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Sagot :
Answer:
a) [tex]Z = -2.88[/tex]
b) [tex]Z = -0.96[/tex]
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 [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.
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 [tex]\mu = 52.2, \sigma = 2.5[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{45 - 52.2}{2.5}[/tex]
[tex]Z = -2.88[/tex]
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 [tex]\mu = 47.4, \sigma = 2.5[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{45 - 47.4}{2.5}[/tex]
[tex]Z = -0.96[/tex]
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.
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