Answer: 0.0019
Step-by-step explanation:
Let x be the random variable that represents the number of years of employment at this department store.
Given : The number of years of employment at this department store is normally distributed,
Population mean : [tex]\mu=5.7[/tex]
Standard deviation : [tex]\sigma=1.8[/tex]
Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Now, the z-value corresponding to 10 : [tex]z=\dfrac{10-5.7}{1.8}\approx2.39[/tex]
P-value = [tex]P(x>10)=P(Z>2.89)=1-P(z\leq2.89)[/tex]
[tex]=1-0.9980737=0.0019263\approx0.0019\text{ (Rounded to nearest ten thousandth)}[/tex]
Hence, the probability that a cashier selected at random has worked at the store for over 10 years = 0.0019
