Question

Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies above the 95th percentile have a condition called macrosomia (also known as "big baby syndrome"). What birth weight marks the 95th percentile?

Answer 1

A birth weight of 4364.17 grams marks the 95th percentile.

Explanation:

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.

In this problem, we have that:

`\mu = 3466, \sigma = 546`

What birth weight marks the 95th percentile?

This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

`1.645 = (X - 3466)/(546)`

`X - 3466 = 1.645*546`

`X = 4364.17`

A birth weight of 4364.17 grams marks the 95th percentile.