84.13%
1) For this situation at Fry's grocery store we need to find out the Probability of those workers who worked less than 32 hours weekly.
2) We need to find out the Z score, using this formula:
[tex]Z=\frac{x-\mu}{\sigma}\Rightarrow Z=\frac{32-28}{4}=\frac{4}{4}=1[/tex]
Notice that the Z score we start with this and then go check at a Z score table.
[tex]P(Z<1)=0.8413[/tex]
3) So we can state that approximately 84.13% of those workers, worked fewer than 32 hours
