Answer:
66.48% of full-term babies are between 19 and 21 inches long at birth
Step-by-step explanation:
Normal Probability Distribution:
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean length of 20.5 inches and a standard deviation of 0.90 inches.
This means that [tex]\mu = 20.5, \sigma = 0.9[/tex]
What percentage of full-term babies are between 19 and 21 inches long at birth?
The proportion is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19. Then
X = 21
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21 - 20.5}{0.9}[/tex]
[tex]Z = 0.56[/tex]
[tex]Z = 0.56[/tex] has a p-value of 0.7123
X = 19
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{19 - 20.5}{0.9}[/tex]
[tex]Z = -1.67[/tex]
[tex]Z = -1.67[/tex] has a p-value of 0.0475
0.7123 - 0.0475 = 0.6648
0.6648*100% = 66.48%
66.48% of full-term babies are between 19 and 21 inches long at birth
