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Sagot :
Using the normal distribution, it is found that there is a 0.6568 = 65.68% probability that the variable that is between −1.33 and 0.67.
Normal Probability Distribution
In a normal distribution 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]
- It 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, which is the percentile of X.
In this problem, we have a standard normal distribution, hence [tex]\mu = 0, \sigma = 1[/tex].
The probability that the variable is between −1.33 and 0.67 is the p-value of Z when X = 0.67 subtracted by the p-value of Z when X = -1.33, hence:
X = 0.67
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.67 - 0}{1}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a p-value of 0.7486.
X = -1.33
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{-1.33 - 0}{1}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a p-value of 0.0918.
0.7486 - 0.0918 = 0.6568.
0.6568 = 65.68% probability the variable that is between −1.33 and 0.67.
You can learn more about the normal distribution at https://brainly.com/question/24663213
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