Question

Time magazine reported that 60% of young children have blood lead levels that could impair their neurological development. A sample of 10 children is randomly chosen. The probability that at least 8 children out of 10 in the sample may have a blood level that may impair development is:

Answer 1

Answer:

$P(X \geq 8)=P(X=8)+P(X=9)+P(X=10)$

$P(X=8)=(10C8)(0.6)^8 (1-0.6)^{10-8}=0.1209$

$P(X=9)=(10C9)(0.6)^9 (1-0.6)^{10-9}=0.0403$

$P(X=10)=(10C10)(0.6)^{10} (1-0.6)^{10-10}=0.006$

$P(X \geq 8)=0.1209+0.0403+0.006=0.1672$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest (number of children that may have a blood level that may impair development), on this case we know that:

$X \sim Binom(n=10, p=0.6)$

Since we select a sample of 10 and the probability of success for each trial is 0.6

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X \geq 8)=P(X=8)+P(X=9)+P(X=10)$

$P(X=8)=(10C8)(0.6)^8 (1-0.6)^{10-8}=0.1209$

$P(X=9)=(10C9)(0.6)^9 (1-0.6)^{10-9}=0.0403$

$P(X=10)=(10C10)(0.6)^{10} (1-0.6)^{10-10}=0.006$

$P(X \geq 8)=0.1209+0.0403+0.006=0.1672$