Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
