albiguguiismx

## Answered question

2022-09-11

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= 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= =0.3 ??
Many thanks in advance!

### Answer & Explanation

shosautesseleol

Beginner2022-09-12Added 16 answers

One is a normal distribution SD (zero.9) and the other binomial (zero.3). The hassle has taken a ordinary distribution, the quantity of cars, and is asking a query of a binomial nature, yes or no to ownership.

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