A simple random sample of 500 household in California found that 121 had no car, 172 had onecar, and 207 had two or more cars. Estimate the percentage of households in California with oneor more cars and attach a standard error to the estimate. If this is not possible, explain why not.

We know that out of 500 households;

172 had one car.

207 had two or more cars.

Then the total number that has one or more cars is:

172 + 207 = 379.

Then if 500 is our 100%, we have that 379 represents:

(376/500)*100% = 75.8%

Then the estimate of the percentage of households in California that have one or more cars is 75.8%.

Now, to calculate the standard error of this estimate we need the number of the sample, that we know, and the mean deviation.

The problem here is that we can not really calculate the mean deviation, as we do not really know how many cars have a lot of these people. So we really can not calculate the standard error.

Let's make an estimation, and see why it does not work:

But we can think in the next system.

379 have at least one car.

121 do not have a car.

Then, the "mean" is:

if having one or more cars counts as a 1

not having a car counts as a 0.

M = (379*1 + 121*0)/500 = 0.758

The standard deviation will be:

σ =√((1/500)*(1 - 0.758)^379*(0 - 0.758)^121) = 4*10^-126

This is a really small number, and the standard error is smaller than this (because we divide by square root of 500) Then we can conclude that this estimation of the standard error does not add any useful information to our math.

If we knew the exact number of cars that the people with two or more cars have, we could calculate the standard error of the estimation.

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet