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Chapter 6: Problem 35

The pucks used by the National Hockey League for ice hockey must weigh between \(5.5\) and \(6.0\) ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of \(5.75\) ounces and a standard deviation of \(.11\) ounce. What percentage of the pucks produced at this factory cannot be used by the National Hockey League?

Short Answer

Expert verified
Approximately 2.38% of pucks produced at the factory cannot be used by the NHL.

Step by step solution

01

Calculate the Z-Score for the lower limit

First, calculate the z-score for the lower limit of the acceptable weight range using the formula: \( Z = (X - 渭) / 蟽 \), where X is the value (5.5), 渭 is the mean (5.75), and 蟽 is the standard deviation (0.11). The detailed calculation would be: \( Z = (5.5 - 5.75) / 0.11 = -2.27 \)
02

Calculate the Z-Score for the upper limit

Next, calculate the z-score for the upper limit of acceptable weight range using the same formula: \( Z = (X - 渭) / 蟽 \), where X is the value (6.0), 渭 is the mean (5.75), and 蟽 is the standard deviation (0.11). The detailed calculation would be: \( Z = (6.0 - 5.75) / 0.11 = 2.27 \)
03

Compute the probabilities

Now, lookup these z-scores in the standard normal distribution table, or use a calculator. From the table, a Z-score of -2.27 corresponds to a left tail probability of 0.0119 and a Z-score of 2.27 corresponds to a left tail probability of 0.9881. Subtracting the lower probability from the higher gives the percentage of pucks which have an acceptable weight. That is: \(0.9881 - 0.0119 = 0.9762 \) or 97.62%. Therefore, 100% - 97.62% = 2.38% of pucks produced at the factory cannot be used by NHL.

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Key Concepts

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

Z-Score Calculation
The Z-score is a crucial statistic in understanding normal distribution. It measures the number of standard deviations a data point is from the mean. In this exercise, we are calculating how far the weight of a hockey puck is from the average weight. To calculate the Z-score, the formula used is:
  • \( Z = \frac{(X - \mu)}{\sigma} \)
Here:
  • \(X\) is the value of the data point being measured; in our case, the lower and upper limits of acceptable puck weights.
  • \(\mu\) is the mean, which in this example is \(5.75\) ounces.
  • \(\sigma\) is the standard deviation, given as \(0.11\) ounces.
A Z-score tells us how "unusual" a value is in the context of the normal distribution. For instance, a Z-score of -2.27 indicates that the weight \(5.5\) is 2.27 standard deviations below the mean.
Standard Deviation
Standard deviation is a measurement of the spread or dispersion of a set of values within a data set. In normal distribution, it indicates how much variation there is from the average (mean). A smaller standard deviation indicates that the values are closer to the mean. Conversely, a larger standard deviation indicates a wider spread.
In our scenario, the standard deviation of puck weights is \(0.11\) ounces. This means most pucks' weights will be within \(0.11\) ounces of the mean weight of \(5.75\) ounces.
Standard deviation is essential for calculating Z-scores and understanding the likelihood of certain values occurring within a normal distribution.
Probability
Probability in statistics helps us quantify the chance of a specific event occurring. When data is normally distributed, we can use Z-scores to determine the probability of a data point falling within a particular range. In this example, we're interested in the probability of a puck being within the NHL's required weight range, between \(5.5\) and \(6.0\) ounces.
  • A Z-score table (also known as a standard normal distribution table) translates Z-scores into probabilities.
  • A Z-score of -2.27 translates to a probability of \(0.0119\), representing the proportion of data below \(5.5\) ounces.
  • A Z-score of 2.27 corresponds to \(0.9881\), signifying the proportion of data below \(6.0\) ounces.
The probability of a puck being within the desired range is calculated as the difference between these two probabilities.
Mean Value
The mean value, often referred to as the average, is a central measure in statistics that provides us with a single value that represents the central tendency of a data set. It's calculated by adding all values and dividing by the number of values.
In this problem, the mean weight of the pucks is \(5.75\) ounces, signaling the center point of our normally distributed data.
This mean tells us the default or expected value. While individual puck weights may vary, typically, they will hover around this mean. Understanding the mean is foundational in stats, as it aids in identifying patterns and calculating statistical measures like standard deviation and Z-scores.

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Most popular questions from this chapter

Find the following areas under a normal distribution curve with \(\mu=20\) and \(\sigma=4\) a. Area between \(x=20\) and \(x=27\) b. Area from \(x=23\) to \(x=26\) c. Area between \(x=9.5\) and \(x=17\)

A machine at Keats Corporation fills 64 -ounce detergent jugs. The machine can be adjusted to pour, on average, any amount of detergent into these jugs. However, the machine does not pour exactly the same amount of detergent into each jug; it varies from jug to jug. It is known that the net amount of detergent poured into each jug has a normal distribution with a standard deviation of \(.35\) ounce. The quality control inspector wants to adjust the machine such that at least \(95 \%\) of the jugs have more than 64 ounces of detergent. What should the mean amount of detergent poured by this machine into these jugs be?

Hurbert Corporation makes font cartridges for laser printers that it sells to Alpha Electronics Inc. The cartridges are shipped to Alpha Electronics in large volumes. The quality control department at Alpha Electronics randomly selects 100 cartridges from each shipment and inspects them for being good or defective. If this sample contains 7 or more defective cartridges, the entire shipment is rejected. Hurbert Corporation promises that of all the cartridges, only \(5 \%\) are defective. a. Find the probability that a given shipment of cartridges received by Alpha Electronics will be accepted. b. Find the probability that a given shipment of cartridges received by Alpha Electronics will not be accepted.

Alpha Corporation is considering two suppliers to secure the large amounts of steel rods that it uses. Company A produces rods with a mean diameter of \(8 \mathrm{~mm}\) and a standard deviation of \(.15 \mathrm{~mm}\) and sells 10,000 rods for $$\$ 400$$ Company B produces rods with a mean diameter of \(8 \mathrm{~mm}\) and a standard deviation of \(.12 \mathrm{~mm}\) and sells 10,000 rods for $$\$ 460$$ A rod is usable only if its diameter is between \(7.8 \mathrm{~mm}\) and \(8.2 \mathrm{~mm}\). Assume that the diameters of the rods produced by each company have a normal distribution. Which of the two companies should Alpha Corporation use as a supplier? Justify your answer with appropriate calculations.

Under what conditions is the normal distribution usually used as an approximation to the binomial distribution?
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