Question

Childhood lead poisoning is a public health concern in the U.S. In a certain population, 1 child in 9 has a high blood lead level (defined as 30 μg/dl or more). Suppose a researcher randomly chose 15 children from the population consecutively.

1. Find the probability that more than 1 has high blood lead by using the binomial probability formula.

2.Find the probability that less than 5 has high blood lead by using dbinom or pbinom in
RStudio.

3 .Find the mean and standard deviation of the binomial random variable, X.

Answer 1

n = 15, p = 1/9 = 0.1111, q = 1 - p = 8/9 = 0.8889; P(x) = C(n, x) p^x q^(n - x)

x	P(x) = C(15, x) (1/9)^x (8/9)^(15 - x)
0	0.170888235
1	0.32041544
2	0.28036351
3	0.151863568
4	0.056948838
5	0.01566093
6	0.003262694
7	0.000524362
8	0.0001
9	0
10	0
11	0
12	0
13	0
14	0
15	0

(a) P(x > 1) = 1 - P(x ≤ 1) = 1 - [P(0) + P(1)] = 0.5087

(b) P(x < 5) = P(0) + P(1) + P(2) + P(3) + P(4) = 0.9805

(c) Mean = np = 15(1/9) = 1.6667 and Standard deviation = √(npq) = √(15 * (1/9) * (8/9)) = 1.2172