Question

3. Submit your answers to the following binomial questions. You may use the appendix table B #5 to answer parts (a) and (b). According to a government study, 15% of all children live in a household that has an income below the poverty level. If a random sample of 15 children is selected:

a) what is the probability that 5 or more live in poverty?

b) what is the probability that 5 live in poverty?

c) what is the expected number (mean) that live in poverty? What is the variance? What is the standard deviation?

Answer 1

Here, n = 15, p = 0.15, (1 - p) = 0.85 and x 5 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 5).
P(X >= 5) = 1 - P(x< =4)
P(x>=5) = 1 - (15C0 * 0.15^0 * 0.85^15) + (15C1 * 0.15^1 * 0.85^14) + (15C2 * 0.15^2 * 0.85^13) + (15C3 * 0.15^3 * 0.85^12) + (15C4 * 0.15^4 * 0.85^11)
P(X >= 5) = 1 - (0.087 + 0.231 + 0.286 + 0.218 + 0.116)
P(X >= 5) = 0.0620

b)
Here, n = 15, p = 0.15, (1 - p) = 0.85 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 5)
P(X = 5) = 15C5 * 0.15^5 * 0.85^10
P(X = 5) = 0.0449

c)

mean = np
= 0.15 * 15
= 2.2500

std.dev = sqrt(npq)
= sqrt(15 * 0.15 *0.85)
= 1.3829

varian ce = npq
= (15 * 0.15 *0.85)
=1.9125