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Chapter 4: Problem 57

The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter (a) What is the probability that a sample's strength is less than \(6250 \mathrm{Kg} / \mathrm{cm}^{2} ?\) (b) What is the probability that a sample's strength is between 5800 and \(5900 \mathrm{Kg} / \mathrm{cm}^{2}\) (c) What strength is exceeded by \(95 \%\) of the samples?

Short Answer

Expert verified
(a) 99.38%, (b) 13.59%, (c) 5853.5 kg/cm虏

Step by step solution

01

Identify the Given Parameters

We are given the mean \( \mu = 6000 \) kg/cm虏 and the standard deviation \( \sigma = 100 \) kg/cm虏 for the normal distribution of cement strength.
02

Convert to Z-score for Part (a)

For part (a), we need to find the probability of a sample's strength being less than 6250 kg/cm虏. We convert this value into a Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X = 6250 \) kg/cm虏. Thus, \( Z = \frac{6250 - 6000}{100} = 2.5 \).
03

Calculate Probability for Part (a)

Using the standard normal distribution table, we find the probability that \( Z < 2.5 \). This corresponds to approximately 0.9938, meaning the probability is about 99.38%.
04

Convert to Z-scores for Part (b)

For part (b), find the probability that a sample's strength is between 5800 and 5900 Kg/cm虏. First, we calculate the Z-scores: \( Z_1 = \frac{5800 - 6000}{100} = -2 \) and \( Z_2 = \frac{5900 - 6000}{100} = -1 \).
05

Calculate Probability for Part (b)

Using the Z-table, find the probabilities for \( Z_1 = -2 \) (about 0.0228) and \( Z_2 = -1 \) (about 0.1587). The probability of strength between 5800 and 5900 kg/cm虏 is the difference: \( 0.1587 - 0.0228 = 0.1359 \), or 13.59%.
06

Determine Z-score for Part (c)

For part (c), determine the strength exceeded by 95% of samples (where only 5% exceed). This corresponds to a Z-score of approximately -1.645 (from Z-table, which provides P(Z < z) = 0.05).
07

Solve for Strength in Part (c)

Using the Z-score, solve for the strength \( X \) using \( X = Z \cdot \sigma + \mu = -1.645 \cdot 100 + 6000 = 5853.5 \) kg/cm虏.

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Key Concepts

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

Compressive Strength
Compressive strength is a critical measure in engineering. It refers to the ability of a material, in this case, cement, to withstand loads tending to reduce size. When we talk about compressive strength of cement, we're referring to how much load the cement can bear before it fails.
In our exercise, the mean compressive strength is given as 6000 kg/cm虏 with a standard deviation of 100 kg/cm虏. This means:
  • The average cement sample can endure up to 6000 kg/cm虏.
  • The variation in strength between samples is typically around 100 kg/cm虏.
Understanding the compressive strength helps engineers design structures that can sustain anticipated loads without experiencing failure.
Z-score Calculation
The Z-score is a statistical measure that describes a value's position within a distribution. It is particularly handy when you need to determine how far away a given data point is from the mean, measured in units of standard deviation.
To calculate the Z-score, you use the formula:

\[ Z = \frac{X - \mu}{\sigma} \]
  • \(X\) is the value you are examining.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
In our exercise, to find the probability of a sample having a compressive strength less than 6250 kg/cm虏, we first convert 6250 kg/cm虏 into a Z-score. Calculating this gives us a Z-score of 2.5. This tells us that 6250 kg/cm虏 is 2.5 standard deviations above the mean.
Probability Distribution
A probability distribution depicts all possible outcomes and their associated probabilities. For continuous data like compressive strength, we often use the normal distribution, which is bell-shaped and symmetric around the mean.
In the context of normal distribution, the Z-score provides insight into where a particular value falls with respect to the overall distribution.
In our example:
  • To find the probability of a strength less than 6250 kg/cm虏, we look for the area to the left of the Z-score of 2.5 in the standard normal distribution table. This area represents the probability.
  • Similarly, for strengths between 5800 and 5900 kg/cm虏, the calculation involves finding probabilities between two Z-scores: \(-2\) and \(-1\).
  • The area between them provides the likelihood of a sample's strength falling within that range.
Understanding probability distributions, especially normal distribution, is essential to predicting outcomes and making informed engineering decisions.
Statistics for Engineers
Statistics offers invaluable tools for engineers, providing ways to make data-driven decisions and predictions. In engineering, understanding the variability and expected performance of materials, like cement, is crucial.
This is why:
  • Statistics help in determining the probability of a material meeting specific criteria, as seen when calculating the probability of different compressive strengths.
  • Z-scores and probability distributions allow engineers to quantify and manage risks related to material performance.
  • For this exercise, using these statistical principles, an engineer can decide, with a certain level of confidence, whether a material will perform adequately under given conditions.
Overall, statistics not only guide engineers in design and analysis but also help in ensuring safety and reliability in construction and manufacturing.

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