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

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺

it would be a on edgenuit y

the diagram below is your final answer