In a local university, 70% of the students live in the dormitories. A random sample of 75 students is selected for a particular study. The probability that the sample proportion (the proportion living in the dormitories) is at least 0.60 is
A) 0.0294.
B) 0.9706.
C) 0.0228.
D) 0.9772.

Question

Quizplus · Accepted Answer

The sample proportion of students living in dormitories follows a normal distribution (by the Central Limit Theorem) with mean $p = 0.70$ and standard error $SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.70(0.30)}{75}}$. To find the probability that the sample proportion is at least 0.60, we calculate the Z-score for 0.60: $Z = \frac{0.60 - 0.70}{SE}$, and then find the probability to the right of this Z-score in the standard normal distribution. This calculation gives a high probability, indicating that it is very likely to observe a sample proportion of at least 0.60 when the true proportion is 0.70, hence option B is correct.

In a Local University, 70% of the Students Live in the Dormitories