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Chapter 4: Problem 48

Suppose Appendix Table A.3 contained \(\Phi(z)\) only for \(z \geq 0\). Explain how you could still compute a. \(P(-1.72 \leq Z \leq-.55)\) b. \(P(-1.72 \leq Z \leq .55)\) Is it necessary to tabulate \(\Phi(z)\) for \(z\) negative? What property of the standard normal curve justifies your answer?

Short Answer

Expert verified
Symmetry of normal curve allows use of positive \(z\) values for negative \(z\).

Step by step solution

01

Understanding the Standard Normal Distribution

The standard normal distribution is symmetric around the mean, which is 0. This means that for a standard normal variable \(Z\), the probability \(P(Z \leq z)\) is the same as \(1 - P(Z \leq -z)\). Thus, the table for \(\Phi(z)\) with \(z \geq 0\) can be used to derive probabilities for negative \(z\) values.
02

Solving for Probability in a. (-1.72 \leq Z \leq -0.55)

To find \(P(-1.72 \leq Z \leq -0.55)\), use the symmetry property: \(P(-1.72 \leq Z \leq -0.55) = P(0.55 \leq Z \leq 1.72)\). To compute this, we use the table for positive \(z\) values: \(\Phi(1.72) - \Phi(0.55)\). Look up those values in the table to calculate the probability.
03

Solving for Probability in b. (-1.72 \leq Z \leq 0.55)

To find \(P(-1.72 \leq Z \leq 0.55)\), split it into two parts using symmetry: \(P(-1.72 \leq Z \leq 0.55) = P(-1.72 \leq Z \leq 0) + P(0 \leq Z \leq 0.55)\). For \(P(-1.72 \leq Z \leq 0)\), use symmetry and the table: \(\Phi(1.72) - 0.5\). For \(P(0 \leq Z \leq 0.55)\), use \(\Phi(0.55) - 0.5\). Add these two results together.
04

Conclusion on Negative Values Table Necessity

It is not necessary to tabulate \(\Phi(z)\) for negative \(z\) due to the symmetry of the standard normal curve. The property used is that the normal distribution is symmetric around the mean; hence, it allows using negative \(z\) values through positive table values.

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Key Concepts

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

Symmetry Property
The symmetry property is an essential concept for understanding the Standard Normal Distribution. The Standard Normal Distribution is a type of continuous probability distribution that is symmetric around its mean.
This distribution has a mean (\(\mu\)) of 0 and a standard deviation (\(\sigma\)) of 1. Because of its symmetry, the curve is a perfect mirror image around the mean of zero.
When it comes to probabilities, this symmetry allows us to use positive ones for both sides.
  • For any value \(z\), the probability \(P(Z \leq z)\) is equal to \(1 - P(Z \leq -z)\).
  • This means if you have a \(z\) value of -1, it is mirrored to a \(z\) value of 1, making it \(P(Z \leq 1) = 1 - P(Z \leq -1)\).
This property eliminates the need for tables to tabulate negative \(z\) values, as you can always refer to their positive counterparts. This makes calculations easier and more efficient.
Standard Normal Table
The Standard Normal Table, also known as the Z-table, is an essential tool to determine probabilities related to the Standard Normal Distribution.
This table helps find the area (probability) under the curve to the left of any \(z\)-score for \(z\geq0\). What this table offers is a standardized way to get probabilities for any normal distribution through conversion.
  • Since the normal distribution is symmetric, knowing the probability for a positive \(z\) score tells us the corresponding negative \(z\) probability.
  • For example, using the table for 1.72 gives \(\Phi(1.72)\) and symmetry tells us \(\Phi(-1.72) = 1 - \Phi(1.72)\).
Hence, with the Standard Normal Table, one can easily access probabilities without needing different tables for negative \(z\) values, leveraging the symmetry property.
Probability Calculation
Calculating probabilities with the Standard Normal Distribution involves using the available standard normal table and the symmetry property.
Let's say, you are required to find the probability for the statement \(P(-1.72 \leq Z \leq -0.55)\).
  • First, apply the symmetry property to translate this to a positive \(z\) score probability: \(P(0.55 \leq Z \leq 1.72)\).
  • Look up both \(\Phi(1.72)\) and \(\Phi(0.55)\) in the table. The probability is \(\Phi(1.72) - \Phi(0.55)\).
For a mixed range, \(P(-1.72 \leq Z \leq 0.55)\), split it further:
  • First half: \(P(-1.72 \leq Z \leq 0)\) which is \(\Phi(1.72) - 0.5\).
  • Second half: \(P(0 \leq Z \leq 0.55)\) is \(\Phi(0.55) - 0.5\).
  • Add these two results.
Using the standard normal table and understanding the properties, anyone can efficiently calculate complex probabilities without the need for negative \(z\) values.

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Most popular questions from this chapter

Stress is applied to a 20 -in. steel bar that is clamped in a fixed position at each end. Let \(Y=\) the distance from the left end at which the bar snaps. Suppose \(Y / 20\) has a standard beta distribution with \(E(Y)=10\) and \(V(Y)=\frac{100}{7}\). a. What are the parameters of the relevant standard beta distribution? b. Compute \(P(8 \leq Y \leq 12)\). c. Compute the probability that the bar snaps more than 2 in. from where you expect it to.

Suppose the force acting on a column that helps to support a building is a normally distributed random variable \(X\) with mean value \(15.0\) kips and standard deviation \(1.25\) kips. Compute the following probabilities by standardizing and then using Table A.3. a. \(P(X \leq 15)\) b. \(P(X \leq 17.5)\) c. \(P(X \geq 10)\) d. \(P(14 \leq X \leq 18)\) e. \(P(|X-15| \leq 3)\)

Let \(X\) have a binomial distribution with parameters \(n=25\) and \(p\). Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases \(p=.5, .6\), and \(.8\) and compare to the exact probabilities calculated from Appendix Table A.1. a. \(P(15 \leq X \leq 20)\) b. \(P(X \leq 15)\) c. \(P(20 \leq X)\)

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(.5 \%\) of all transistors would still be operating at termination?

Find the following percentiles for the standard normal distribution. Interpolate where appropriate. a. \(91 \mathrm{st}\) b. 9 th c. 75 th d. 25 th e. 6 th
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