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

answer: 6 strips

step-by-step explanation:

54

step-by-step explanation:

6x9