Question

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?

Answer 1

0% probability that more than 28 children become sick.

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:

`√(V(X)) = √(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 = (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 = √(V(X))`.

In this problem, we have that:

`n = 43, p = 0.36`

So

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

`\sigma = √(V(X)) = √(np(1-p)) = √(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 = (X - \mu)/(\sigma)`

`Z = (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.