Answer: 0.0384
Explanation:
Given : Population mean = [tex]\mu=100[/tex]
Population standard deviation = [tex]\sigma=20[/tex]
Sample size : n= 50
Let [tex]\overline{X}[/tex] be the sample mean.
Then, the probability that the sample mean is more than 105 is given by :_
[tex]P(\overline{x}>105)=1=P(\overline{x}\leq105)\\\\=1-P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\leq\dfrac{105-100}{\dfrac{20}{\sqrt{50}}})\\\\=1-P(z\leq1.77)\ \ [\because\ z=(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-0.96164\ \ [\text{Using z-table}]\\\\=0.03836\approx0.0384[/tex]
Hence, the probability that the sample mean is more than 105= 0.0384
