b). For exactly 15 are employed outside 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. The probability of at least 10 employed outside the home means does not directly fit our formula,

We would have to calculate P(x 10) as P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10). That means that we would have to use the above formula 11 times and add up ALL the results. Instead we will use the calculator solution below.

Calculator 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. We also have that x = 10.

There is a program on the TI83/84 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 10).

Press 2nd VARS [DISTR], arrow down to A:binomcdf( and press ENTER. Now we enter the parameters n, p, x. Type 20,.6, 10) and then press enter.

Thus, P(x 10 ) .245