Answer:
0.4207
Step-by-step explanation:
Determine z-score
[tex]z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\ \\z=\frac{52-50}{10/\sqrt{70}}\\ \\z\approx1.6733[/tex]
Find P(Z>1.6733)
[tex]P(Z > 1.6733)=normalcdf(1.6733,\infty)\approx0.4207[/tex]
Therefore, the probability that the mean stress level for a sample of 70 employees will fall above 52 is 0.4207
