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

Find the $15$th percentile for the sums.

According to the provided details, 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 lowest 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 $50$ values will be $300$ and the sum of the maximum $50$ values will be$500$.

To calculate ${15}^{\text{th}}$percentile for the sums, use $Ti-83$calculator. For this, click on $2^{\text{nd}}$, then DISTR, and then scroll down to the invnorm option and enter the provided details. After this, click on ENTER button of the calculator to have the desired result. The screenshot is given as below: