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Chapter 4: Problem 40

In each case, determine the value of the constant \(c\) that makes the probability statement correct. a. \(\phi(c)=.9838\) b. \(P(0 \leq Z \leq c)=.291\) c. \(P(c \leq Z)=.121\) d. \(P(-c \leq Z \leq c)=.668\) e. \(P(c \leq|Z|)=.016\)

Short Answer

Expert verified
(a) 2.13, (b) 0.82, (c) 1.17, (d) 0.99, (e) 2.41

Step by step solution

01

Understanding Normal Distribution

The problem involves determining constants using normal distribution properties. The standard normal distribution, denoted as \(Z\), is a normal distribution with a mean of 0 and a standard deviation of 1. \(\phi(c)\) represents the cumulative distribution function (CDF) for \(Z\).
02

Sub-problem (a) Conversion to CDF

For \(\phi(c) = 0.9838\), we are given the cumulative probability, which means we need \(c\) such that this area under the standard normal curve is equal to 0.9838. We look up the standard normal distribution table or use a calculator to find \(c\).
03

Solution to (a) Using Distribution Table

Looking up 0.9838 in the \(Z\)-table, we find that \(c = 2.13\). Hence, \(\phi(2.13) \approx 0.9838\).
04

Sub-problem (b) Find Cumulative CDF on Right Side

For \(P(0 \leq Z \leq c) = 0.291\), this probability is the cumulative probability from 0 to \(c\). We know \(P(-\infty \leq Z \leq 0) = 0.5\), so our aim is to find \(P(0 \leq Z \leq c) = 0.291\).
05

Solution to (b) Using Distribution Table

From symmetry and the table, \(c = 0.82\) because \(\phi(0.82) - 0.5 = 0.291\).
06

Sub-problem (c) Conversion to CDF and Complement

For \(P(c \leq Z) = 0.121\), this implies \(\phi(c) = 1 - 0.121 = 0.879\).
07

Solution to (c) Using Distribution Table

Looking up 0.879 in the \(Z\)-table, \(c = 1.17\) since \(\phi(1.17) \approx 0.879\).
08

Sub-problem (d) Symmetric Probability Division

For \(P(-c \leq Z \leq c) = 0.668\), use symmetry of the normal distribution. This means \(\phi(c) - [1 - \phi(-c)] = 0.668\), or simply \(2 \cdot \phi(c) - 1 = 0.668\).
09

Solution to (d) Solving the Probability Equation

Solving \(\phi(c) = 0.834\) gives \(c = 0.99\) from the table, since \(2\phi(0.99) - 1 = 0.668\).
10

Sub-problem (e) Absolute Value Transformation

For \(P(c \leq |Z|) = 0.016\), this means two tails (one on each side). Thus, \(2 \cdot \phi(-c) = 0.016\), or \(\phi(-c) = 0.008\).
11

Solution to (e) Using Symmetry

Since \(\phi(-c) = 0.008\), \(c = 2.41\) by looking up the table, since \(\phi(2.41) \approx 1 - 0.008 = 0.992\).

