So that the means of the sample falls into the dangerous region then [tex]{\displaystyle {\overline {Y}}}-\mu_{{\displaystyle {\overline {Y}}}} > 3[/tex]
Where [tex]{\displaystyle {\overline {Y}}}[/tex] is normally distributed with mean [tex]= \mu[/tex] and standar desviation [tex]\sigma_{{\displaystyle {\overline {Y}}}} =\frac{\sigma}{\sqrt{n}}[/tex]
Then [tex]n = 49[/tex]
[tex]\mu_{{\displaystyle {\overline {Y}}}}=\mu= 10[/tex]
[tex]\sigma_{{\displaystyle {\overline {Y}}}} = \frac{1}{\sqrt{49} }[/tex]
So:
[tex]P(\frac{{\displaystyle {\overline {Y}}}-\mu}{\frac{\sigma}{\sqrt{n}}})> \frac{3}{\frac{1}{\sqrt{49}}}[/tex]
Finally
[tex]P (Z> 21) = 0[/tex].
The probability that the sample mean falls in the dangerous region is approximately zero.
This makes sense, since as the sample size increases, it is less likely to deviate more than 2 standard deviations from the mu population mean.
