1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normally distrbution with Probability

  1. Dec 7, 2013 #1
    Assume the length of waiting at supermarket is approximately normally distributed with mean 6 minutes and standard deviation 1.5 minutes.
    (1) Fund the probability that waiting time is longer than 8 minutes
    My way is P(z > (8-6)/1.5) = P(z > 4/3) = .0918
    (2) What length of the waiting time constitutes the 99th percentile ($Φ_.99$)?
    My way is
    For 99th percentile, find p(z > 2.327) = .01
    find (y-6)/1.5 = 2.237 <=> y = 9.4905. For 0th percentile, y should be negative infinity. However, it does not make sense since we talk about time here. We should take 0 instead.
    The waiting time length is between 0 and 9.4905
    (3) Find probability that waiting time lasted longer than 6 minutes given that it lasted longer than 5 minutes
    My way is P(Y>6|Y>5) = P(Y>6,Y>5)/P(Y>5) = P(Y>6)/P(Y>5) = P(z > (6-6)/1.5)/P(z > (5-6)/1.5) = P(z > 0) = P(z > -2/3) = .5/(1-.2514)=.67.

    Could anyone please check (1), (2), (3) for me?
     
    Last edited: Dec 7, 2013
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?