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Chapter 3: Problem 33

Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year of operation. The pdf that characterizes the proportion \(Y\) that make a profit is given by $$ f(x)=\left\\{\begin{array}{ll} k y^{4}(1-y)^{3}+ & 0 \leq y \leq 1, \\ 0_{1} & \text { elsewhere. } \end{array}\right. $$ (a) What is the value of \(k\) that renders the above a valid density function? (b) Find the probability that at most \(50 \%\) of the firms make a profit in the first year. (c) Find the probability that at least \(80 \%\) of the firms make a profit in the first year.

Short Answer

Expert verified
To summarise, the constant value \(k\) for the PDF is found by solving the first integral. The probability of at most 50% of firms making a profit in the first year is calculated from the second integral, and the probability of at least 80% of firms making a profit in the first year is calculated from the third integral. The actual numerical values depend on the value of \(k\) and on the results of the integrations.

Step by step solution

01

Finding the value of \(k\)

To find the value of \(k\), it's essential to understand that a function is a valid probability density function (PDF) if it integrates to 1 over its entire domain. Here, the domain of the function is [0, 1]. Therefore, we have to solve the integral to find \(k\):\[\int_0^1 ky^{4}(1-y)^{3} dy = 1\]This \(k\) will be our normalization constant.
02

Calculating the probability that at most 50% of firms made a profit

To calculate this probability, we need to integrate the PDF from 0 to 0.5. This is because the percentage of firms making a profit is represented by \(y\) and 'at most 50%' corresponds to \(y \leq 0.5\). Thus, we form the following integral:\[p(y\leq 0.5) = \int_0^{0.5} ky^{4}(1-y)^{3} dy\]Evaluate this integral to calculate the probability.
03

Calculating the probability that at least 80% of firms made a profit

Similarly, to calculate the probability that at least 80% of firms made a profit, we need to integrate the PDF from 0.8 to 1. This is because 'at least 80%' corresponds to \(y \geq 0.8\). Thus, the following integral is formed:\[p(y\geq 0.8)=\int_{0.8}^1 ky^{4}(1-y)^{3} dy\]Evaluate this integral to find the probability.

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Key Concepts

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

Normalization Constant
In probability theory, the concept of a normalization constant is essential for ensuring that a probability density function (PDF) correctly represents a probability distribution. For a PDF to be valid, the total probability of all possible outcomes must equate to 1. This is a fundamental rule in probability, reflecting the axiom that a certainty (the occurrence of some outcome) has a probability of 1.

A normalization constant modifies the function so that this criterion is met. In the context of the exercise, the constant 'k' functions as this normalization constant. To find 'k', integral calculus is employed to integrate the function over its domain, which is from 0 to 1 in this case. We set this integral equal to 1 because the total probability must be 1, and solve for 'k'. This process ensures that the function 'f(x)' is legitimately normalized and thus can be used for accurate probability estimation.
Integral Calculus
Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the areas under and between curves. In the context of probability, integral calculus is used to determine the sum of the probabilities across a continuum of outcomes, particularly when dealing with continuous random variables.

When performing probability estimations using a continuous PDF, the integral of this PDF across a given interval represents the probability that a random variable falls within that interval. In the exercise, integral calculus is used to find the normalization constant 'k' by setting up the integral of 'f(x)' over the range of [0, 1] to 1. It is also used to estimate probabilities for varying proportions of firms making a profit by integrating 'f(x)' over the desired intervals: [0, 0.5] for at most 50% and [0.8, 1] for at least 80%.
Probability Estimation
Probability estimation is a key concept in statistics and data analysis. It involves calculating the likelihood of different outcomes based on a given set of parameters. In the exercise, probability estimation is used to compute the chances of a specific event occurring within the first year of operation for data processing firms, based on the PDF 'f(x)'.

The integral calculus applied over the domain of the PDF tells us these probabilities. For instance, when the exercise asks for the probability that at most 50% of firms make a profit, the integral of 'f(x)' from 0 to 0.5 with the normalization constant 'k' applied gives the answer. Similarly, for the calculation of the probability that at least 80% of firms are profitable, the integral from 0.8 to 1 is used. These estimations are fundamental in helping stakeholders make informed decisions based on hypothetical scenarios.

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Most popular questions from this chapter

Suppose it is known from large amounts of historical data that \(X\), the number of cars that arrive at a specific intersection during a 20 second time period, is characterized by the following discrete probability function $$ /(x)=e^{-6} \frac{b^{x}}{x !}, \quad x=0,1.2, \ldots $$ (a) Find the probability that in a specific 20 -second time period, more than 8 cars arrive at the intersection. (b) Find the probabilitythat only 2 cars arrive.

Determine the values of \(c\) so that the following functions represent joint probability distributions of the random variables \(A^{\prime \prime}\) and \(Y\) : (a) \(f(x, y)-c x y,\) for \(x=1,2,3 ; y=1,2,3\) (b) \(f(x, y)=c|x-y|,\) for \(x=-2,0,2 ; y=-2,3\).

A service facility operates with two service lines. On a randomly selected day, let \(X\) be the proportion of time that the first line is in use whereas \(Y\) is the proportion of time that the second line is in use. Suppose that the joint probability density function for \((\mathrm{A}, \mathrm{V})\) is $$ f i x, y)=\left\\{\begin{array}{l} \frac{3}{2}\left(x^{2}+y^{2}\right),-0 \leq x, y \leq 1, \\ 0, \quad \text { elsewhere. } \end{array}\right. $$ (a) Compute the probability that neither line is busy more than half the time. (b) Find the probability that the first line is busy more than \(75 \%\) of the time.

Let \(X\) denote the diameter of an armored electric cable and \(Y\) denote the diameter of the ceramic mold that makes the cable. Both \(X\) and \(Y\) are scaled so that they range between 0 and \(1 .\) Suppose that \(X\) and \(Y\) have the joint density $$ f(x, y)=\left\\{\begin{array}{ll} \frac{1}{y}, & 0< x < y<1; \\ 10, & \text { elsewhere. } \end{array}\right. $$ Find \(P(X+Y>1 / 2)\).

Let the number of phone calls received by a switchboard during a 5 -minute interval be a random variable \(X\) with probability function $$ f(x)=\frac{e^{-2} 2^{x}}{x !}, \quad \text { for } x=0,1,2, \ldots $$ (a) Determine the probability that \(X\) equals \(0,1,2,3,\) \(4,5,\) and 6 (b) Graph the probability mass function for these values of \(x\). (c) Determine the cumulative distribution function for these values of \(X\).
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