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Homework Help: Statistic Question

  1. Nov 1, 2006 #1
    Statistic Question!!

    I have these four questions I'm having trouble solving...
    Could anyone write the solutions, but with steps on how they did it so I can possibly learn! Thanks a lot

    I'm pretty sure they use the binomial or poisson distribution formulas... can't get proper answers

    1. In a large shipment of chips, 5% are defective. What is the probability that exactly two out of a sample of ten are defective?

    2. On average, a system breaks down every 50hrs. Find the probability of more than two break-downs in a 24hr period.

    3. Show that there are more families of 6 children split 4-2 than those with 3 boys and 3 girls.

    4. People arrive at a bank at the rate of 60 per hour. Find the chance of getting 0,1,2 or 3 in the next minute.
  2. jcsd
  3. Nov 2, 2006 #2
    This is how I would to do it.
    1. Let p=0.05 q=.95. Do a binomial of (10 choose 2)x p^2 x q^8
    You need the number of ways to arrange 2 defects in 10 trials multiplied by the probabilty of each event, ie. two defects.
    2. Need to convert the Lamba given to one that represents a breakdown per 24 hr period. Here Lamda is 1 per 50 hour period. That represents a lamda of 12/25 per 24 hr period. Now find the probability of no breakdowns, 1 breakdown, and 2 breakdowns in a 24 hr period and then subtract that from 1 leaves probability of more than 2.
    3. No sure, but on first glance seems like showing that 6 choose 4 is larger than 6 choose 3. I would check on this though.
    4. Same as 2. Convert the lambda to a mean for 1 minute and use Poisson distribution.
  4. Nov 2, 2006 #3
    4. Same as 2. Convert the lambda to a mean for 1 minute and use Poisson distribution.

    so would lambda be 1/60 and after I find the chance of getting 3 in the next minute... thats the chance of 3 not coming in the next minute? so I subtract that from 1 to get 3 coming in the next minute?

    Thanks a lot for the help
  5. Nov 2, 2006 #4
    Yeah use lamda as 1/60. Then it is just a case of finding probability for X=0 X=1 X=2 X=3. 1-P(x=3) is the chance of any number of people in the next minute other than 3 people, but they are not asking you for that. Just plug in x=i i=0,1,2,3 and youre done. You should have four different probabilities corresponding to each part.
  6. Nov 2, 2006 #5
    I get it...
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