Answer:
option 0.8%
Step-by-step explanation:
Data provided in the question:
Mean = 5.7 years
Standard deviation, s = 1.8 years
Now,
P(the employee has worked at the store for over 10 years)
= P(X > 10 years)
=
or
=
= P (Z > 2.389 )
or
= 0.008447 [from standard z table]
or
= 0.008447 × 100% = 0.84% ≈ 0.8%
Hence,
the correct answer is option 0.8%