Mathematics, 29.05.2020 08:58 jamesnaquan132

Suppose at random 36% of school children develop nausea and vomiting following holiday parties and that you conduct a study to examine this phenomenon, with a sample size of n=43. What is the probability that more than 28 children become sick?

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Answer from: jgcugucgucg

0% probability that more than 28 children become sick.

Step-by-step explanation:

I am going to use the normal approximation to the binomial 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 = 43, p = 0.36

$\mu = E(X) = np = 43*0.36 = 15.48$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{43*0.36*0.64} = 3.1476$

What is the probability that more than 28 children become sick?

Using continuity correction, this is P(X > 28+0.5) = P(X > 28.5), which is 1 subtracted by the pvalue of Z when X = 28.5. So

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

$Z = \frac{28.5 - 15.48}{3.1476}$

Z = 4.14

Z = 4.14 has a pvalue of 1

1 - 1 = 0

0% probability that more than 28 children become sick.

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Suppose at random 36% of school children develop nausea and vomiting following holiday parties and t...