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 probability that the mean actual weight for the $100$ weights is greater than $25.2$.

Short answer

The value of probability that $P(\Sigma X>25.2)$ is approximately$.0003\%$.

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:

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

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

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

$=0.577$

Final Answer

To calculate the probability that $P\left(\sum X>25.2\right)$, use $Ti-83$calculator. For this, click on $2^{\text{nd}}$, then DISTR and then scroll down to the normal CDF option and enter the provided details. After this, click on ENTER button on the calculator to have the desired result.

Therefore, the value of probability that $P\left(\sum X>25.2\right)$is given as:

$=1-.99973$

$=.00027$