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Chapter 4: Problem 19

The article "Characterization of Room Temperature Damping in Aluminum-Indium Alloys" (Metallurgical Trans., 1993: 1611-1619) suggests that Al matrix grain size \((\mu \mathrm{m})\) for an alloy consisting of \(2 \%\) indium could be modeled with a normal distribution with a mean value 96 and standard deviation \(14 .\) a. What is the probability that grain size exceeds 100 ? b. What is the probability that grain size is between 50 and 80 ? c. What interval \((a, b)\) includes the central \(90 \%\) of all grain sizes (so that \(5 \%\) are below \(a\) and \(5 \%\) are above \(b\) )?

Short Answer

Expert verified
a. 0.387, b. 0.1266, c. (72.97, 119.03)

Step by step solution

01

Understand the Distribution

Grain sizes are modeled with a normal distribution. The parameters for this normal distribution are: mean \( \mu = 96 \) and standard deviation \( \sigma = 14 \).
02

Convert Threshold to Z-Score for Part a

To find the probability that the grain size exceeds 100, convert 100 to a z-score using the formula: \[ z = \frac{x - \mu}{\sigma} \]Substituting \( x = 100 \), \( \mu = 96 \), and \( \sigma = 14 \), we get:\[ z = \frac{100 - 96}{14} = \frac{4}{14} \approx 0.29 \].
03

Calculate Probability for Part a

Using standard normal distribution tables or a calculator, find the probability corresponding to the z-score of 0.29. The probability that \( Z < 0.29 \) is approximately 0.613.Thus, the probability that the grain size exceeds 100 is \( P(X > 100) = 1 - P(Z \leq 0.29) = 1 - 0.613 = 0.387 \).
04

Convert Limits to Z-Scores for Part b

For the probability that the grain size is between 50 and 80, calculate the z-scores:- For \( x = 50 \): \[ z = \frac{50 - 96}{14} = \frac{-46}{14} \approx -3.29 \]- For \( x = 80 \): \[ z = \frac{80 - 96}{14} = \frac{-16}{14} \approx -1.14 \].
05

Calculate Probability for Part b

Look up z-scores in the standard normal table or use a calculator:- \( P(Z < -3.29) \approx 0.0005 \)- \( P(Z < -1.14) \approx 0.1271 \)Hence, the probability that the grain size is between 50 and 80:\[ P(50 < X < 80) = P(-3.29 < Z < -1.14) = P(Z < -1.14) - P(Z < -3.29) = 0.1271 - 0.0005 = 0.1266 \].
06

Find z-Scores for Central 90% Interval for Part c

To find the interval \((a, b)\) that contains the central 90% of the distribution:- The left tail is 5% and the right tail is 5%.Using z-tables, the z-scores for 5% and 95% are approximately -1.645 and 1.645, respectively.
07

Calculate Interval Values for Part c

Use the z-scores to find \(a\) and \(b\):- \( a = \mu + z \cdot \sigma = 96 + (-1.645 \cdot 14) \approx 72.97 \)- \( b = \mu + z \cdot \sigma = 96 + (1.645 \cdot 14) \approx 119.03 \)So the interval \((a, b) = (72.97, 119.03)\).

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Key Concepts

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

Probability Calculation
Probability calculation in the realm of statistics often involves the use of the normal distribution. In our exercise, we are interested in determining how likely it is for a particular event to occur, given that event follows a normal distribution For a normal distribution, we use the mean (average) and the standard deviation (a measure of spread) to understand where most of the data falls. In this exercise, the mean grain size is 96 with a standard deviation of 14. To establish the probability of finding a grain size greater than 100, we first convert this value into a Z-score. After obtaining the Z-score, we consult a standard normal distribution table or use a statistical calculator to find the corresponding probability. This tells us the likelihood of finding a grain size exceeding 100. Don鈥檛 forget, this process applies to any value within a normally distributed dataset.
Z-Score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is a crucial component when working with normal distributions because it standardizes different datasets, making them comparable.To calculate the Z-score, the formula is:\[z = \frac{x - \mu}{\sigma} \]where:
  • \(x\) is the value from the dataset (e.g., 100 in our example)
  • \(\mu\) is the mean of the dataset
  • \(\sigma\) is the standard deviation
After finding the Z-score, use a Z-table to find the probability associated with this Z-score. This will give you the probability of a value being below a particular threshold in a standard normal distribution. In the exercise, Z-scores help determine the probability of various grain sizes, aiding in a thorough understanding of the dataset characteristics.
Central Interval
The central interval in the context of a normal distribution is an interval around the mean that contains a certain percentage of the data. For instance, to find an interval that includes the central 90% of all grain sizes, you need to exclude the 5% smallest and the 5% largest values. To calculate this interval:1. Determine the Z-scores that correspond to the 5th and 95th percentiles. Typically, these Z-scores are approximately -1.645 and 1.645.2. Use the formula:\[a = \mu + z\cdot \sigma \]3. Plugging in these values, you calculate the values for \(a\) and \(b\):\[a = 96 + ( -1.645 \times 14) \approx 72.97 \]\[b = 96 + (1.645 \times 14) \approx 119.03 \]This interval tells us where the grain sizes will fall 90% of the time given a normal distribution, offering valuable insights into the distribution's dispersion.

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Most popular questions from this chapter

Suppose a particular state allows individuals filing tax returns to itemize deductions only if the total of all itemized deductions is at least \(\$ 5000\). Let \(X\) (in 1000 s of dollars) be the total of itemized deductions on a randomly chosen form. Assume that \(X\) has the pdf $$ f(x, \alpha)=\left\\{\begin{array}{cc} k / x^{\alpha} & x \geq 5 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Find the value of \(k\). What restriction on \(\alpha\) is necessary? b. What is the cdf of \(X\) ? c. What is the expected total deduction on a randomly chosen form? What restriction on \(\alpha\) is necessary for \(E(X)\) to be finite? d. Show that \(\ln (X / 5)\) has an exponential distribution with parameter \(\alpha-1\).

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