Suppose the prices of new snacks are normally distributed. If the population standard deviation is 14 dollars, what minimum sample size is needed to be 95% confident that the sample mean is within 4 dollars of the true population mean

The minimum sample size needed to be 95% confident that the sample mean is within 4 dollars of the true population mean is 47.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

If the population standard deviation is 14 dollars, what minimum sample size is needed to be 95% confident that the sample mean is within 4 dollars of the true population mean

This is n when . So

The minimum sample size needed to be 95% confident that the sample mean is within 4 dollars of the true population mean is 47.

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

answer

false

explanation

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