Question

Lengths of full-term babies in the US are Normally distributed with a mean length of 20.5 inches and a standard deviation of 0.90 inches. (Each question is worth 3 points) What percentage of full-term babies are between 19 and 21 inches long at birth

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 20.5, \sigma = 0.9$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 20.5}{0.9}$

$Z = 0.56$

$Z = 0.56$ has a p-value of 0.7123

X = 19

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{19 - 20.5}{0.9}$

$Z = -1.67$

$Z = -1.67$ 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