Question

A manufacturer produces $25$-pound lifting weights. The lowest actual weight is $24$pounds, and the highest is $26$pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of $100$weights is taken.

Find the ${90}^{\text{th}}$percentile for the mean weight for the $100$ weights.

Short answer

The${90}^{\text{th}}$ percentile for the mean weight for role="math" localid="1648563187122" $100$ weights is$25.07$.

Step-by-step solution

Given Information

The distribution for weights of one $25$pound lifting weight will follow the uniform distribution because the weights are equally likely. Thus, the uniform distribution of the lowest $24$pounds and the highest $26$pounds weight are given as:

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

$X~\mathrm{U}(24,26)$

Explanation

The mean of the uniform distribution is given as:

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

$=\frac{24+26}{2}$

$=25$

And the standard deviation is given as:

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

$=\sqrt{\frac{(26-24{)}^2}{12}}$

$=\sqrt{\frac{4}{12}}$

$=0.577$

The distribution of the mean weight of $100$$,25$pounds is given as:

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

$\overline{X}-N(25,.5777/\sqrt{100})$

$X-N(25,0577)$

Final Answer

To calculate ${90}^{\text{th}}$percentile for the mean weight for $100$weights, 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.

Therefore, the ${90}^{th}$percentile for the mean weight for the 100 weights is$25.0739$