Answer:
a) ![Z = -2.88](https://tex.z-dn.net/?f=Z%20%3D%20-2.88)
b) ![Z = -0.96](https://tex.z-dn.net/?f=Z%20%3D%20-0.96)
c) 40 weeks gestation babies
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.
Here, we use
. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{45 - 52.2}{2.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B45%20-%2052.2%7D%7B2.5%7D)
![Z = -2.88](https://tex.z-dn.net/?f=Z%20%3D%20-2.88)
b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.
Here, we use
. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{45 - 47.4}{2.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B45%20-%2047.4%7D%7B2.5%7D)
![Z = -0.96](https://tex.z-dn.net/?f=Z%20%3D%20-0.96)
c. For which group is a birth length of 45 cm more common?
For each group, the probability is 1 subtracted by the pvalue of Z.
Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.