Probability with the Z-score?
Suppose in a recent national election 40% of the voters were men. If a random sample of 25 voters were selected, what is the probability that more than 10 but less than 20 were women?
I think I know my z-score is 1.53, but how do I get that z-score by going through the formulas? Would I use the sample mean z-score or population?
Show me please :-)
I really, Reealllly appreciate your assistance on this question. I have all of these formulas and can't seem to figure out, when to use the darn things! The calculations as well as showing your work was magtastic!
- GuyLv 77 years agoFavorite Answer
Are you supposed to use the normal approximation to the binomial?
60% of the voters were women. The question asks for "more than 10 but less than 20", which translates to "between 11 and 19, inclusive".
The actual probability would be (using a TI-83): binomcdf(25,.6,19) - binomcdf(25,.6,10) = .9362462771
Now let's do the normal approximation. Remember to use the continuity correction.
mean = .6(25) = 15
standard deviation = sqrt(25*.6*.4) = 2.45
P(10.5 < x < 19.5)
= P[(10.5 - 15)/2.45 < z < (19.5 - 15)/2.45]
= P(-1.84 < z < 1.84)
= normalcdf(-1.84,1.84) = .9342318831
As you can see, the normal approximation of .9342 is very close to the actual probability of .9362.