SOLUTION: On a certain run of a commuter train, the average number of passengers is 476 and the standard deviation is 22. Assume the variable is normally distributed. If the train makes t

Algebra ->  Probability-and-statistics -> SOLUTION: On a certain run of a commuter train, the average number of passengers is 476 and the standard deviation is 22. Assume the variable is normally distributed. If the train makes t      Log On


   

Question 1161600: On a certain run of a
commuter train, the average number of passengers is 476
and the standard deviation is 22. Assume the variable is
normally distributed. If the train makes the run, find the
probability that the number of passengers will be
a. Between 476 and 500 passengers
b. Less than 450 passengers
c. More than 510 passengers
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
a. Between 476 and 500 passengers

z1 = %28476-476%29%2F22 = 0%2F22 = 0

Look up 0 on a z-table. The result is 0.5000.

z2 = %28500-476%29%2F22 = 24%2F22 = 1.09

Look up 1.09 on a z-table. The result is 0.8621.

Subtract 0.5000 from 0.8621 to get a result of 0.3621.

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b. Less than 450 passengers

z = %28450-476%29%2F22 = %28-26%29%2F22 = -1.18

Look up -1.18 on a z-table. The result is 0.1190.

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b. More than 510 passengers

z = %28510-476%29%2F22 = 34%2F22 = 1.55

Look up 1.55 on a z-table. The result is 0.9394. This is the probability that FEWER than 510 passengers are on the train. So, the probability that MORE than 510 passengers are on the train is 1 - 0.9394...or, 0.0606.