suppose that the average annual income of a defense attorney is $150,000 with a standard deviation of $40,000. Assume that the income distribution is normal. What is the probability that the average annual income of a sample of 100 defense attorneys is more than $120,000?

P (X > 120000)Normal Distribution, µ = 150000, σ = 40000, n = 100x − µxx−µ√we convert this to standard normal using z ==σxσ/ n120000 − (150000)√z== −7.5040000/…