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Sagot :
Using the normal distribution, it is found that the relative frequency of classical CDs with playing time X between 49 and 69 minutes is 0.8403 = 84.03%.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 54, \sigma = 5[/tex]
The relative frequency of classical CDs with playing time X between 49 and 69 minutes is the p-value of Z when X = 69 subtracted by the p-value of Z when X = 49, hence:
X = 69:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{69 - 54}{5}[/tex]
Z = 3
Z = 3 has a p-value of 0.9987.
X = 49:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{49 - 54}{5}[/tex]
Z = -1
Z = -1 has a p-value of 0.1584.
0.9987 - 0.1584 = 0.8403.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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