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 89% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 10% of children in the United States are at risk of high blood-lead levels. (a) In a random sample of 206 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.) Incorrect: Your answer is incorrect. (b) In a random sample of 206 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.) Incorrect: Your answer is incorrect.

Answer 1

a)

X ~ Bin ( n , p)

Where n = 206 , p = 0.89

Mean = np = 206 * 0.89 = 183.34

Standard deviation = sqrt [ 206 * 0.89 ( 1 - 0.89) ] = 4.4908

Using normal approximation,

P(X < x) = P(Z < (x - mean) / SD )

With continuity correction,

P(X >= 50) = P(X > 49.5)
P ( X > 49.5 ) = P(Z > (49.5 - 183.34 ) / 4.4908 )
= P ( Z > -29.8 )
= 1 - P ( Z < -29.8 )
= 1 - 0

= 1

b)

X ~ Bin ( n , p)

Where n = 206 , p = 0.10

Mean = np = 206 * 0.10 = 20.6

Standard deviation = sqrt [ 206 * 0.20 ( 1 - 0.20) ] = 5.7411

Using normal approximation,

P(X < x) = P(Z < (x - mean) / SD )

With continuity correction,

P(X >= 50) = P(X > 49.5)

P ( X > 49.5 ) = P(Z > (49.5 - 20.6 ) / 4.3058 )
= P ( Z > 6.71 )
= 1 - P ( Z < 6.71 )
= 1 - 1

= 0