According to New Jersey Transit, the 8:00 a.m. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let Y = the number of days on which the train arrives on time. What is P(Y = 4)? Interpret this value. (0.90)¹ (0.10)². There is a 0.66% probability that exactly 4 of the 6 trains arrive on time. (0.90)4 (0.10)². There is a 0.66% probability that at least 4 of the 6 trains arrive on time. This probability cannot be calculated because this is not a binomial setting. (9)(0.90)+(0.10)². There is a 9.84% probability that at least 4 of the 6 trains arrive on time. O()(0.90)¹ (0.10)². There is a 9.84% probability that exactly 4 of the 6 trains arrive on time.
According to New Jersey Transit, the 8:00 a.m. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let Y = the number of days on which the train arrives on time. What is P(Y = 4)? Interpret this value. (0.90)¹ (0.10)². There is a 0.66% probability that exactly 4 of the 6 trains arrive on time. (0.90)4 (0.10)². There is a 0.66% probability that at least 4 of the 6 trains arrive on time. This probability cannot be calculated because this is not a binomial setting. (9)(0.90)+(0.10)². There is a 9.84% probability that at least 4 of the 6 trains arrive on time. O()(0.90)¹ (0.10)². There is a 9.84% probability that exactly 4 of the 6 trains arrive on time.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:According to New Jersey Transit, the 8:00 a.m. weekday train from Princeton to New York City has a 90% chance of arriving on
time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let Y = the number of days on which the
train arrives on time.
What is P(Y = 4)? Interpret this value.
(0.90)4 (0.10)². There is a 0.66% probability that exactly 4 of the 6 trains arrive on time.
(0.90) ¹ (0.10)². There is a 0.66% probability that at least 4 of the 6 trains arrive on time.
This probability cannot be calculated because this is not a binomial setting.
O()(0.90)+ (0.10)². There is a 9.84% probability that at least 4 of the 6 trains arrive on time.
O()(0.90)4 (0.10)². There is a 9.84% probability that exactly 4 of the 6 trains arrive on time.
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