For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
it would be y = -12/5. Hes the work. hope this helps!
Well off the top of my head, (700,001),(700,002),(700,003), and (700,004) are 4 numbers that round to 700,00 when rounded to the nearest hundred thousand. But if you're looking for something a little more unique, then any number from 650,000 to 749,999 would round to to 700,000 when rounded to the nearest hundred thousand. I hope this helps!
The answer to the problem is 220