Answer:
0.02
Explanation:
Given a normal distribution :
Mean income (m) = 25000
Standard deviation of income (s) = 6000
X ≥ 12000
Using the relation to fund the standardized score :
Zscore =(x - m) / s
Zscore = ( 12000 - 25000) / 6000
Zscore = -13000 / 6000
Zscore = - 2.167
Using a z probability calculator :
P(Z ≤ - 2.167) = 0.015117
= 0.02