Respuesta :
Answer:
A birth weight of 4364.17 grams marks the 95th percentile.
Step-by-step explanation:
Problems of normally distributed samples 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.
In this problem, we have that:
[tex]\mu = 3466, \sigma = 546[/tex]
What birth weight marks the 95th percentile?
This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 3466}{546}[/tex]
[tex]X - 3466 = 1.645*546[/tex]
[tex]X = 4364.17[/tex]
A birth weight of 4364.17 grams marks the 95th percentile.
