Question

In part (a), show work by writing out the individual probability formulas for P(X = x) and letting C(n,x) = "n choose x" .

Based on 2000 census data, the median annual household income was \$39,000.

(a) Among five randomly selected U.S. households, find the probability that four or more have incomes *exceeding* \$39,000 per year.

Consider a random sample of 16 U.S households for the next 3 questions.

(b) What is the probability of seeing at least 10 of the 16 households with annual incomes under \$39,000? You may use distribution functions in `R` for this one.

(c) What is the expected number of households with annual incomes under \$39,000?

(d) What is the standard deviation of the number of households with annual incomes under \$39,000?

Answer 1

Suppose, random variable Y denotes the amount of income in $.

Based on 2000 census data we have,

(a)

Suppose we define getting a household with income exceeding $39000 as success.

Let, random variable X denotes the number of households with income exceeding $39000.

So, probability of success in each trial .

Required probability

For problem (b) to (c)-

Suppose we define getting a household with income under $39000 as success.

Let, random variable Z denotes the number of households with income under $39000.

So, probability of success in each trial .

(b)

Required probability

(c)

We know, for mean is given by .

Hence, expected number of households with annual income under $39000 is 16*0.5 = 8

(d)

We know, for variance is given by .

So, variance = 16*0.5*0.5 = 4

Hence, standard deviation of the number of households with annual incomes under $39000 is 4^0.5 = 2