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
The same would be the right one
There ar 4persons so each will pay
Only two persons have spent more money than everyone average i.e mia and Jasmin.
Now
- Mia has-28.47
- Jasmin has=20.99
Mia left more money so she saved most money.
A ^ 0 = 1.
This is the final solution.
Because the discriminant is less than zero, there are no real solutions in the equation.
