Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
0.1587 = 15.87% probability that a person will wait for more than 9 minutes.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The mean waiting time is 6 minutes and the variance of the waiting time is 9.
This means that [tex]\mu = 6, \sigma = \sqrt{9} = 3[/tex]
Find the probability that a person will wait for more than 9 minutes.
This is 1 subtracted by the p-value of Z when X = 9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9 - 6}{3}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.8413.
1 - 0.8413 = 0.1587
0.1587 = 15.87% probability that a person will wait for more than 9 minutes.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
