A worker in the automobile industry works an average of 43.7 hours per week. If the distribution is approximately normal with a standard deviation of 1.6 hours what is the probability that a randomly selected automobile worker works less than 40 hours per week

1.04%

Step-by-step explanation:

Using the z-score formula, we have:

z = (X - P) / σ

z = score

x = given value

P = average

σ = standard deviation

Z = (40 - 43.7) / 1.6 = 2.31 > 0.9896 (tabulated in the z-table)

Now, with that number, we find the probability that a randomly selected worker works less than 40 hours per week.

1 - 0.9896 = 0.0104 * 100 = 1.04%