Answer: 0.3241
Step-by-step 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. [tex]\mu=20\ \ \sigma=6[/tex]
Sample size = 30
The probability that the employees in your sample have taken, on average, more than 20.5 hours i will be :
[tex]P(x>20.5)=P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{20.5-20}{\dfrac{6}{\sqrt{30}}})\\\\=P(z>0.4564)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{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[/tex]
∴ The probability that the employees in your sample have taken, on average, more than 20.5 hours is 0.3241 .
