Answer:
The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is 68%.
The probability that a randomly selected infant has a birth weight between 110 and 130 is 38%.
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 of 120 ounces and a standard deviation of 20 ounces.
This means that ![\mu = 120, \sigma = 20](https://tex.z-dn.net/?f=%5Cmu%20%3D%20120%2C%20%5Csigma%20%3D%2020)
The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is
p-value of Z when X = 140 subtracted by the p-value of Z when X = 100.
X = 140
![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{140 - 120}{20}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B140%20-%20120%7D%7B20%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
has a p-value of 0.84
X = 100
![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{100 - 120}{20}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B100%20-%20120%7D%7B20%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a p-value of 0.16
0.84 - 0.16 = 0.68
The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is 68%.
The probability that a randomly selected infant has a birth weight between 110 and 130
This is the p-value of Z when X = 130 subtracted by the p-value of Z when X = 110.
X = 130
![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{130 - 120}{20}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B130%20-%20120%7D%7B20%7D)
![Z = 0.5](https://tex.z-dn.net/?f=Z%20%3D%200.5)
has a p-value of 0.69
X = 110
![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{110 - 120}{20}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B110%20-%20120%7D%7B20%7D)
![Z = -0.5](https://tex.z-dn.net/?f=Z%20%3D%20-0.5)
has a p-value of 0.31
0.69 - 0.31 = 0.38 = 38%.
The probability that a randomly selected infant has a birth weight between 110 and 130 is 38%.