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The width of bolts of fabric is normally distributed with mean \(950 \mathrm{mm}\) (millimeters) and standard deviation 10 mm. a. What is the probability that a randomly chosen bolt has a width of between 947 and \(958 \mathrm{mm}\) ? b. What is the appropriate value for \(C\) such that a randomly chosen bolt has a width less than \(C\) with probability .8531?

Short Answer

Expert verified
a. 0.4060 b. 960.5 mm

Step by step solution

01

Understand the Normal Distribution

The problem specifies that the width of the fabric bolts follows a normal distribution with a mean \(\mu = 950\, \text{mm}\) and a standard deviation \(\sigma = 10\, \text{mm}\). This information forms the basis for solving both parts.
02

Convert Widths to Z-Scores for Part a

For part (a), we need to find the probability that a width is between 947 mm and 958 mm. First, convert these widths to Z-scores using the formula: \(Z = \frac{X - \mu}{\sigma}\). For 947 mm, \(Z = \frac{947 - 950}{10} = -0.3\). For 958 mm, \(Z = \frac{958 - 950}{10} = 0.8\).
03

Use Z-Table to Find Probability for Part a

Look up the Z-scores in the standard normal distribution table (Z-table). \(P(Z < -0.3) \approx 0.3821\) and \(P(Z < 0.8) \approx 0.7881\). The probability of the width being between 947 mm and 958 mm is \(P(-0.3 < Z < 0.8) = P(Z < 0.8) - P(Z < -0.3) = 0.7881 - 0.3821 = 0.4060\).
04

Find Z-Score for Probability 0.8531 for Part b

For part (b), find the Z-score corresponding to a cumulative probability of 0.8531. Using the Z-table, we find that \(P(Z < 1.05) \approx 0.8531\). Thus, the Z-score for this cumulative probability is 1.05.
05

Convert Z-Score to Width for Part b

Use the Z-score formula to find the corresponding width \(C\). Since \(Z = \frac{C - 950}{10}\) and \(Z = 1.05\), solve \(1.05 = \frac{C - 950}{10}\). This gives: \(C - 950 = 10.5\), thus \(C = 960.5\, \text{mm}\).

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Key Concepts

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

Z-score
The Z-score is a simple yet powerful tool in statistics, particularly useful when working with normal distributions. It allows you to transform data from one distribution into a standardized scale with a mean of zero and a standard deviation of one. This process is called "standardizing" the data and it's critical when comparing individual data points to the overall distribution.

To calculate a Z-score, use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \( X \) is the value of the individual data point we are interested in, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.

For example, if you have a fabric bolt width of 958 mm, a mean width of 950 mm, and a standard deviation of 10 mm, the Z-score would be:
  • \( Z = \frac{958 - 950}{10} = 0.8 \)
A positive Z-score indicates the data point is above the mean, while a negative Z-score shows it is below the mean. Understanding how to convert data points into Z-scores is crucial for further calculations involving the standard normal distribution.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution that serves as the basis for calculations involving Z-scores. It has a mean of 0 and a standard deviation of 1. This distribution allows statisticians and data analysts to easily determine probabilities and make inferences about data.

In practice, when we convert our data into Z-scores, we're effectively mapping it onto the standard normal distribution. This allows us to use the Z-table, a tool that provides the cumulative probability for any given Z-score. For instance, if you're working with a Z-score of 0.8, the Z-table shows you that the probability of a value being less than \( Z = 0.8 \) is approximately 0.7881. This probability corresponds to the area under the curve to the left of this Z-score.
  • Z-scores and the standard normal distribution are interconnected, facilitating the understanding of probabilities and data comparisons.
  • By standardizing data and using this distribution, you can handle complex problems with simplicity and clarity.
Learning to navigate the standard normal distribution is key to interpreting and solving statistical problems efficiently.
Probability Calculation
Probability calculation in a normal distribution scenario often involves determining the likelihood of an event by using Z-scores and the standard normal distribution. It allows us to predict how likely it is for a given data point or range of data points to occur.

When solving for probabilities like in part (a) of the exercise, you start by converting data points (947 mm and 958 mm in this case) into Z-scores. Once you have these Z-scores (-0.3 and 0.8), you can use the Z-table to find the probabilities associated with these scores:
  • Probability that Z < -0.3 is approximately 0.3821.
  • Probability that Z < 0.8 is approximately 0.7881.
To find the probability of the fabric width being between 947 mm and 958 mm, calculate the difference:
  • \( P(-0.3 < Z < 0.8) = P(Z < 0.8) - P(Z < -0.3) = 0.7881 - 0.3821 = 0.4060 \)
Similarly, for part (b), you identify the Z-score of 1.05 for a cumulative probability of 0.8531 from the Z-table, and solve for the corresponding width \( C \). The capacity to translate probabilities into tangible values, both predicted and observed, is a rewarding aspect of mastering probability in normal distributions.

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