/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 165 Determine the constant \(c\) so ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the constant \(c\) so that the following function is a probability mass function: \(f(x)=c x\) for \(x=\) 1,2,3,4.

Short Answer

Expert verified
The constant \( c \) is \( \frac{1}{10} \).

Step by step solution

01

Write the Probability Mass Function

We are given that the probability mass function is defined as \( f(x) = c x \) for \( x = 1, 2, 3, 4 \). Our task is to determine the constant \( c \) such that \( f(x) \) is a valid probability mass function.
02

Apply Probability Mass Function Condition

For \( f(x) \) to be a probability mass function, the sum of probabilities for all possible \( x \) must equal 1. Therefore, we must have: \[ f(1) + f(2) + f(3) + f(4) = 1. \]
03

Substitute Values into the Function

Substitute \( x = 1, 2, 3, 4 \) into the function \( f(x) \) and sum them up: \[ f(1) = c \cdot 1, \]\[ f(2) = c \cdot 2, \]\[ f(3) = c \cdot 3, \]\[ f(4) = c \cdot 4. \] Thus, the equation becomes: \[ c \cdot 1 + c \cdot 2 + c \cdot 3 + c \cdot 4 = 1. \]
04

Simplify the Equation

Simplify the equation from step 3: \[ c\cdot (1 + 2 + 3 + 4) = 1. \] The sum inside the parenthesis is 10, so the equation simplifies to: \[ 10c = 1. \]
05

Solve for the Constant c

Solve the equation \( 10c = 1 \) for \( c \): \[ c = \frac{1}{10}. \]
06

Verify the Solution

Verify that the found \( c \) satisfies the probability condition: \( f(1) = \frac{1}{10}, \) \( f(2) = \frac{2}{10}, \) \( f(3) = \frac{3}{10}, \) \( f(4) = \frac{4}{10} \). The sum of these is: \[ \frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} = 1. \] The solution is verified.

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Key Concepts

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

Understanding Discrete Probability Distributions
Discrete probability distributions are one of the fundamental concepts in probability and statistics. Essentially, they are functions that assign probabilities to discrete outcomes, such as integers. Imagine you are rolling a die; the possible outcomes (1, 2, 3, 4, 5, and 6) are all discrete.

In the context of this exercise, we see that the function given is \( f(x) = cx \) where \( x \) can take values 1, 2, 3, and 4. This is a classic example of a discrete probability distribution. Each different \( x \) value has an associated probability value, which is determined by the constant \( c \).

Key Characteristics of Discrete Probability Distributions:
  • The set of possible outcomes must be countable.
  • Each outcome is assigned a probability value.
  • The sum of all probabilities must equal 1.
These aspects ensure that the function \( f(x) \) is a valid probability mass function, allowing us to determine the constant \( c \) through further analysis.
Finding the Constant: Constant Determination
Determining the constant \( c \) in a probability mass function is a crucial step to ensure that the function is valid and useful for statistical analysis. For the function \( f(x) = cx \), the determination of \( c \) involves the use of the probability mass function condition.

To find the correct value of \( c \), we substitute each possible \( x \) value (1, 2, 3, 4) into the function and sum up the probabilities. This requires using the equation\[\sum_{i=1}^{4} f(x_i) = 1\]By substituting the function values into this equation as shown in the original solution, we perform the following steps:
  • Calculate \( f(1) = c \cdot 1 \)
  • Calculate \( f(2) = c \cdot 2 \)
  • Calculate \( f(3) = c \cdot 3 \)
  • Calculate \( f(4) = c \cdot 4 \)
The sum of these probabilities must equal 1, which gives us a straightforward equation to solve: \( 10c = 1 \). Solving for \( c \) gives \( c = \frac{1}{10} \).

This determined constant ensures that the sum of probabilities adheres to the rules of a valid probability mass function.
Ensuring Validity: Probability Conditions
To verify and validate a probability mass function, it must adhere to certain probability conditions. These conditions are the guiding principles that maintain the integrity of the distribution.

The primary condition is that the sum of all probabilities must equal 1. This ensures that all possible outcomes of the random variable are accounted for. It's like making sure that all pieces of a pie add up to the whole pie. For our function \( f(x) = cx \), substituting the calculated \( c \) for each value of \( x \) yields:
  • \( f(1) = \frac{1}{10} \)
  • \( f(2) = \frac{2}{10} \)
  • \( f(3) = \frac{3}{10} \)
  • \( f(4) = \frac{4}{10} \)
When these probabilities are added together, we validate that they indeed sum to 1:
\[\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} = 1\]
Another important condition is that each probability must be between 0 and 1, inclusive. This confirms that none of the probabilities are negative or exceed certainty. By confirming these conditions, we can ascertain that our probability mass function is both valid and reliable.

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Most popular questions from this chapter

A company employs 800 men under the age of 55 . Suppose that \(30 \%\) carry a marker on the male chromosome that indicates an increased risk for high blood pressure. (a) If 10 men in the company are tested for the marker in this chromosome, what is the probability that exactly one man has the marker? (b) If 10 men in the company are tested for the marker in this chromosome, what is the probability that more than one has the marker?

Derive the formula for the mean and standard deviation of a discrete uniform random variable over the range of integers \(a, a+1, \ldots, b\).

Show that for a discrete uniform random variable \(X\), if each of the values in the range of \(X\) is multiplied by the constant \(c,\) the effect is to multiply the mean of \(X\) by \(c\) and the variance of \(X\) by \(c^{2}\). That is, show that \(E(c X)=c E(X)\) and \(V(c X)=c^{2} V(X).\)

Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 140 cards, and 20 are selected without replacement for functional testing. (a) If 20 cards are defective, what is the probability that at least 1 defective card is in the sample? (b) If 5 cards are defective, what is the probability that at least one defective card appears in the sample?

Heart failure is due to either natural occurrences \((87 \%)\) or outside factors \((13 \%) .\) Outside factors are related to induced substances or foreign objects. Natural occurrences are caused by arterial blockage, disease, and infection. Suppose that 20 patients will visit an emergency room with heart failure. Assume that causes of heart failure between individuals are independent. (a) What is the probability that three individuals have conditions caused by outside factors? (b) What is the probability that three or more individuals have conditions caused by outside factors? (c) What are the mean and standard deviation of the number of individuals with conditions caused by outside factors?

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