Respuesta :
Answer:
Babies that weigh more than 2409 grams or less than 4341 grams will be included in the study.
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 = 3375, \sigma = 493[/tex]
Top 2.5%
More than X when Z has a pvalue of 1-0.025 = 0.975. So more than X when Z = 1.96.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.96 = \frac{X - 3375}{493}[/tex]
[tex]X - 3375 = 493*1.96[/tex]
[tex]X = 4341[/tex]
More than 4341 grams is the top 2.5%.
Bottom 2.5%.
Less than X when Z has a pvalue of 0.025. So less than X when Z = -1.96.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.96 = \frac{X - 3375}{493}[/tex]
[tex]X - 3375 = 493*(-1.96)[/tex]
[tex]X = 2409[/tex]
Less than 2409 grams is in the bottom 2.5%.
Babies that weigh more than 2409 grams or less than 4341 grams will be included in the study.
