Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Using the normal distribution, it is found that the score needed to qualify is of 88.32.
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, the mean and the standard deviation are given, respectively, by [tex]\mu = 80[/tex] and [tex]\sigma = 8[/tex].
The top 15% of the applicants are selected, hence the score needed to qualify is the 85th percentile, which is X when Z = 1.04, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.04 = \frac{X - 80}{8}[/tex]
X - 80 = 1.04(8)
X = 88.32.
The score needed to qualify is of 88.32.
More can be learned about the normal distribution at https://brainly.com/question/24663213
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
