Mathematics, 15.11.2019 18:31 AADJ4
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. 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
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![P(x20.5)=P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}\dfrac{20.5-20}{\dfrac{6}{\sqrt{30}}})\\\\=P(z0.4564)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(z\leq0.4564)\ \ [\because P(Zz)=1-P(Z\leq z)]\\\\=1-0.6759\ \ [\text{ by using p-value table for z}]=0.3241](/tpl/images/0376/1080/4f0c7.png)
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