The employees in certain division of Cybertronics Inc.need to complete a certification online.On average, it takes 20 hours to complete the coursework and successfully pass all tests, with a standard deviation of 6 hours.If you select a random sample of size 30, the probability that the employees in your sample have taken, on average, more than 20.5 hours is ______.
A)0.47
B)0.45
C)0.41
D)0.32
E)0.01

Question

Quizplus · Accepted Answer

Using the Central Limit Theorem, we can calculate the probability that the sample mean is more than 20.5 hours. The standard error (SE) of the sample mean is the standard deviation divided by the square root of the sample size, which is $SE = \frac{6}{\sqrt{30}}$. To find the Z-score for 20.5 hours, we use the formula $Z = \frac{\bar{x} - \mu}{SE}$, where $\bar{x}$ is the sample mean (20.5 hours), and $\mu$ is the population mean (20 hours). Plugging in the values, we get a Z-score. Looking up this Z-score in the standard normal distribution table gives us the probability of observing a sample mean greater than 20.5 hours. The correct probability corresponds to option D, which is 0.32, indicating that there's a 32% chance the sample mean is more than 20.5 hours.

The Employees in Certain Division of Cybertronics Inc