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. 1) More than a decade ago, high levels of lead in the blood put 84% 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) 100% probability that 50 or more had high blood-lead levels in the sample taken a decade ago.

b) 0% probability that 50 or more had high blood-lead levels in the sample taken now.

Explanation:

For each children, there are only two possible outcomes. Either they have high levels of lead blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

(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?

Here we have
`n = 216`.

84% of children at risk. This means that
`p = 0.84`

So we have

`\mu = E(X) = np = 216*0.84 = 181.44`

`\sigma = √(V(X)) = √(np(1-p)) = √(216*0.84*0.16) = 5.39`

The probability is 1 subtracted by the pvalue of Z when
`X = 50`. So

`Z = (X - \mu)/(\sigma)`

`Z = (50-181.44)/(5.39)`

`Z = -24.38`

`Z = -24.38` has a pvalue of 0. This means that there was a 100% probability that 50 or more had high blood-lead levels in the sample taken a decade ago.

(b) In a random sample of 216 children taken now, what is the probability that 50 or more have high blood-lead levels?

Here we have
`n = 216`.

8% of children at risk. This means that
`p = 0.08`

So we have

`\mu = E(X) = np = 216*0.o8 = 17.28`

`\sigma = √(V(X)) = √(np(1-p)) = √(216*0.08*0.92) = 3.99`

The probability is 1 subtracted by the pvalue of Z when
`X = 50`. So

`Z = (X - \mu)/(\sigma)`

`Z = (50-17.28)/(3.99)`

`Z = -8.20`

`Z = -8.20` has a pvalue of 1. This means that there is a 0% probability that 50 or more had high blood-lead levels in the sample taken now.