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, and the distribution is approximately normal. if an employee is picked at random, what is the probability that the employee has worked at the store for over 10 years? 99.2% 0.8% 49.2% 1.7%

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%

answer is 144

step-by-step explanation:

answer:

125 yards

explanation:

75+50=125. that is my answer