Asset income across individuals has a mean of $500 with standard deviation $400. In a simple random sample of 30 individuals, what is the probability that the total amount of asset income will exceed $18,000?

P (X > 18,000) = 0.0853

Step-by-step explanation:

Mean income = $500

Standard deviation = $400

Number of samples = 30

Probability of income > $18,000?

Let X denotes variable income

The expect mean is

E (X) = 30*500

E (X) = 15,000

Var (X) = 30 * (400) ²

Var (X) = 4,800,000

As we know

z = (X - E (X)) / √Var (X)

so

P (X > 18,000) = (18000 - 15000) / √4,800,000

P (X > 18,000) = 1.37

P (X > 1.37) = 1 - P (X < 1.37)

From the z-table

P (X > 1.37) = 1 - 0.9147

P (X > 1.37) = 0.0853

Therefore, the probability that the total amount of asset income will exceed $18,000 is 0.0853