Mathematics, 26.03.2020 05:17 itzia00

More than a decade ago, high levels of lead in the blood put 85% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 16% of children in the United States are at risk of high blood-lead levels.

(a) In a random sample of 184 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.)

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Answer from: jbcupcake

1 = 100% probability that 50 or more had high blood-lead levels

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

n = 184, p = 0.85

p is 0.85 because we are working with a sample from a decade ago.

$\mu = E(X) = np = 184*0.85 = 156.4$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{184*0.85*0.15} = 4.84$

(a) In a random sample of 184 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.)

This is 1 subtracted by the pvalue of Z when X = 50. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{50 - 156.4}{4.84}$

Z = -22

Z = -22 has a pvalue of 0.

1 - 0 = 1

1 = 100% probability that 50 or more had high blood-lead levels

Answer from: Quest

It is 94.7% that uses the metric system

Answer from: Quest

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

Which are the possible side lengths of a triangle?

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More than a decade ago, high levels of lead in the blood put 85% of children at risk. A concerted ef...