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Welcome to the post class activity for probability, developed as part of the
MTStatPAL project at Middle Tennessee State University.
If you did the in-class activity for the binomial random variable, you dropped candies
on paper plates to model a problem about nursing retention. 75% of nurses leave
their job within the first year. If a hospital hires ten new nurses, it would like to know
what kind of nurse turnover numbers are to be expected. Let’s let the random variable X
be the number of nurses who leave after one year.
We know that X is a binomial random variable because it counts the number of successes
in a binomial experiment. A binomial experiment is a sequence of trials where the
number of trials is fixed, each trial is independent, each trial has only two outcomes,
and the probability of success doesn’t change. In the nursing example, we have 10 trials,
since there are 10 nurses. We will assume that the nurses make their job decisions
independently of each other. The two outcomes we are interested in are nurses
leaving their job and not leaving their job. And the probability of one nurse making
the decision to leave is 0.75.
During the in-class activity, you saw this table and graph, which shows that the
probability of exactly 6 nurses leaving the hospital is .146, or 14.6%. Let’s talk about
where the table came from.
The formula for the probability of getting exactly x number of successes in n trials of a
binomial experiment is this: Let’s look at each piece of this formula separately. Notice
the p raised to the x power. Because the trials are independent, and the
probability of getting one success is p, the probability of getting x successes is p
times p times p, x times, or p raised to the power x. Now look at the (1-p) to the n-x term.
Because the probability of a success is p, the probability of a failure is (1-p). And
because there are x successes, and n total trials, there are n-x failures. Finally, what
about this nCx term? This is the number of ways to choose which trials are the successes
and which are the failures.
If we want to calculate the probability that exactly 6 nurses leave within the first
year, we can use the formula, or we can use a graphing calculator. Press 2nd Vars to
get to the distributions menu. Scroll down to select binompdf . Then enter n, p, and x. For our
example, n is 10, p is .75, and x is 6. The probability that exactly 6 nurses leave in the
first year is 14.6%.
What if we want the probability that no more than 6 nurses leave the first year? That is, what
is the probability that the number of nurses who leave is less than or equal to 6?
Finding this probability requires adding the probability that exactly 0 leave, that
exactly 1 leave, that exactly 2 leave, etc, until we get to the probability that exactly 6 leave.
There are a couple of ways to do this on your calculator. We will use this way. Select the
binompdf choice and enter n and p as before. This time, though, instead of entering one number
for x, we enter a whole list of numbers: 0, 1, 2, 3, 4, 5, and 6. You can enter a curly
bracket by pressing 2nd and the parentheses key. We want to store the output of this
command in a list, so press store and L1 to enter the output in the list L1.
You can see the list of values in L1 that were stored there by our command. The first value,
9.5 times 10 ^ -7 is the probability that X is 0, the second is the probability that
X is 1, etc, until the last, which is the probability that X is 6. To sum these numbers,
press STAT and choose the CALC menu. Select 1-Var Stats, then press 2nd I for L1 and press
enter. The sum is the second item on the list. So the probability that no more than 6
nurses leave is 22.4 percent.