Answer :
Answer:
0.8643 = 86.43% probability that the weight will be less than 4640 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 [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:
The standard deviation is the square root of the variance, so
[tex]\mu = 3907, \sigma = \sqrt{444.889} = 667[/tex]
Of a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4640 grams.
This probability if the pvalue of Z when X = 4640. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4640 - 3907}{667}[/tex]
[tex]Z = 1.10[/tex]
[tex]Z = 1.10[/tex] has a pvalue of 0.8643.
0.8643 = 86.43% probability that the weight will be less than 4640 grams.