Question

Weightlifting. The Atlas BodyBuilding Company (ABC) sells "starter sets" of barbells that consist of one bar, two 20-pound weights, and four 5 -pound weights. The bars weigh an average of 10 pounds with a standard deviation of \(0.25\) pounds. The weights average the specified amounts, but the standard deviations are \(0.2\) pounds for the 20-pounders and \(0.1\) pounds for the 5 -pounders. We can assume that all the weights are normally distributed. a) \(\mathrm{ABC}\) ships these starter sets to customers in two boxes: The bar goes in one box and the six weights go in another. What's the probability that the total weight in that second box exceeds \(60.5\) pounds? Define your variables clearly and state any assumptions you make. b) It costs \(\mathrm{ABC} \$ 0.40\) per pound to ship the box containing the weights. Because it's an odd-shaped package, though, shipping the bar costs \(\$ 0.50\) a pound plus a \(\$ 6.00\) surcharge. Find the mean and standard deviation of the company's total cost for shipping a starter set. c) Suppose a customer puts a 20-pound weight at one end of the bar and the four 5 -pound weights at the other end. Although he expects the two ends to weigh the same, they might differ slightly. What's the probability the difference is more than a quarter of a pound?

Short answer

a) Probability: 18.74%. b) Mean cost: $35; Std Dev: $0.226. c) Probability difference > 0.25 lbs: 57.54%.

Step-by-step solution

Define Variables and Distributions (Part a)

Let's define random variables for the weights in the second box: - Let \( X \) be the weight of a 20-pound plate. Given \( E(X) = 20 \) and \( \sigma(X) = 0.2 \).- Let \( Y \) be the weight of a 5-pound plate. Given \( E(Y) = 5 \) and \( \sigma(Y) = 0.1 \).The second box contains two 20-pound plates and four 5-pound plates, so the total weight \( W \) of the second box can be expressed as:\[ W = 2X + 4Y \].Since \(X\) and \(Y\) are normally distributed and independent, \( W \) is also normally distributed.

Compute Mean and Standard Deviation of Second Box (Part a)

To find \( E(W) \) and \( \sigma(W) \), use the properties of expectation and variance:- \( E(W) = 2E(X) + 4E(Y) = 2(20) + 4(5) = 40 + 20 = 60 \).- \( \sigma^2(W) = (2^2)(\sigma^2(X)) + (4^2)(\sigma^2(Y)) = 4(0.2^2) + 16(0.1^2) = 0.16 + 0.16 = 0.32 \).- \( \sigma(W) = \sqrt{0.32} \approx 0.566 \).

Calculate Probability for Second Box (Part a)

We need to find the probability that \( W > 60.5 \) pounds.Convert \( W = 60.5 \) into a standard normal variable \( Z \):\[ Z = \frac{W - E(W)}{\sigma(W)} = \frac{60.5 - 60}{0.566} \approx 0.883 \].Using standard normal distribution tables, \( P(Z > 0.883) \approx 1 - 0.8126 = 0.1874 \).

Compute Shipping Cost Calculations (Part b)

Define the cost variables:- Shipping the weights: \( E(C_{weights}) = 0.40E(W) = 0.40 \times 60 = 24 \) dollars.- \( \sigma(C_{weights}) = 0.40 \times \sigma(W) \), therefore \( \sigma(C_{weights}) = 0.40 \times 0.566 \approx 0.226 \).For the bar:- \( E(C_{bar}) = 0.50E(Bar) + 6 = 0.50 \times 10 + 6 = 11 \) dollars. - \( \sigma(C_{bar}) = 0.50 \times \sigma(Bar) = 0.50 \times 0.25 = 0.125 \).Total mean cost \( E(T) = E(C_{weights}) + E(C_{bar}) = 24 + 11 = 35 \) dollars.Total variance of cost \( \sigma^2(T) = \sigma^2(C_{weights}) + \sigma^2(C_{bar}) \approx (0.226)^2 + (0.125)^2 \approx 0.051076 \).Thus, \( \sigma(T) = \sqrt{0.051076} \approx 0.226 \).

Weight Distribution Calculation (Part c)

Define the weight difference variable, \( D = X - 4Y \), where \( X \) is one 20-pound weight, and \( 4Y \) is the total for the four 5-pound weights.- \( E(D) = E(X) - E(4Y) = 20 - 20 = 0 \).- \( \sigma^2(D) = \sigma^2(X) + 16\sigma^2(Y) = 0.2^2 + 16 \times 0.1^2 = 0.04 + 0.16 = 0.2 \).- \( \sigma(D) = \sqrt{0.2} \approx 0.447 \).To find \( P(|D| > 0.25) \), use \( Z = \frac{D}{\sigma(D)} \):- \( P(D > 0.25) \approx 1 - \text{P}(Z < 0.559) \) where \( Z = \frac{0.25}{0.447} \approx 0.559 \).- Look up \( P(Z < 0.559) \approx 0.7123 \), thus \( P(Z > 0.559) \approx 0.2877 \).- Double this probability for both tails: \( 2 \times 0.2877 = 0.5754 \).

Key concepts

Normal Distribution

When we talk about a normal distribution in statistics, we're referring to a probability distribution that’s symmetric. Imagine a bell curve peaking at the mean. This shape means most data points cluster around the mean, and fewer lie at the outskirts.
A key feature of normal distributions is how they allow us to model real-world behaviors, like heights or test scores.
For example:

• The average weight of the weights in our exercise example follows a normal distribution.
• This distribution helps us determine probabilities of weights or costs falling within certain ranges.

Normal distribution charts often utilize a standard deviation which lets us calculate probabilities. With the normal distribution, the curve's spread is affected by this deviation, allowing for more accurate estimations.

Standard Deviation

Standard deviation is a key statistical tool for quantifying variability or dispersion in a set of numbers. In simple terms, it tells us how spread out the numbers are in a dataset from the mean.
For our weightlifting example:

• The standard deviation for the weights and bars varies. For instance, the bar has a standard deviation of 0.25 pounds.
• Each weight type has its own deviation, indicating differences in how closely the weights conform to their mean values.

The smaller the standard deviation, the closer the data points are to the mean. So when setting weights, knowing the standard deviation helps manufacturers ensure consistency. It's also crucial when calculating the likelihood of different outcomes, such as whether the second box’s weight exceeds a certain limit.

Mean

In statistics, the mean is perhaps the most straightforward concept: it’s the average. To get it, sum up all values and divide by the number of values. But it's vital in understanding data tendencies.
In the weightlifting problem:

• The mean weight of the bar is 10 pounds.
• For weights, even though they’re labeled predictably, knowing their actual mean confirms that the labels are accurate on average.

A deeper understanding of the mean helps predict outcomes like shipping costs. By combining means of different components, like weights and bars, we can forecast overall averages more effectively for larger operations. Comprehending mean is essential for analyzing tendencies and making informed predictions in probability and statistics.