At a large department store, the average number of years of employment for a cashier is 5.7 with a standard deviation of 1.8 years. If the number of years of employment at this department store is normally distributed, what is the probability that a cashier selected at random has worked at the store for over 10 years?

Answer:   0.0019Step-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