# Normal Approximation

1. Jun 7, 2006

### Ceresu

Hello ~

I be in dire need of help with this problem because I fell asleep in math class. Could anyone be so kind as to thoroughly guide me through the following problem?

"A school has enrolled the same number of boys and girls. Two hundred students are selected at random to participate in a survey. Use a normal approximation to the binomial distribution to estimate the probability that at least 110 girls will be selected."

Thanks!
-Mathematically-challenged student

2. Jun 7, 2006

### HallsofIvy

If you are selecting 200 students and boys and girls both have probability 1/2 of being selected then the expected number of girls is (1/2)(200)= 100 (in general, with a binomial distribution with probabilities p and 1-p and number n, the expected value is np). The standard deviation is
$\sqrt{200(\frac{1}{2})(\frac{1}{2})}= 5\sqrt{2}$ (in general, it is
$\sqrt{np(1-p)}$).
"At least 110 girls" means, since the normal distribution does not apply only to integers, "greater than or equal to 109.5" (that's the "half integer correction"). Use a table of normal distribution values to determine the probability that the standard "z" score is greater than $\frac{109.5- 100}{5\sqrt{2}}=1.34$.

Last edited by a moderator: Jun 10, 2006
3. Jun 9, 2006

If I am not mistaken, the standard deviation used in this question should be $\sqrt{np(1-p)}=\sqrt{200(\frac{1}{2})(\frac{1}{2})}$.