A uniform distribution has a minimum of six and a maximum of ten. A sample of $50$is taken.

Find the third quartile for the sum.

According to the provided attributes, a uniform distribution has the lowest value is 6 and the highest value is $10$. The sample size taken is$50$.

The uniform distribution of the lower value is $6$and the highest value is $10$given as:

$X~\mathrm{U}(\mathrm{a},\mathrm{b})$

$X~\mathrm{U}(6,10)$

The mean of the uniform distribution is given as:

${\mu }_x=\frac{a+b}{2}$

$=\frac{6+10}{2}$

$=8$

And the standard deviation is given as:

${\sigma }_X=\sqrt{\frac{(b-a{)}^2}{12}}$

$=\sqrt{\frac{(10-6{)}^2}{12}}$

$=\sqrt{\frac{16}{12}}$

$=1.154$

The distribution of mean is given as:

$\overline{X}~N\left({\mu }_X,{\sigma }_x/\sqrt{n}\right)$

$\overline{X}~N(8,1.1547/\sqrt{50})$

$\overline{X}~N(8,.1632)$

And, the distribution of sum of $50$values is given as:

$\sum X~N\left(n{\mu }_X,n{\sigma }_X/\sqrt{n}\right)$

$\overline{X}~N(8\times 50,1.1547\times 50/\sqrt{50})$

$\overline{X}~N(400,8.16)$

The sum of the minimum of $50$ values will be $300$, and the sum of the maximum $50$ values will be $500$.

To compute the third quartile for the sums, use the $Ti-83$calculator. For this, click on $2^{\text{nd}}$, then DISTR, and then scroll down to the invnorm choice and enter the provided details. After this, click on ENTER button of the calculator to have the expected result.