A certain town has 25,000 families. These families own 1.6 cars, on the average; the SD is 0.90. And 10% of them have no cars at all. As part of an opinion survey, a simple random sample of 1,500 families is chosen. What is the chance that between 9% and 11% of the sample families will not own cars?

So I'm looking at the solution of this question it shows that to get the Standard Error of the Sum is SD= $\sqrt{0.9\cdot \text{}0.1}$ However, I thought the formula to get SE is:

SE= SD of the Box $\sqrt{NumberofDraws}$

In this case, isn't the Box of the SD given at 0.9? Why do we still use the bootstrap method to get SD= $\sqrt{0.9\cdot \text{}0.1}$=0.3 ??

Many thanks in advance!

So I'm looking at the solution of this question it shows that to get the Standard Error of the Sum is SD= $\sqrt{0.9\cdot \text{}0.1}$ However, I thought the formula to get SE is:

SE= SD of the Box $\sqrt{NumberofDraws}$

In this case, isn't the Box of the SD given at 0.9? Why do we still use the bootstrap method to get SD= $\sqrt{0.9\cdot \text{}0.1}$=0.3 ??

Many thanks in advance!