CLT question

lllll · #1 $Old$ November 22nd, 2008, 10:25 PM

A researcher samples 50 fish in each of two lakes. She sets a trap and records the number of fish with the specific species that get trapped. Past experience shows that this species gets trapped 10% of the time in lake A, while it is 20% in lake B. Calculate the probability that the difference between the sample portions will be within 0.1 of the difference between the true proportions.

$\overline{Y}-\mu = 0.1$

for the variance it would be:

$V[Y_A] = np(1-p) = 50\cdot 0.1 \cdot 0.9 = 4.5$
$V[Y_B] = np(1-p) = 50\cdot 0.2 \cdot 0.8 = 8$

$V[Y_A]+V[Y_B] = 4.5+8=12.5$ which I based on Normal distribution - Wikipedia, the free encyclopedia under Property 2

putting it all together I get:

$\frac{\overline{Y}-\mu}{\sqrt{\sigma^2/n}} = \frac{0.1}{\sqrt{12.5/100}} = 0.2828$

so $P(-0.2828 <Z <0.2828) = 0.7794$

but the answer in the back of the book is 0.8414, which would correspond to $P(-0.141 <Z <0.141)$