Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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
#SPJ1
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
