Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. More than a decade ago, high levels of lead in the blood put 83% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 13% of children in the United States are at risk of high blood-lead levels.

(a) In a random sample of 216 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to three decimal places.)

(b) In a random sample of 216 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.)

Answer 1

a)
np = 216*0.83 = 179.28
n(1-p) = 216*0.17 = 36.72

It is appropriate to use normal approximation as np and n(1-p) are greater than 5

mean = np = 216*0.83 = 179.28
sd = sqrt(p*(1-p)*n) = sqrt(0.83*0.17*216) = 5.5207

P(X >= 50)
= P(X > 49.5) .. continuity correction
= P(z > (49.5 - 179.28)/5.5207)
= P(z > -23.5)
= 1.000

b)
np = 216*0.13 = 28.08
n(1-p) = 216*0.87 = 187.92

It is appropriate to use normal approximation as np and n(1-p) are greater than 5

mean = np = 216*0.13 = 28.08
sd = sqrt(p*(1-p)*n) = sqrt(0.13*0.87*216) = 4.9426

P(X >= 50)
= P(X > 49.5) .. continuity correction
= P(z > (49.5 - 28.08)/4.9426)
= P(z > 4.3)
= 0.000