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a fundamental concept in probability and statistics. It's a smooth, bell-shaped curve that describes data that clusters around a mean. Specifically, it is a normal distribution with a mean of 0 and a standard deviation of 1. This makes it a powerful tool, as it standardizes different datasets, allowing them to be compared easily.
  • The mean (0) and standard deviation (1) simplify the calculation of probabilities.
  • It is symmetrical, which means the shape of the curve is identical on both sides of the mean.
  • Data within one standard deviation of the mean covers approximately 68% of the total data.
Understanding this concept allows us to analyze and interpret real-world phenomena that follow normal spreading patterns.
Cumulative Distribution Function
The cumulative distribution function (CDF) is crucial when dealing with probability distributions like the standard normal distribution.
The CDF gives the probability that a random variable is less than or equal to a certain value. In essence, it accumulates the probabilities up to a point, providing a complete picture of a distribution's behavior.
  • It starts at 0 and approaches 1 as you move further along the x-axis, indicating 100% probability.
  • For a standard normal distribution, denoted as \( \phi(c) \), it represents the area under the curve to the left of the value \( c \).
  • The CDF is non-decreasing, meaning it never decreases as it accumulates probabilities.
By understanding the CDF, one can determine potential outcomes and their likelihoods in a given probability distribution setting.
Symmetry in Probability
The concept of symmetry in probability simplifies the calculation of probabilities for symmetric distributions, like the standard normal distribution.
The left and right sides of the mean are mirror images, allowing for easier probability assessments and calculations.
  • This symmetry is why probabilities are often calculated from zero to a positive value, then doubled for both sides of the mean.
  • Such symmetry also allows us to easily reason about the tails of the distribution.
  • Generally, it implies that \(-c \leq X \leq c\) has the same probability as the rest on the positive side.
Using this symmetry in calculations reduces complexity and provides intuitive insights into how data behaves around its central value.
Z-Score Tables
Z-score tables, or standard normal distribution tables, are one of the key tools in statistics for determining probabilities and identifying where a value sits within a distribution relative to the mean.
These tables list \( Z \) values against their cumulative probabilities, linking a \( Z \) score to the area under the standard normal curve to its left.
  • The \( Z \) score measures how many standard deviations a particular value is from the mean.
  • Using a \( Z \)-table, you can find probabilities more quickly without manually calculating each time.
  • These scores convert different datasets into a comparable format, allowing analyses across varied datasets.
By mastering Z-score tables, students gain powerful insights into probability distributions and data standardization practices.

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Most popular questions from this chapter

Let \(X\) have the chi-squared distribution with 2 degree of freedom, so \(f_{X}(x)=\frac{1}{2} e^{-x / 2}, \quad x>0\). Find the pdf of \(Y=\sqrt{X}\). Suppose you choose a point in two dimensions randomly, with the horizontal and vertical coordinates chosen independently from the standard normal distribution. Then \(X\) has the distribution of the squared distance from the origin and \(Y\) has the distribution of the distance from the origin. Because \(Y\) is the length of a vector with normal components, there are lots of applications in physics, and its distribution has the name Rayleigh.

The article "Determination of the MTF of Positive Photoresists Using the Monte Carlo Method" (Photographic Sci. Engrg., 1983: 254-260) proposes the exponential distribution with parameter \(\lambda=.93\) as a model for the distribution of a photon's free path length \((\mu \mathrm{m})\) under certain circumstances. Suppose this is the correct model. a. What is the expected path length, and what is the standard deviation of path length? b. What is the probability that path length exceeds \(3.0 ?\) What is the probability that path length is between \(1.0\) and \(3.0\) ? c. What value is exceeded by only \(10 \%\) of all path lengths?

The weight distribution of parcels sent in a certain manner is normal with mean value \(12 \mathrm{lb}\) and standard deviation \(3.5 \mathrm{lb}\). The parcel service wishes to establish a weight value \(c\) beyond which there will be a surcharge. What value of \(\mathrm{c}\) is such that \(99 \%\) of all parcels are at least \(1 \mathrm{lb}\) under the surcharge weight?

The article "The Statistics of Phytotoxic Air Pollutants" (J. Roy. Statist Soc., 1989: 183-198) suggests the lognormal distribution as a model for \(\mathrm{SO}_{2}\) concentration above a forest. Suppose the parameter values are \(\mu=1.9\) and \(\sigma=.9 .\) a. What are the mean value and standard deviation of concentration? b. What is the probability that concentration is at most 10 ? Between 5 and 10 ?

Suppose that the pdf of \(X\) is $$ f(x)=\left\\{\begin{array}{cc} 5-\frac{x}{8} & 0 \leq x \leq 4 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Show that \(E(X)=\frac{4}{3}, V(X)=\frac{8}{9}\). b. The coefficient of skewness is defined as \(E\left[(X-\mu)^{3}\right] / \sigma^{3}\). Show that its value for the given pdf is \(.566\). What would the skewness be for a perfectly symmetric pdf?
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