Answer:
The probability that the total amount of asset income will exceed $18,000 is 0.0853
Step-by-step explanation:
Given;
Individual mean, μ = $500
Individual standard deviation, S.D = $400
Sample number = 30
Raw score, X = $18,000
Mean for 30 individuals, μ = 30($500) = $15,000
If S.D = $400, then Variance = (S.D)² = ($400)², for 30 individual = 30($400)²
= $4,800,000
S.D for 30 individual [tex]=\sqrt{4,800,000} =2,190.89[/tex]
Z- score [tex]= \frac{X-\mu}{S.D} = \frac{18,000-15,000}{2190.89}[/tex]
Z = 1.369
P (X > $18,000) = P (Z > 1.37)
P(Z=1.37) = 0.9147
P(Z>1.37) = 1 - 0.9147 = 0.0853
Therefore, the probability that the total amount of asset income will exceed $18,000 is 0.0853
