Respuesta :
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 [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.
Mean of 120 ounces and a standard deviation of 20 ounces.
This means that [tex]\mu = 120, \sigma = 20[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 120}{20}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.84
X = 100
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 120}{20}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 120}{20}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a p-value of 0.69
X = 110
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{110 - 120}{20}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] 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%.
