Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The
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Answer:
And we can find this probability with this difference:
And we can find the real value with the following excel code using the binomial distribution:
"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"
And we got 0.118 a very close value from the value obtained using the normal approximation
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
We need to check the conditions in order to use the normal approximation.
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
So then we can approximate the random variable as
And we want this probability:
We can use the z score formula given by:
And replacing we got:
And we can find this probability with this difference:
And we can find the real value with the following excel code using the binomial distribution:
"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"
And we got 0.118 a very close value from the value obtained using the normal approximation