Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
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.
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 2.8, \sigma = 0.7[/tex].
The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is one subtracted by the p-value of Z when X = 4, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 2.8}{0.7}[/tex]
Z = 1.71
Z = 1.71 has a p-value of 0.9564.
1 - 0.9564 = 0.0436.
0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.
More can be learned about the normal distribution at https://brainly.com/question/24663213
#SPJ1
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.