Question

A news company reported that 6% of adult Americans have a food allergy. Consider selecting 15 adult Americans at random. Define the random variable x as

x = number of people in the sample of 15 that have a food allergy.

Find the following probabilities. (Round your answers to three decimal places.)

(a)

P(x < 3)

(b)

P(x ≤ 3)

(c)

P(x ≥ 4)

(d)

P(1 ≤ x ≤ 3)

Answer 1

$x\sim Bin(n, p)$

Given $n=15, p=0.06, q=1-p=1-0.06=0.94.$

(a)

$$P(x<3)=P(x=0)+P(x=1)$$

$$+P(x=2)=\dbinom{15}{0}(0.06)^0(0.94)^{15-0}$$

$$+\dbinom{15}{1}(0.06)^1(0.94)^{15-1}+\dbinom{15}{2}(0.06)^2(0.94)^{15-2}$$

$$\approx0.3953+0.3785+0.1691\approx0.943$$

(b)

$$P(x\leq 3)=P(x<3)+P(x=3)$$

$$\approx0.9429+\dbinom{15}{3}(0.06)^3(0.94)^{15-3}$$

$$\approx0.9429+0.0468\approx0.990$$

(c)

$$P(x\geq 4)=1-P(x\leq 3)$$

$$\approx1-0.9897\approx0.010$$

(d)

$$P(1\leq x\leq 3)=P(x=1)+P(x=2)$$

$$+P(x=3)=\dbinom{15}{1}(0.06)^1(0.94)^{15-1}$$

$$+\dbinom{15}{2}(0.06)^2(0.94)^{15-2}+\dbinom{15}{3}(0.06)^2(0.94)^{15-3}$$

$$\approx0.3785+0.1691+0.0468\approx0.594$$