Question

A random sample of 100 U.S. adults were selected from a population. In the population, the probability of having hypertension is about 24%. What is the probability that there are at most 30 people in the

asked Jan 4, 2021 100k views

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Itwarilal · Answer 1 · 2021-01-09T20:28:32+0000

Answer:

For this case we want to calculate this probability:

$P(X\leq 30)$

And for this case if we use the normal approximation we can use the z score formula given by:

$z = (X -\mu)/(\sigma)$

And using this formula with the correction factor of continuity we got:

$P(X\leq 30) =P(X <30.5)=P(Z< (30.5-24)/(4.27)) = P(Z \leq 1.522)$

And using the normal standard table or excel we got:

$P(Z \leq 1.522)= 0.936$

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".

Let X the random variable of interest "number of people with hypertension", on this case we now that:

$X \sim Binom(n=100, p=0.24)$

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=(n!)/((n-x)! x!)

We need to check the conditions in order to use the normal approximation.

$np=100*0.24=24 \geq 10$

$n(1-p)=100*(1-0.24)=76 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=100*0.24=24

$\sigma=√(np(1-p))=√(100*0.24(1-0.24))=4.27$

Solution to the problem