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| quote: | Originally posted by DJ_Science
I hope this makes sense...
> 1. A normal distribution has a mean of 80 and a standard deviation
> of 14. Determine the value above which 80 percent of the values will > occur.
You need a table for this one. Its about 68.3. Basically you want to look at z&sigma for which a one tailed test gives 20 percent outside. They key to this is a one tailed test. then &mu - z&sigma is the value above which 80% will fall.
> 2. Assume a binomial probability distribution with n=40 and p=.55.
> Compute the following:
> a) the mean and standard deviation of the random variable.
mean = &mu = n*pi
&sigma^2 = n*p*q = n*p*(1-p)
> b) the probability that x is 25 or greater.
You have to add a bunch of terms
P(x >= 25) = sum{n=25 to 40} {40!/[(n!)(15!)]}(0.55)^n(0.45)^(40-n)
= 0.2142
> c) the probability that x is 15 or less.
See "b", just sum from 0 to 15, P(x <= 15) = 0.0196
> d) the probability that x is between 15 and 25 inclusive.
P(15 <= x <= 25) = 0.8588 Again, same as above, just sum from 15 to 25.
> 3. Two-litre plastic bottles used for bottling cola are shipped in
> lots of 100. Suppose the lots are 5 percent defective. Some bottles > leak, some are too small, and so forth.
> a) In the sample of 100, how many of the bottles contains 8 or more > defectives?
They have essentially told you the mean is 5 or the probability of
the bottle being bad is 5% (mu = N*P, N = 100 => P = 0.05%).
You can use the diff. of the binomial from here on with P = 0.05.
(There i am defining a pass as a broken bottle).
P(8 or more) = 1 - P(7 or less)
= 1 - sum(n=0,7) [(100!/(n!*(100-n)!)](0.05)^0.05(0.95)^0.95
= 0.2340
> b) Tell why this situation meets the binomial assumptions.
Its a pass/fail test. Either the bottle has a defect or not. In
these cases, you want a distribution that is binary in nature, ie
the binomial.
> c) What is the probability that a shipment of plastic bottles
> contains 8 or more defectives?
Isn't this the same as a?
> d) What is the probability that 8 and 10 bottles are defectives?
at the Prob for 8 and 10 only, P = 0.08158677208886
> e) What is the probability that there are exactly 8 defectives?
0.10602553736479 same as above, but only the n = 8 term counts
> f) What is the probability that there are no defectives?
0.00592052922033 n = 0 term only |
Buddy!!! You just made me miss finite & business math. Awesome job 
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