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

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Answer:

[tex]Z = -2.2[/tex]

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 3600, \sigma = 500[/tex]

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

This is Z when X = 2500. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{2500 - 3600}{500}[/tex]

[tex]Z = -2.2[/tex]

Answer:

Step-by-step explanation:

-2

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