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.

a. What is the distribution for the weights of one $25$-pound lifting weight? What is the mean and standard deviation?

b. What is the distribution for the mean weight of $100,25$-pound lifting weights?

c. Find the probability that the mean actual weight for the $100$ weights is less than $24.9$.

Short answer

(a) The mean and standard deviation are $25$and $0.577$

(b) The distribution for the mean weight is $\overline{X}~N(25,.0577)$.

(c) The Probability is$4.15\%$.

Step-by-step solution

Given Information Part(a)

According to the provided details, a manufacture produces $25$-pound weights. The lowest weight is $24$pounds and the highest weight is $26$ pounds. Each weight is equally likely and the sample size considered $100$ weights.

Explanation Part(a)

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:

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

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

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:

localid="1648803309552" ${\sigma }_x=\sqrt{\frac{(b-a{)}^2}{12}}$

localid="1648803315274" $=\sqrt{\frac{(26-24{)}^2}{12}}$

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

$=0.577$

Given Information Part(b)

According to the provided details, a manufacture produces $25$-pound weights. The lowest weight is $24$pounds and the highest weight is $26$ pounds. Each weight is equally likely and the sample size considered $100$ weights.

Explanation Part(b)

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

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

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

$\overline{X}~N(25,.0577)$

Given Information Part(c)

According to the provided details, a manufacture produces $25$-pound weights. The lowest weight is $24$pounds and the highest weight is $26$ pounds. Each weight is equally likely and the sample size considered $100$ weights.

Explanation Part(c)

To calculate the probability that $P(\Sigma X<24.9)$, 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 of the calculator to have the desired result.

Therefore, the probability that the mean actual weight for the 100 weights is less than 24,9 is approximately$4.15\%$