Respuesta :
Answer:
The proportion of baby boys in the United States that are born with low birth weight is 0.0495.
Step-by-step explanation:
Problems of normal distributions can be 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.
3.41 kg with a standard deviation of 0.55 kg.
This means that [tex]\mu = 3.41, \sigma = 0.55[/tex]
What proportion of baby boys in the United States are born with low birth weight?
This is the pvalue of Z when X = 2.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.5 - 3.41}{0.55}[/tex]
[tex]Z = -1.65[/tex]
[tex]Z = -1.65[/tex] has a pvalue of 0.0495
The proportion of baby boys in the United States that are born with low birth weight is 0.0495.
