Question

A newborn who weighs 2,500 g or less has a low birth weight. Use the information on the right to find the z-score of a 2,500 g baby. In the United notes, birth weights of newborn babies are approximately normally distributed with a mean of mu = 3,600 grams and a standard deviation of sigma = 500 grams. Z = StartFraction x minus mu Over sigma EndFraction

Answer 1

Answer:

Step-by-step explanation:

-2

ur welcome ;)

Answer 2

Answer:

$Z = -2.2$

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 3600, \sigma = 500$

Use the information on the right to find the z-score of a 2,500 g baby.

This is Z when X = 2500. So

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

$Z = \frac{2500 - 3600}{500}$

$Z = -2.2$