10000C225 (.02)^225 (.98)^9775
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Using a TI-89 calculator, I got 0.005804750003.
This can also be done using the normal approximation to the binomial distribution (for large numbers of trials), if you do not have as advanced of a calculator.
Note that 10000C225 (.02)^225 (.98)^9775 is the binomial distribution probability of exactly 225 successes out of 10000 independent trials, when the success probability is 0.02.
In the normal approximation, think of 225 as the interval (224.5, 225.5).
The mean is np = 10000(0.02) = 200.
The standard deviation is sqrt(np(1-p)) = sqrt(10000(0.02)(0.98)) = 14.
Let X = number of successes. The normal approximation of the probability of exactly 225 successes is
P(224.5 < X < 225.5)
= P((224.5 - 200)/14 < Z < (225.5 - 200)/14), where Z is the z-score
= P(Z < (225.5 - 200)/14) - P(Z < (224.5 - 200)/14)
= P(Z < 1.82) - P(Z < 1.75)
= 0.9656 - 0.9599
= 0.0057, which is close to the calculator result 0.005804750003.
Lord bless you today!
This can also be done using the normal approximation to the binomial distribution (for large numbers of trials), if you do not have as advanced of a calculator.
Note that 10000C225 (.02)^225 (.98)^9775 is the binomial distribution probability of exactly 225 successes out of 10000 independent trials, when the success probability is 0.02.
In the normal approximation, think of 225 as the interval (224.5, 225.5).
The mean is np = 10000(0.02) = 200.
The standard deviation is sqrt(np(1-p)) = sqrt(10000(0.02)(0.98)) = 14.
Let X = number of successes. The normal approximation of the probability of exactly 225 successes is
P(224.5 < X < 225.5)
= P((224.5 - 200)/14 < Z < (225.5 - 200)/14), where Z is the z-score
= P(Z < (225.5 - 200)/14) - P(Z < (224.5 - 200)/14)
= P(Z < 1.82) - P(Z < 1.75)
= 0.9656 - 0.9599
= 0.0057, which is close to the calculator result 0.005804750003.
Lord bless you today!