Question

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?

Answer 1

0.0082 or 0.82%.

Step-by-step explanation:

We can use the standard normal distribution to solve this problem. First, we need to calculate the z-score for an employee who has worked at the store for 10 years or more.

z = (x - μ) / σ

where x is the number of years, μ is the mean, and σ is the standard deviation.

z = (10 - 5.7) / 1.8 = 2.39

Using a standard normal table or calculator, we can find the probability that a z-score is greater than 2.39. This is equivalent to finding the area under the standard normal curve to the right of 2.39.

The probability can be found as follows:

P(Z > 2.39) = 1 - P(Z ≤ 2.39)

Using a standard normal table or calculator, we can find that the probability of a z-score being less than or equal to 2.39 is approximately 0.9918. Therefore:

P(Z > 2.39) = 1 - 0.9918 = 0.0082

So the probability that a randomly chosen employee has worked at the store for over 10 years is approximately 0.0082 or 0.82%.