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 90% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 8% 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)

n=

216

p=

0.9000

since np and n(1-p) both are greater than 10, we can use normal approximation of binomial distribution:

here mean of distribution=μ=np=		194.40
and standard deviation σ=sqrt(np(1-p))=		4.41
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability that 50 or more had high blood-lead levels:

probability =P(X>49.5)=P(Z>(49.5-194.4)/4.409)=P(Z>-32.86)=1-P(Z<-32.86)=1-0=1.0000

b)

n=

216

p=

0.0800

since np and n(1-p) both are greater than 10, we can use normal approximation of binomial distribution:

here mean of distribution=μ=np=		17.28
and standard deviation σ=sqrt(np(1-p))=		3.99
for normal distribution z score =(X-μ)/σx

probability that 50 or more have high blood-lead levels

probability =P(X>49.5)=P(Z>(49.5-17.28)/3.987)=P(Z>8.08)=1-P(Z<8.08)=1-1=0.0000