Question

46. Draw the graph from Exercise 7.45

Short answer

The graph from Exercise 7.45 is given below:

Step-by-step solution

Given Information

Because the weights are equally likely, the distribution for 100, and 25-pound lifting weights will follow the uniform distribution. As a result, the uniform distribution of the lowest 2400 pound weight and the highest 2600 pound weight is:

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

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

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 of 100,25 -pounds is given as:

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

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

$X~N(2500,5.7735)$

Calculation

Use the Ti-83 calculator to find the 90th percentile for the mean weight for 100,25-pound weights. To do so, go to 2nd, then DISTR, and then scroll down to the Invnorm option and fill in the required information. After that, press the calculator's ENTER button to get the desired result. The following is a screenshot:

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