Let X be the random variable for the number of women to be selected.
Binomial parameters:
p = 0.5
q = 1 - 0.5 = 0.5
n = 20
We are asked for P(X
> 12), and we can use P(X
> 12) = 1 - P(X < 12).
We're approximating a discrete distribution (binomial) with a continuous one (normal), so applying a continuity correction means we're looking for 1 - P(X < 11.5).
Normal distribution parameters:
\(\displaystyle \mbox{ \mu = np = (20)(0.5) = 10 }\) (np
> 5 and nq
> 5 requirements for the approximation are satisfied)
\(\displaystyle \mbox { \sigma = \sqrt{npq} = \sqrt{(20)(0.5)(0.5)} = \sqrt{5}}\)
\(\displaystyle \mbox{ Z = \frac{ X - \mu}{\sigma} = \frac{11.5 - 10}{\sqrt{5}} \approx 0.671}\)
A Z-score of 0.671 corresponds to a probability of 0.2489 in the normal tables.
Sketch the information on a bell curve to get the correct probability:
Code:
0.2489
*** /
0.5 : |/_
\ * : *
_\| : :
* : : *
: :
* - - - - - - - - - - - - - - - - - *
10 11.5
So P(X < 11.5) = 0.5 + 0.2489 = 0.7489
And we have P(X
> 12) = 1 - P(X < 12) = 1 - 0.7489 = 0.2511
If you have binomial tables which go up to n=20, although ultimately making the exercise pointless, you can compare the approximation with the actual.