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A regulation hockey puck must weigh between 5.5 and 6 ounces. The weights \(X\) of pucks made by a particular process are normally distributed with mean 5.75 ounces and standard deviation 0.11 ounce. Find the probability that a puck made by this process will meet the weight standard.

Short Answer

Expert verified
The probability is approximately 0.9768.

Step by step solution

01

Identify the problem

To find the probability that a puck weighs between 5.5 and 6 ounces, we must calculate the area under the normal distribution curve for these limits.
02

Standardize the variable

Convert the weight limits to z-scores using the formula for the standard normal variable: \( Z = \frac{X - \mu}{\sigma} \). For 5.5 ounces, \( Z = \frac{5.5 - 5.75}{0.11} \). For 6 ounces, \( Z = \frac{6 - 5.75}{0.11} \).
03

Calculate the z-scores

For 5.5 ounces: \( Z = \frac{5.5 - 5.75}{0.11} = -2.27 \).For 6 ounces: \( Z = \frac{6 - 5.75}{0.11} = 2.27 \).
04

Find the probabilities associated with the z-scores

Using a standard normal distribution table or calculator, find the probabilities for \( Z = -2.27 \) and \( Z = 2.27 \). The probability for \( Z = -2.27 \) is approximately 0.0116, and for \( Z = 2.27 \) is approximately 0.9884.
05

Calculate the probability of the puck meeting the standard

The probability that a puck's weight is between 5.5 and 6 ounces is the difference in probabilities: \( P(5.5 < X < 6) = P(Z < 2.27) - P(Z < -2.27) = 0.9884 - 0.0116 = 0.9768 \).

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Key Concepts

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

Understanding Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It gives us an idea of how much variation exists from the average (mean) value. In simpler terms, it helps us understand if the data are clustered closely around the mean or widely dispersed.

When data points are close to the mean, the standard deviation is low. On the other hand, if data points are spread out over a larger range of values, the standard deviation is high. In the context of the hockey puck problem, we have a mean weight of 5.75 ounces and a standard deviation of 0.11 ounces, indicating that most pucks weigh close to 5.75 ounces. However, some variation exists due to the manufacturing process.

The standard deviation is crucial for probability calculations because it allows us to standardize different sets of data. This means converting them into a comparable scale which is critical for analyzing data and making informed decisions in a variety of fields.
Exploring Z-Scores
Z-scores are a way of standardizing individual data points relative to the mean and standard deviation of a dataset. They tell us how many standard deviations an element is from the mean. This is important because it allows for the comparison of data from different distributions.

For example, if you have a puck that weighs 5.5 ounces, you can calculate the z-score to find out how far the puck deviates from the average weight. The formula for calculating a z-score is:
  • \( Z = \frac{X - \mu}{\sigma} \)
where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Understanding z-scores helps us to assess the probability of whether certain data points fall within a specific range when dealing with normally distributed data. For the hockey pucks, we calculated z-scores of -2.27 and 2.27 for weights of 5.5 and 6 ounces respectively, indicating how many standard deviations these weights are from the mean.
Probability Calculations Made Easy
Probability calculations allow us to determine the likelihood of specific outcomes within a given set of data. When working with normal distributions, we can use z-scores to find probabilities associated with specific data points. This involves understanding the area under a bell curve between given z-score limits.

Once you have calculated the z-scores, you can use a standard normal distribution table or calculator to find associated probabilities. These probabilities indicate how much of the dataset falls within certain limits. In our case, after determining the z-scores for weights of 5.5 and 6 ounces, we looked up or calculated the probabilities for these z-values.

For the hockey puck problem, finding the probability of a puck meeting the weight standard involved calculating the difference between the probabilities for z-scores of 2.27 and -2.27. This difference tells us that around 97.68% of pucks made by this process would weigh between 5.5 and 6 ounces. By mastering these calculations, you can effectively manage and interpret the randomness and variability within normally distributed datasets.

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