Question

In a large university, 70% of the students live in the dormitories. A random sample of 110 students is selected for a particular study.The probability that the sample proportion of students living in the dormitories falls in between 0.6 and 0.8 equals a.0.9780 b.0.9318 c.0.9365 d.1.6450

Answer 1

The correct option is (a) 0.9780.

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:

`\mu_(\hat p)=p`

The standard deviation of this sampling distribution of sample proportion is:

`\sigma_(\hat p)=\sqrt{(p(1-p))/(n)}`

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:

`\mu_(\hat p)=p=0.70\\\\\sigma_(\hat p)=\sqrt{(p(1-p))/(n)}=\sqrt{(0.70(1-0.70))/(110)}=0.044`

Compute the probability that the sample proportion of students living in the dormitories falls in between 0.60 and 0.80 as follows:

`P(0.60<\hat p<0.80)=P((0.60-0.70)/(0.044)<(\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.80-0.70)/(0.044))`

`=P(-2.27<Z<2.27)\\\\=P(Z<2.27)-P(Z<-2.27)\\\\=0.98840-0.01160\\\\=0.9768\\\\\approx0.9780`

*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).