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About 25% of the people called upon for jury duty give an excuse not to serve. If 12 ppl are called for duty...
a) What is the prob. that 8 or more will not be able to serve on the jury?
this is wat i did..
1-(12|0 ,.25^0, .75^12)+ ....and do the binomial for 0,1,2,3,4,5,6,7 and add then up. For my answer, i got .649.
__Is this right?
b)How many people must the jury commissioner contact to be 95% sure of finding at least 12 ppl who are available to serve?
__How would I do this?
a) Your answer is wrong.Quote:
Originally Posted by Morgan82
Let X be the random variable number of people who can't serve.
Why not just calculate Pr(X = 8) + Pr(X = 9) + Pr(X = 10) + Pr(X = 11) + Pr(X = 12) .......? Less to calculate than how I think you've tried to do it .....
b) Let Y be the random variable number of people who can serve.
Then Y ~ Binomial(n = ?, p = 3/4).
You require the smallest integer value of n such that.
The easiest approach to solving this inequality is to get the technology available to you to do trial-and-error - perhaps setting up a table of probabilities with n.
Using my TI-89 and defining Y1 = binomcdf(x, 0.75, 12, x) and using a table of values for Y1, I easily get that n = 20 ........
Certainly I think it will be a hopeless task trying to solve it without using technology that lets you easily calculate the cumulative probability => you ought to have access to such technology => you should apply that technology to part a ......... !