Newsweek in 1989 reported that 60% of young children have blood lead levels that could impair their neurological development. Assuming a random sample from the population of all school children at risk, find the probability that at least 5 children out of 10 in a sample taken from a school may have a blood level that may impair development.

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Answer:

The answer is 83.38%.

Explanation:

The probability of young children having blood levels that impair their neurological development is given as 60% in the question. To find the probability of at least 5 children out of 10 in a sample having said blood levels, we need to use the binomial probability.

n represents the total number of children in the sample so n = 10 and p is the probability of the children having blood levels causing the problem which is 60% so p = 0.6.

We want the probability of it being observed for 5 or more children and that is [tex]P(x\geq 5)[/tex]. If we subtract the [tex]P( x\leq 4)[/tex] from 1, which is the probability of not observing the condition in 5 or more children, we will have the [tex]P(x\geq 5)[/tex].

[tex]1 - P(x\leq 4) = 1 - pbinom(4, 10, 0.6)[/tex] which are the probability range in the function, the sample size and the probability of observing the condition respectively.

The result is 1 - 0.1662 = 0.8338 which means that the probability that is asked is 83.38%.

I hope this answer helps.