Chapter 5 - Normal Probability Distributions - Section 5.4 Sampling Distributions and the Central Limit Theorem - Exercises - Page 271: 27

0.15% of sample will have mean salary less than 60000

To calculate P(x̄ < 60000), we need to find the z-score: $z = \frac{x̄ - μ_{x̄}}{σ/\sqrt n}$ $= \frac{60000 - 66000}{12000/\sqrt 35}$ = -2.96 P(z < -2.96) = 0.0015, hence: P(x̄ 60000) = 0.0015