Answer: 0.3241
Explanation:
Let x be the random variable that represents the time to complete the coursework and successfully pass all tests.
Given : 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, and the standard deviation is 6 hours.
i.e.
![\mu=20\ \ \sigma=6](https://img.qammunity.org/2020/formulas/mathematics/college/iuk27e63ricbhboya5obworjc5a3u3x7su.png)
Sample size = 30
The probability that the employees in your sample have taken, on average, more than 20.5 hours i will be :
![P(x>20.5)=P((x-\mu)/((\sigma)/(√(n)))>(20.5-20)/((6)/(√(30))))\\\\=P(z>0.4564)\ \ [\because\ z=(x-\mu)/((\sigma)/(√(n)))]\\\\=1-P(z\leq0.4564)\ \ [\because P(Z>z)=1-P(Z\leq z)]\\\\=1-0.6759\ \ [\text{ by using p-value table for z}]=0.3241](https://img.qammunity.org/2020/formulas/mathematics/college/zvq52n5acff152xd2c55v3qtdjln983h7m.png)
∴ The probability that the employees in your sample have taken, on average, more than 20.5 hours is 0.3241 .