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 90th percentile for the total weight of the 100 weights.

Short answer

The${90}^{\text{th}}\text{percentile for the mean weight for}100,25\text{-pounds weights is}2507.39\text{.}$

Step-by-step solution

Given Information

The distribution for weights of 100, 25 -pound lifting weight will follow the uniform distribution because the weights are equally likely. Thus, the uniform distribution of the lowest 2400 pounds and the highest 2600 pounds weight is given as:

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

$X-\mathrm{U}(2400,2600)$

Explanation

The mean of the uniform distribution is given as:

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

$=\frac{2400+2600}{2}$

=2500

And the standard deviation is given as:

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

$=\sqrt{\frac{(2600-2400{)}^2}{12}}$

$=\sqrt{\frac{40000}{12}}$

=57.735

The Distribution of Mean weight

The distribution of mean weight of 100,25 -pounds is given as follow:

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

$\overline{X}~N(2500,57.735/\sqrt{100})$

$X~N(2500,5.7735)$

Calculation

To calculate ${90}^{\text{th}}$percentile for the mean weight for 100, 25 -pounds 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 a calculator to have the desired result. The screenshot is given as below