Answer:
0.999 = 99.9% probability that the baby weighs less than 150 Oz.
Step-by-step explanation:
When the distribution is normal, we use 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.
The weights of newborn babies are distributed normally, with a mean of approximately 105 Oz and a standard deviation of 15 Oz.
This means that [tex]\mu = 105, \sigma = 15[/tex]
What if the probabability that the baby weighs less than 150 Oz?
This is the pvalue of Z when X = 150. SO
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{150 - 105}{15}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.999
0.999 = 99.9% probability that the baby weighs less than 150 Oz.
