Problem 71 - Applied Statistics and Probability for Engineers

Question

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?

Step-by-Step Solution

Answer

(a) Probability is 0.9938. (b) Probability is 0.1359. (c) Strength exceeded by 95% of samples is 5840.5 kg/cm².

Step 1: Understand the Problem

The compressive strength is normally distributed with mean \( \mu = 6000 \ \text{kg/cm}^2 \) and standard deviation \( \sigma = 100 \ \text{kg/cm}^2 \). We have to find the probabilities related to this distribution.

Step 2: Find Z-score for Part (a)

For question (a), which requires the probability of a sample strength being less than 6250 kg/cm², calculate the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X = 6250 \), \( \mu = 6000 \), \( \sigma = 100 \). Hence, \( Z = \frac{6250 - 6000}{100} = 2.5 \).

Step 3: Calculate Probability for Part (a)

Using the standard normal distribution table, find the probability that \( Z \le 2.5 \). This is approximately 0.9938. Therefore, the probability that a sample's strength is less than 6250 kg/cm² is 0.9938.

Step 4: Find Z-scores for Part (b)

For question (b), calculate the Z-scores for the two values: 5800 kg/cm² and 5900 kg/cm².\[ \begin{align*} Z_{5800} &= \frac{5800 - 6000}{100} = -2.0, \ Z_{5900} &= \frac{5900 - 6000}{100} = -1.0. \end{align*} \]

Step 5: Calculate Probability for Part (b)

Use the standard normal distribution table to find the probabilities: \[ P(Z \le -2.0) \approx 0.0228 \] and \[ P(Z \le -1.0) \approx 0.1587. \] The probability that the sample strength is between 5800 and 5900 kg/cm² is \( P(-2.0 < Z < -1.0) = 0.1587 - 0.0228 = 0.1359 \).

Step 6: Determine Z-score for Part (c)

For question (c), determine the Z-score such that 95% of samples exceed a certain strength. Use standard normal distribution properties, where 5% is left in the upper tail. From the Z-table or using standard distribution, \( P(Z < z) = 0.05 \) gives \( z \approx -1.645 \).

Step 7: Calculate Strength Exceeded by 95% for Part (c)

Use the Z-score of -1.645 (since it's the lower tail, we negate it to find the cutoff point), and solve for \( X \) using the formula \( Z = \frac{X - \mu}{\sigma} \):\[ \begin{align*} -1.645 &= \frac{X - 6000}{100}, \ X &= 6000 - 1.645 \times 100 = 5840.5. \end{align*} \] Thus, 95% of the samples have strength greater than 5840.5 kg/cm².

Key Concepts

Normal Distribution

A normal distribution is a type of continuous probability distribution for a real-valued random variable. It is one of the most important concepts in statistics and is often referred to as a Gaussian distribution. The graph of a normal distribution is bell-shaped and symmetric around its mean.
The normal distribution is defined by two parameters:

• Mean (\( \mu \)) - This is the central value around which the distribution is centered.
• Standard deviation (\( \sigma \)) - This measures the spread or variability of the distribution.

In a practical scenario like compressive strength analysis of cement, understanding that the data follows a normal distribution helps in predicting and making informed decisions about quality and safety.

Z-Score Calculation

Z-scores are a standardized way of expressing values from a normal distribution. To calculate a Z-score, you use the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where

• \( X \) is the value in the dataset,
• \( \mu \) is the mean, and
• \( \sigma \) is the standard deviation.

The Z-score tells you how many standard deviations an element is from the mean.
A positive Z-score indicates the value is above the mean, while a negative Z-score indicates it is below the mean.
For instance, when calculating the Z-score for a cement sample strength of 6250 kg/cm², you can identify how far from the average strength this value lies. This helps assess the probability of such a strength occurring within the given distribution.

Probability Concepts

Probability is a fundamental concept in statistics, representing the likelihood of an event occurring. In the context of normal distributions, probability is used to determine how likely a given Z-score (or range of scores) is.
This is done by using a standard normal distribution table.

• For a single Z-score, the table can show the probability that the observed value is below that Z-score.
• For a range, like between two Z-scores, you measure the probability of the value falling between those scores by finding the probabilities for each and subtracting.

Understanding these concepts allows us to calculate specific probabilities, such as the likelihood of a cement sample having a particular compressive strength or falling within a specified strength range.

Compressive Strength Analysis

Compressive strength analysis involves evaluating the ability of a material, like cement, to withstand loads tending to reduce size. It is crucial in fields like civil engineering, where material strength determines construction stability and safety.
In statistical terms, for samples that follow a normal distribution, this analysis can highlight the expected range of strengths.

• Determining probabilities for specific strength ranges helps in quality control, ensuring that produced materials meet safety norms.
• Applying Z-score calculations aids in assessing how rare or common a specific strength is within the distribution.

By understanding and analyzing compressive strength data using these statistical tools, engineers and quality assurance professionals can more effectively monitor and enhance construction materials and processes.