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Chapter 23: Problem 5

A p.d.f. for the continuous variable \(X\) is given by $$ f(x)=\mathrm{e}^{-x}, \quad x>0 $$ If \(P(0

Short Answer

Expert verified
Answer: α = -ln(0.5)

Step by step solution

01

Compute the CDF

To compute the cumulative distribution function, we will integrate \(f(x)\) from \(0\) to \(x\). Since the p.d.f. is non-negative and continuous, we can safely integrate it. $$ F(x) = \int_{0}^x f(t) dt = \int_{0}^x \mathrm{e}^{-t} dt $$
02

Evaluate the integral

To evaluate the integral above, we will use the anti-derivative of the exponential function. The anti-derivative of \(\mathrm{e}^{-t}\) is \(-\mathrm{e}^{-t}+C\), where C is the constant of integration. In our case, since we're integrating between 0 and x, we don't need to worry about C. $$ F(x) = -\mathrm{e}^{-x} + \mathrm{e}^{0} $$ Simplifying, we get: $$ F(x) = 1 - \mathrm{e}^{-x} $$
03

Find the value of \(\alpha\) for which \(F(x) = 0.5\)

Now that we have the CDF \(F(x)\), we can solve for \(\alpha\) by setting \(F(\alpha) = 0.5\): $$ 0.5 = 1 - \mathrm{e}^{-\alpha} $$ Rearranging to isolate \(\alpha\): $$ \mathrm{e}^{-\alpha} = 1 - 0.5 = 0.5 $$ Taking the natural logarithm of both sides, we get: $$ -\alpha = \ln{0.5} $$ Finally, we find the value of \(\alpha\): $$ \alpha = -\ln{0.5} $$ So, the value of \(\alpha\) for which \(P(0<X<\alpha)=0.5\) is \(\alpha = -\ln{0.5}\).

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Key Concepts

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

Cumulative Distribution Function
When we talk about the cumulative distribution function (CDF), what we're referring to is a crucial concept in the study of probability and statistics. It gives us the probability that a continuous random variable, such as a time measure or a weight, will take on a value less than or equal to a specific value. In simpler terms, it helps you understand 'how much' of the probability has 'accumulated' up to a certain point.

To find a CDF for a given probability density function (pdf), we perform integration. The process involves calculating the area under the probability curve from the lowest possible value of the variable up to the value of interest. For a continuous random variable represented by the pdf \( f(x) \), the CDF \( F(x) \) is given by integrating \( f(x) \) from its lower bound up to \( x \). So if you are told that \( P(0
Exponential Distribution
Exponential distribution is quite special among the various probability distributions available, often associated with the time until an event occurs, like waiting times between buses or the lifespan of an electronic component. Think of it as a way to describe the time we wait for some continuous process that has no memory, meaning past events don't affect the future ones.

The pdf of an exponential distribution, represented as \( f(x) = \mathrm{e}^{-x} \) when \( x > 0 \), decreases exponentially as \( x \) increases. This characteristic means that the probability of our event taking longer to happen reduces as time goes by — it's more likely to happen sooner rather than later. When you're asked to find \( \alpha \) in the context of this distribution given that \( P(0
Integration of Exponential Function
The integration of the exponential function might initially come off as a challenge, but it's a procedure grounded in basic calculus. When we integrate the function \( \mathrm{e}^{-x} \), we're essentially seeking the antiderivative, which is a function whose derivative would give us \( \mathrm{e}^{-x} \).

To find the integral of \( \mathrm{e}^{-x} \), you would perform a simple antiderivative process that results in \( -\mathrm{e}^{-x} + C \), with \( C \) being the constant of integration. However, when calculating definite integrals, like those needed for a CDF, the constant \( C \) cancels out because it's subtracted as part of the process that evaluates the antiderivative at the upper and lower bounds of integration.

It is by integrating the exponential function that we can derive the CDF, and from there, solve for specific probabilities such as finding the desired \( \alpha \) that meets our given condition \( P(0

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Most popular questions from this chapter

A firm has 1400 employees. The probability that an employee is absent on any day is \(0.006\). Use the Poisson approximation to the binomial distribution to calculate the probability that the number of absent employees is (a) eight (b) nine

Components are manufactured by three machines, \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). Machine A makes \(30 \%\) of the components, machine B makes \(25 \%\) of the components and machine \(\mathrm{C}\) makes the rest. Of those components made by machine A, \(6 \%\) are substandard; when made by machine B, \(3 \%\) are substandard; and when made by machine C, \(5 \%\) are substandard. A component is picked at random. Calculate the probability that it is (a) substandard (b) made by machine B given it is substandard (c) made by either machine \(\mathrm{A}\) or machine \(\mathrm{B}\) (d) substandard and made by machine B (e) substandard, given it is made by machine \(\mathrm{A}\) (f) made by machine A or is substandard.

The standard deviation of the values \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\) is \(\sigma .\) Calculate the standard deviation of the values \(k x_{1}, k x_{2}, k x_{3}, \ldots, k x_{n}\) where \(k\) is a constant.

Precision components are made by machines A, B and C. Machines A and C each make \(30 \%\) of the components with machine \(\mathrm{B}\) making the rest. The probability that a component is acceptable is \(0.91\) when made by machine A, \(0.95\) when made by machine B and \(0.88\) when made by machine \(\mathrm{C}\). (a) Calculate the probability that a component selected at random is acceptable. (b) A batch of 2000 components is examined. Calculate the number of components you expect are not acceptable.

The lifespans, \(L\), of 2500 components were monitored and recorded in Table 2 . $$ \begin{array}{rr} \hline \text { Lifespan, } L \text { (hours) } & \text { Frequency } \\ \hline 0 \leq L \leq 5000 & 16 \\ 500012000 & 145 \\ \hline \end{array} $$ Calculate the probability that a component picked at random has a lifespan (a) of between 8000 and 11000 hours (b) of more than 11000 hours (c) of more than 10000 hours given that it has already lasted for at least 8000 hours (d) of between 11000 and 12000 hours given that it has already lasted for at least 9000 hours.
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