Answer:
0.9861 = 98.61% probability that the weight will be less than 4884 grams.
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.
Mean weight of 3903 grams and a standard deviation of 446 grams.
This means that ![\mu = 3903, \sigma = 446](https://tex.z-dn.net/?f=%5Cmu%20%3D%203903%2C%20%5Csigma%20%3D%20446)
If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4884 grams.
P-value of z when X = 4884. 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{4884 - 3903}{446}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B4884%20-%203903%7D%7B446%7D)
![Z = 2.2](https://tex.z-dn.net/?f=Z%20%3D%202.2)
has a p-value of 0.9861
0.9861 = 98.61% probability that the weight will be less than 4884 grams.