Question

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

Find the $90$th percentile for the sums.

Short answer

The $90$th percentile for the sum is$410.4512$.

Step-by-step solution

Given Information

Given a sample of $50$ variables with uniform distribution $U(6,10)$.

Calculation

If variable $Z$has a uniform distribution on the interval $[6,10]$, then the mean of $Z$is $8$and the standard deviation is $1.1547$.

Thus the allocation of the sums is:

$N(50·8,\sqrt{50}·1.1547)=N(400,8.165)$

Finding the $90$th percentile:

$P(\Sigma x<z)=0.9$

$P\left(\frac{\Sigma x-400}{8.165}<\frac{z-400}{8.165}\right)=0.9$

If variable $Z$has a standard normal distribution, then

$P(Z<1.28)=0.9$

Thus,

$\frac{z-400}{8.165}=1.28$

$z-400=10.4512$

$z=410.4512$.