Respuesta :
Using the normal distribution, we find that 32% of the data that is less than 150 or greater than 190.
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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.
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The data in a data set are normally distributed with a mean of 170 and a standard deviation of 20.
This means that [tex]\mu = 170, \sigma = 20[/tex]
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Less than 150:
As a proportion, it is the p-value of Z when X = 150, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{150 - 170}{20}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.16.
0.16x100% = 16%
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More than 190:
As a proportion, it is 1 subtracted by the p-value of Z when X = 190, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{190 - 170}{20}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.84.
1 - 0.84 = 0.16
0.16x100% = 16%
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Estimate the percent of the data that are less than 150 or greater than 190.
We add both percentages, thus:
16% + 16% = 32%
32% of the data that is less than 150 or greater than 190.
A similar question is given at https://brainly.com/question/14694610
