Respuesta :
Answer:
0.8413 = 84.13% probability that the weight will be less than 4570 grams.
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(which is the square root of the variance) [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 = 3915, \sigma = \sqrt{429025} = 655[/tex]
If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4570 grams.
This is the pvalue of Z when X = 4570. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4570 - 3915}{655}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.8413 = 84.13% probability that the weight will be less than 4570 grams.
