(-2.27<Z<2.27)\\\\=P(Z<2.27)-P(Z<-2.27)\\\\=0.98840-0.01160\\\\=0.9768\\\\\approx0.9780[/tex]
*Use a z-table.
Thus, the probability that the sample proportion of students living in the dormitories falls in between 0.60 and 0.80 is approximately equal to 0.9780.
The correct option is (a).
