Probability distribution function

In summary, the conversation discusses finding the value of the constant k in a technical specification for an electrical product, which is given by the function f(t) = 1 - ke^(-t/t0) if 0 < t < tmax and f(t) = 0 if t > tmax. The solution involves finding the integral of the function between 0 and 10 years and setting it equal to 1, which results in the equation 10 - k*e^(-10/t0)*(-t0) = 1. One solution provided in a past exam paper is k = -9/(t0(e^(-10/t0) - 1)), while the other solution is incorrect due to a mistake in
  • #1
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Homework Statement


The technical specification of a particular electrical product states that the probability of its failure with time is given by the function:

f(t) = 1 - ke^(-t/t0) if 0 < t < tmax
f(t) = 0 if t > tmax

where t is the time of service in years, and the constant t0 = 100 years defines the characteristic deterioration time. If the maximal life span of the product tmax is 10 years,
find the value of constant k , explaining its meaning,

Homework Equations


i thought i was supposed to find the integral of 1 - ke^(-t/t0) with respect to t between 0 and 10 years and put the answer equal to 1. then find k from that. [/B]

The Attempt at a Solution



now this is a past exam paper question so i actually have the partial solution to the question which says that:

k = -9/(t0(e^(-10/t0) - 1))

but what i get is k = -9/(t0e^(-10/t0))

I'm pretty sure I'm doing the integral correctly so there must be some step I'm missing that accounts for that extra -1 in the denominator.

[/B]
 
Last edited:
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  • #2
Did the past exam question have exactly the same problem statement?
The second solution looks odd. It gives wrong results for very large t0, for example. How did you get it?

Why do both solutions depend on t? They should not do that. Is that tmax?
 
  • #3
yeah sorry that t should be tmax = 10... I've edited it
 
  • #4
i just did the integral 0 and 10 of the function given and put it equal to 1 because the probability of it failing within a 10 year period is 100%. once I've done the integral I get:

10 - k*e^(-10/t0)*(-t0) = 1 (int. between 0 and 10 of 1 is 10 and int. between 0 and 10 of k*e^(-t/t0) is k*e^(-10/t0)*(-t0) )

then i find k from that. is this wrong?
 
  • #5
OMG I just realized what I did! I'm really sorry! never mind! please ignore!
 
  • #6
i considered e^0 = 0 I've made this mistake quite a few times especially in integrlas! o0)
 

What is a probability distribution function (PDF)?

A probability distribution function (PDF) is a mathematical function that describes the likelihood of a random variable taking on different values. It maps all possible outcomes of a random variable to their corresponding probabilities.

What is the difference between a PDF and a probability mass function (PMF)?

While both a PDF and a PMF describe the probability of a random variable taking on certain values, a PDF is used for continuous random variables while a PMF is used for discrete random variables. This means that a PDF assigns probabilities to ranges of values, while a PMF assigns probabilities to specific values.

What is the area under a PDF curve?

The area under a PDF curve represents the total probability of all possible outcomes. This area is equal to 1, as the sum of all possible probabilities must equal 1.

What is the relationship between a PDF and a cumulative distribution function (CDF)?

A CDF is the cumulative sum of the probabilities in a PDF. It represents the probability that a random variable will take on a value less than or equal to a given value. In other words, the CDF is the integral of the PDF.

How is a PDF used in statistics?

In statistics, a PDF is used to analyze and understand the behavior of random variables. It allows for the calculation of probabilities for different outcomes, as well as the determination of important measures such as mean, median, and variance.

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