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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

2 Answers

3 votes

Answer:

Explanation:

-2

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User Mustapha GANGA
by
6.1k points
4 votes

Answer:


Z = -2.2

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


Z = (2500 - 3600)/(500)


Z = -2.2

User Collin James
by
6.7k points
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