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.

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  1. 100 trials
    probability of success = .9
    probability of failure = .1
    we need P(86)
    etc down to

    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!)]
    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

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  2. I get .5393

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