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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. 0.81, c. 1.17, d. 0.97, e. 2.41.

Step by step solution

01

Understanding the Problem

We need to find the constant \(c\) that satisfies different probability statements using the properties of the standard normal distribution \(Z\). This involves using the cumulative distribution function \(\Phi\) and probability properties of \(Z\).
02

Solving Part a: \(\Phi(c)=.9838\)

The cumulative distribution function \(\Phi(c)\) gives the probability that \(Z\) is less than or equal to \(c\). Using standard normal distribution tables or calculators, find \(c\) such that \(\Phi(c) = 0.9838\). The value of \(c\) is approximately 2.13.
03

Solving Part b: \(P(0 \leq Z \leq c)=.291\)

The probability \(P(0 \leq Z \leq c) = 0.291\) means we need \(\Phi(c) - \Phi(0) = 0.291\). Since \(\Phi(0) = 0.5\), solve \(\Phi(c) = 0.791\). The value of \(c\) is approximately 0.81.
04

Solving Part c: \(P(c \leq Z)=.121\)

The probability \(P(c \leq Z) = 0.121\) translates to \(1 - \Phi(c) = 0.121\). Solve \(\Phi(c) = 0.879\) to find \(c\). The value of \(c\) is approximately 1.17.
05

Solving Part d: \(P(-c \leq Z \leq c)=.668\)

This probability is symmetrical around 0. So, \(2\Phi(c) - 1 = 0.668\). Solving \(\Phi(c) = 0.834\), the value of \(c\) is approximately 0.97.
06

Solving Part e: \(P(c \leq |Z|)=.016\)

Here, \(P(|Z| \geq c) = 0.016\), or equivalently, \(2P(Z \geq c) = 0.016\). So, \(P(Z \geq c) = 0.008\), meaning \(\Phi(c) = 0.992\). The value of \(c\) is approximately 2.41.

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function, often abbreviated as CDF, is a fundamental concept in statistics that describes the probability that a random variable takes on a value less than or equal to a specific amount.
For the standard normal distribution, the CDF is denoted by \(\Phi(c)\).
Unlike a probability density function which displays likelihood across various potential outcomes, the CDF gives the cumulative probability up to and including a given value.
This means when we say \(\Phi(c)\), we are talking about the probability that the standard normal variable \(Z\) is less than or equal to \(c\).
In practice, the CDF helps in determining probabilities for the standard normal distribution when calculating a specific constant \(c\).
  • For example, \(\Phi(c)=0.9838\) implies a probability of 98.38% that \(Z\) will be less than or equal to \(c\).
This makes it a powerful tool in statistical analysis, aiding in the computation of probabilities related to normally distributed data.
Probability Properties of Z
The standard normal distribution, represented by \(Z\), is a special form of the normal distribution with a mean of 0 and a standard deviation of 1.
This leads to some unique properties that make \(Z\) highly useful in statistical calculations.One important property is symmetry around the mean, which means that the distribution of values around 0 looks the same on both sides.
Another key property involves the total area under the curve, which equals 1 since the total probability for all outcomes must account to 100%.
For specific probability computations, such as finding \(P(0 \leq Z \leq c)\), knowing these properties allows for adjustments using the CDF, \(\Phi\), to compute probabilities.
Understanding these properties of \(Z\) enables solving complex probability queries by dissecting them into manageable components, such as when we need to deduce values like \(\Phi(c)\) or adjust probabilities based on symmetrical distribution segments.
Standard Normal Distribution Tables
Standard normal distribution tables, or Z-tables, are an essential tool for finding probabilities and critical values associated with the standard normal distribution.
These tables typically show the area to the left of a given Z-score, essentially providing \(\Phi(c)\) for various values of \(c\).
Using these tables allows students to determine exact probabilities without extensive calculations, simply by locating values corresponding to Z-scores on the table. For instance, if \(\Phi(c) = 0.791\), you would search the table to find that \(c\) is approximately 0.81.
Though calculators and statistical software have largely replaced manual table lookups, understanding how to read and interpret Z-tables remains a crucial skill for foundational statistics courses.
  • Z-tables are valuable for making quick probability and percentile calculations.
  • They provide an easy cross-reference for understanding the relationship between Z-scores and probabilities.
This understanding is applicable to a wide range of statistical applications beyond classroom exercises.
Solving Probability Equations
Solving probability equations involves interpreting the probability expression and rewriting it in terms of a cumulative distribution function or another statistical expression to find the unknown variable.
In exercises involving the standard normal distribution, this often translates into finding a constant \(c\) for which the given probability holds.
A typical approach includes setting up the equation based on given probabilities, transforming these probabilities using properties of Z, and leveraging the CDF or Z-tables to find solutions.
This method might include breaking down the expressions, such as \(P(c \leq Z) = 0.121\), to equivalent forms like \(1 - \Phi(c) = 0.121\), which subsequently helps find \(\Phi(c) = 0.879\).
Once altered into forms that involve \(\Phi\), you can use Z-tables or calculators to extract the corresponding value of \(c\).
  • Always ensure consistency by checking that the sum of associated values matches logical statistical rules.
  • Present the results efficiently with minimal calculation errors by following structured problem-solving steps.
Mastering these techniques is integral for statistical problem solving and converts complex expressions into manageable solutions.

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Most popular questions from this chapter

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