Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weighing less than 2500 grams at birth are considered "low weight." If we randomly select a baby, what is the probability that it has a low birth weight?

Respuesta :

Answer:

3.84% probability that it has a low birth weight

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 = 3466, \sigma = 546[/tex]

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

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

[tex]Z = \frac{2500 - 3466}{546}[/tex]

[tex]Z = -1.77[/tex]

[tex]Z = -1.77[/tex] has a pvalue of 0.0384

3.84% probability that it has a low birth weight