The Problem |
If 60% of all women are employed outside the home, find the probability that in a sample of 20 women, a). Exactly
15 are employed outside
of the home. |
c). More than
5 are not employed outside of the home.
Symbolic Solution
From the text of
the problem, we see that n = 20, the number in the sample. If we
call "success" as "women employed outside the home"
we have p = 60% = 0.6. Thus q =
1 - p = 0.4. At most 5 are NOT employed outside
of the home means more than five are employed outside of the home.
Thus we need to compute P(X > 5) or using complements to simplify the
problem, we get 1 - P( X
5).
The probability of at least 5 employed outside the home means does not directly fit our formula,
We would have to calculate P(x 5) as P(0) + P(1) + P(2) + P(3) + P(4) + P(5) and then subtract that sum from 1. That means that we would have to use the above formula 6 times and add up ALL the results. Instead we will use the calculator solution below.
From the text of the problem, we see that n = 20, the number in the sample. If we call "success" as "women employed outside the home" we have p = 60% = 0.6. We also have that x = 10.
There is a program on the TI84 that does the above formula for us. It is of the form: binomcdf(n, p, x). This program computes the cummulative frequency of
P(X x) or in this case, P( x 5).
On the home screen, type 1- and then press 2nd VARS [DISTR], arrow down to A:binomcdf( and press ENTER. Now we enter the parameters n, p, x. Type 20, for p: type .6, and for x value: type 5, and then press enter three times.
Thus, P( x > 5) = 1 - P(x 5 ) 0.998
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