Answer:
![Sum=4*[\frac{n^2-1}{n^2}]](https://tex.z-dn.net/?f=Sum%3D4%2A%5B%5Cfrac%7Bn%5E2-1%7D%7Bn%5E2%7D%5D)
For n=10:
Sum=3.96
For n=100:
Sum=3.9996
For n=1000:
Sum=3.999996
For n= 10000:
Sum=3.99999996
Step-by-step explanation:
Formula:

Rearranging the above formula:
Eq (1)
According to summation formula:

Putt these in Eq (1), and we will get:
![=\frac{12}{n^3}[\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}]\\Taking\ n\ as\ common\\=n*\frac{12}{n^3}[\frac{(n+1)(2n+1)}{6}-\frac{(n+1)}{2}] \\=\frac{12}{n^2}*[\frac{(n+1)(2n+1)}{6}]-\frac{12}{n^2}*[\frac{(n+1)}{2}] \\=\frac{2*(n+1)(2n+1)}{n^2}-\frac{6(n+1)}{n^2}\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B12%7D%7Bn%5E3%7D%5B%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D-%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%5C%5CTaking%5C%20n%5C%20as%5C%20common%5C%5C%3Dn%2A%5Cfrac%7B12%7D%7Bn%5E3%7D%5B%5Cfrac%7B%28n%2B1%29%282n%2B1%29%7D%7B6%7D-%5Cfrac%7B%28n%2B1%29%7D%7B2%7D%5D%20%5C%5C%3D%5Cfrac%7B12%7D%7Bn%5E2%7D%2A%5B%5Cfrac%7B%28n%2B1%29%282n%2B1%29%7D%7B6%7D%5D-%5Cfrac%7B12%7D%7Bn%5E2%7D%2A%5B%5Cfrac%7B%28n%2B1%29%7D%7B2%7D%5D%20%5C%5C%3D%5Cfrac%7B2%2A%28n%2B1%29%282n%2B1%29%7D%7Bn%5E2%7D-%5Cfrac%7B6%28n%2B1%29%7D%7Bn%5E2%7D%5C%5C)
Taking
as common:

After more simplifying,
Now ,for n=10:
![Sum=4[\frac{(10^{2})-1}{10^{2}}]\\Sum=3.96](https://tex.z-dn.net/?f=Sum%3D4%5B%5Cfrac%7B%2810%5E%7B2%7D%29-1%7D%7B10%5E%7B2%7D%7D%5D%5C%5CSum%3D3.96)
For n=100:
![Sum=4[\frac{(100^{2})-1}{100^{2}}]\\Sum=3.9996](https://tex.z-dn.net/?f=Sum%3D4%5B%5Cfrac%7B%28100%5E%7B2%7D%29-1%7D%7B100%5E%7B2%7D%7D%5D%5C%5CSum%3D3.9996)
For n=1000
![Sum=4[\frac{(1000^{2})-1}{1000^{2}}]\\Sum=3.999996](https://tex.z-dn.net/?f=Sum%3D4%5B%5Cfrac%7B%281000%5E%7B2%7D%29-1%7D%7B1000%5E%7B2%7D%7D%5D%5C%5CSum%3D3.999996)
For n=10000:
![Sum=4[\frac{(10000^{2})-1}{10000^{2}}]\\Sum=3.99999996](https://tex.z-dn.net/?f=Sum%3D4%5B%5Cfrac%7B%2810000%5E%7B2%7D%29-1%7D%7B10000%5E%7B2%7D%7D%5D%5C%5CSum%3D3.99999996)
The best approach to this is to use the formular
Standard error = standard deviation / sqrt( sample size)
standard deviation = 0.75 hours
sample size = 35
Hence, Standard error = 0.75 /sqrt(35) = 0.126 = 0.13