Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Using the normal distribution, there is a 0.1587 = 15.87% probability that’s a car picked at random is traveling at more than 100 km/hr.
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 = 90, \sigma = 10[/tex]
The probability that’s a car picked at random is traveling at more than 100 km/hr is one subtracted by the p-value of Z when X = 100, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 90}{10}[/tex]
Z = 1
Z = 1 has a p-value of 0.8413.
1 - 0.8413 = 0.1587.
0.1587 = 15.87% probability that’s a car picked at random is traveling at more than 100 km/hr.
More can be learned about the normal distribution at https://brainly.com/question/28096232
#SPJ1
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
