100 trials
probability of success = .9
probability of failure = .1
we need P(86)
P(87)
P(88)
etc down to
P(91)
This is a binary distribution problem
Then add them all
I will do one of them
P(86) = 100C86 * .9^86 * .1^(100-86)
100C86 = 100! /[ 86!(14!)]
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HOWEVER
as you get a large number of trials the binomial distribution converges to a normal distribution with
mean = n p
in this case mean = 100*.9 = 90
nd sigma = sqrt (n p (1-p) )
sqrt (9) = 3
well that was easy
now go to a normal distribution table
look for 86 to 91 with a mean of 90 and a sigma of 3
try
http://davidmlane.com/hyperstat/z_table.html
I have gotten the concept of probability but the following question has really stumped me. Help please.
The probability that a tomato seed will germinate is 0.9. Estimate the probability that of 100 randomly selected seeds, at least 86, but no more than 91 of them will germinate.
2 answers
I get .5393
1 answer