91Ó°ÊÓ

Designing Manholes According to the website www.torchmate.com, "manhole covers must be a minimum of 22 in. in diameter, but can be as much as 60 in. in diameter" Assume that a manhole is constructed to have a circular opening with a diameter of 22 in. Men have shoulder breadths that are normally distributed with a mean of 18.2 in. and a standard deviation of 1.0 in. (based on data from the National Health and Nutrition Examination Survey). a. What percentage of men will fit into the manhole? b. Assume that the Connecticut's Ever source company employs 36 men who work in manholes. If 36 men are randomly selected, what is the probability that their mean shoulder breadth is less than 18.5 in. Does this result suggest that money can be saved by making smaller manholes with a diameter of 18.5 in.? Why or why not?

Step by step solution

01

Understanding the Problem

We need to determine the percentage of men that fit through a manhole, given their shoulder breadths follow a normal distribution with a mean of 18.2 in and standard deviation of 1.0 in. The manhole has a diameter of 22 in.

02

Standardizing Shoulder Breadths

Convert the shoulder breadth that will fit through the manhole into a z-score using the formula: o = (x - μ) / σo = (22 - 18.2) / 1o = 3.8

03

Using Standard Normal Distribution

Using the z-table, find the percentage of men corresponding to a z-score of 3.8. The value approaches nearly 1 (or 100%) indicating nearly all men fit.

04

Understanding Part b

We need to determine the probability that the mean shoulder breadth of 36 men is less than 18.5 in. Here, the sample mean follows a normal distribution.

05

Central Limit Theorem

For a sample size of 36, the standard error (SE) is calculated as: o = σ / o = 1 / o = 0.1667

06

Standardizing Sample Mean

Convert the shoulder breadth mean (18.5 in.) to a z-score using the formula for a sample mean: o = (x̄ - μ) / SEo = (18.5 - 18.2) / 0.1667o = 1.8

07

Using Standard Normal Distribution for Sample

Using the z-table, find the probability corresponding to a z-score of 1.8. The cumulative probability for z = 1.8 gives approximately 95.9%.

08

Conclusion

With a 95.9% probability of shoulder breadths being less than 18.5 in, it is not feasible to reduce manhole size to 18.5 in. Nearly 4.1% would not fit, leading to practical issues.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution

In statistics, the normal distribution is a continuous probability distribution. It is sometimes called the 'bell curve' because of its bell-shaped appearance. The mean (average) determines the center of the distribution, while the standard deviation determines the spread. When working with real-world data, many natural phenomena follow this type of distribution. For our manhole example, men's shoulder breadths are distributed normally with a mean of 18.2 inches and a standard deviation of 1.0 inch. This distribution helps us understand and predict the proportion of men who can fit through a manhole of a given size.

Z-Score

A z-score tells us how many standard deviations an element is from the mean. In other words, it helps to standardize different data points for comparison. The formula for calculating a z-score is:
\( z = \frac{X - \mu}{\sigma} \).
Here, X is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For our manhole problem, we standardized the shoulder breadth of 22 inches to find out how it compares to the average shoulder breadth of 18.2 inches with a standard deviation of 1.0 inch.
We calculated a z-score of 3.8 using the formula \( z = \frac{22 - 18.2}{1} = 3.8 \). This high z-score indicates that 22 inches is significantly larger than the average shoulder breadth.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that the sampling distribution of the sample mean will be normally distributed, or approximately normal, if the sample size is sufficiently large. This holds true regardless of the shape of the population distribution. In simpler terms, if we take enough samples and calculate their means, those means will form a normal distribution.
For instance, in our problem, when considering the mean shoulder breadths of 36 men, the CLT allows us to treat the sampling distribution of the mean as normally distributed. This is crucial for making further probability calculations.

Standard Error

The standard error (SE) measures the variability of the sample mean estimate of a population mean. It is calculated using the formula:
\( SE = \frac{\sigma}{\sqrt{n}} \),
where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
In our example, the standard deviation \( \sigma \) is 1.0 inch and the sample size \( n \) is 36. Thus, the standard error is calculated as:
\( SE = \frac{1}{\sqrt{36}} = 0.1667 \).
The small value of 0.1667 indicates that the mean shoulder breadth of groups of 36 men will vary only slightly from the population mean.

Probability Calculation

Probability calculation in this context involves determining the likelihood of sample mean shoulder breadths being below a certain threshold. After establishing the z-score, we use the z-table to find the corresponding probability.
For the shoulder breadth less than 18.5 inches, we calculated a z-score using:\( z = \frac{\bar{x} - \mu}{SE} = \frac{18.5 - 18.2}{0.1667} = 1.8 \).
Next, the z-table gives us the cumulative probability. For a z-score of 1.8, the cumulative probability is approximately 95.9%. This means there is a 95.9% chance that the mean shoulder breadths of the 36 men will be less than 18.5 inches.
This result indicates that a smaller manhole size is not suggested, as it would exclude a significant proportion (4.1%) of men.