Question
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.
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.
Answers
Damon
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!)]
============================
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
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!)]
============================
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
Damon
I get .5393