Question

A newborn baby whose Apgar score is over 6 is classified as normal and this happens in 80% of births. As a quality control check, an auditor examined the records of 100 births. He would be suspicious if the number of normal births in the sample of 100 births fell below the lower limit of "usual." What is that lower limit?

Answer 1

The lower limit is 72.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

If X is more than 2 standard deviations from the mean, it is unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

In this question, we have that:

`n = 100, p = 0.8`

So

`\mu = 0.8, s = \sqrt{(0.8*0.2)/(100)} = 0.04`

He would be suspicious if the number of normal births in the sample of 100 births fell below the lower limit of "usual." What is that lower limit?

2 standard deviations below the mean is the lower limit, so X when Z = -2.

Proportion:

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`-2 = (X - 0.8)/(0.04)`

`X - 0.8 = -2*0.04`

`X = 0.72`

Out of 100:

0.72*100 = 72

The lower limit is 72.