Respuesta :
Answer:
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
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.
In this question, we have that:
[tex]\mu = 110, \sigma = 0.15[/tex]
The proportion of infants with birth weights between 125 oz and 140 oz is
This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So
X = 140
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 110}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
X = 125
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{125 - 110}{15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.9772 - 0.8413 = 0.1359
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
