91Ó°ÊÓ

The Rockwell hardness of a metal is determined by impress= ing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 70 and standard deviatson 3. (Rockwell hardness is measured on a continuous scale.) a. If a specimen is acceptable only if its hardness is between 67 and 75 , what is the probahility that a ran= domaly chosen specimen has an acceptable hardness? b. If the acceptable range of hardness is \((7 D-c, 70+c)\), for what value of \(c\) avould \(95 \%\) of all specimens have acceptable hardncss? c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten? d. What is the probability that at most eight of ten independently selected specimens have a hardness of less than \(73.84 ?\) [Hint: \(Y\) - the number among the ten speci= mens with hartuness less than \(73.84\) is a binomial vari= able: what is \(p\) ?]

Step by step solution

01

Understanding the Normal Distribution

The Rockwell hardness follows a normal distribution with mean \( \mu = 70 \) and standard deviation \( \sigma = 3 \). This means the distribution can be expressed as \( N(70, 3^2) \).

02

Calculate Probability for Part (a)

To find the probability that a specimen has hardness between 67 and 75, calculate the z-scores for both values using the formula: \( z = \frac{x - \mu}{\sigma} \).\[ z_{67} = \frac{67 - 70}{3} = -1 \]\[ z_{75} = \frac{75 - 70}{3} = 1.67 \]Use the standard normal distribution table to find probabilities: \( P(Z < -1) \approx 0.1587 \) and \( P(Z < 1.67) \approx 0.9525 \).The probability of being between 67 and 75 is:\[ P(67 < X < 75) = P(Z < 1.67) - P(Z < -1) \approx 0.9525 - 0.1587 = 0.7938 \].

03

Solve for c in Part (b)

The requirement is for \( 95\% \) of all specimens to have hardness between \( 70-c \) and \( 70+c \). This is equivalent to finding the range that covers the middle \( 95\% \), corresponding to \( z = \pm 1.96 \) for \( 95\% \) confidence in a standard normal distribution.So, \( c = z \times \sigma = 1.96 \times 3 = 5.88 \).Thus the hardness range is \( (70 - 5.88, 70 + 5.88) = (64.12, 75.88) \).

04

Determine Number of Acceptable Specimens (Part (c))

Each specimen follows a normal distribution like in part (a), with a probability of acceptability \( 0.7938 \). Thus, for 10 specimens:Expected number = Total specimens \( \times \) Probability of acceptable = \( 10 \times 0.7938 = 7.938 \).

05

Probability for Binomial Distribution (Part (d))

To find the probability that at most 8 out of 10 have hardness less than 73.84, we first find the probability \( p \) for a single specimen:Calculate the z-score for 73.84:\[ z = \frac{73.84 - 70}{3} = 1.28 \]From the standard normal distribution table, \( P(Z < 1.28) \approx 0.8997 \).\( Y \sim \text{Binomial}(n=10, p=0.8997) \). We find \( P(Y \leq 8) \) using the binomial probability formula:\[ P(Y \leq 8) = \sum_{k=0}^{8} \binom{10}{k} (0.8997)^k (0.1003)^{10-k} \]Use a binomial calculator or software to compute this sum.

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Key Concepts

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

Normal Distribution

In probability theory, the normal distribution is one of the most important continuous probability distributions. It is often called the "bell curve" due to its bell-shaped appearance when graphed. This distribution is determined by two parameters: the mean (average) and the standard deviation (spread or width of the curve). For example, the Rockwell hardness of metal in our exercise is distributed with a mean \( \mu = 70 \) and a standard deviation \( \sigma = 3 \).

One key feature of the normal distribution is that it allows us to calculate probabilities for any given range of values. This is achieved through the use of z-scores, which standardize individual values based on the mean and standard deviation of the distribution. As a result, we can determine the likelihood that a randomly chosen specimen falls within a specific interval, helping us make informed decisions in fields like quality control and risk assessment.

Z-Scores

Z-scores are a standardized measure of how many standard deviations away a specific data point is from the mean of a distribution. The z-score is calculated using the formula:

• \( z = \frac{x - \mu}{\sigma} \)

where \( x \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

For our metal hardness example, we calculated the z-scores for hardness values of 67 and 75, which were \( z_{67} = -1 \) and \( z_{75} = 1.67 \) respectively. Using these z-scores, we can refer to a standard normal distribution table to find the probabilities associated with them. This is a crucial step, as it allows us to understand the proportion of a population that lies within a defined range of values, thereby assisting in solving problems related to the normal distribution.

Z-scores provide an essential framework for comparing data that might originate from different distributions or have different scales.

Binomial Distribution

The binomial distribution is crucial in probability theory when we deal with experiments that result in two possible outcomes: success or failure. This distribution is defined by two parameters: the number of trials \( n \) and the probability of success in a single trial \( p \).

In our exercise, when determining the probability that at most 8 out of 10 specimens have a hardness less than a certain value, we used the binomial distribution because each test on a specimen is an independent trial with a specific probability of success \( p \). Here, "success" could mean a specimen falls below a particular hardness threshold

For a binomial distribution \( Y \sim \text{Binomial}(n, p) \), the probability of observing exactly \( k \) successes is given by:

• \( P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

where \( \binom{n}{k} \) is the binomial coefficient, a way to count combinations.

The binomial distribution is widely applicable in fields like genetics, quality testing, and market research, where frequent experimentation and analysis occur.

Expected Value

The expected value is a fundamental concept in probability and statistics, representing the average outcome if an experiment is repeated a large number of times. It's expressed as the sum of all possible values, each multiplied by its probability of occurrence.

In the context of our hardness problem, we wanted to find the expected number of specimens that meet the acceptability criteria. We first determined the probability of a single specimen being acceptable, which was around 0.7938. Then, for 10 specimens tested independently, the expected number is calculated by multiplying the total number of specimens by the probability of acceptance:

• \( ext{Expected value} = n \times p \)

Thus, for our problem, it is \( 10 \times 0.7938 = 7.938 \).

Understanding the expected value helps in anticipating the number of successful outcomes in various scenarios, contributing significantly to decision-making processes in economics, engineering, and beyond.