The lifetime, in years, of some electronic component is a continuous random variable with the de

step 1 Welcome to this lesson. In this lesson we have the lifetime in years of some electronic componen

The lifetime, in years, of some electronic component is a...

Key Concepts

Tail Probability (Survival Function)

This concept refers to the probability that a continuous random variable exceeds a specific value. It can be computed as one minus the cumulative distribution function at that value. This measure is particularly useful for analyzing the longevity or reliability of components and systems.

Probability Density Function

This concept describes a function that defines the likelihood of a continuous random variable taking on a particular value. The density function, when integrated over a range, gives the probability that the variable falls within that interval, provided the function is non-negative and integrates to one over its domain.

Normalization of a Density Function

This principle involves determining a constant that scales the probability density function so that the total probability over its domain is equal to one. For continuous distributions, this is achieved by integrating the density function over its entire support and setting the result equal to one.

Cumulative Distribution Function

The cumulative distribution function (CDF) gives the probability that a random variable is less than or equal to a certain value. It is defined as the integral of the probability density function from the lower bound of the distribution up to that value. The CDF is a non-decreasing function that approaches one as the variable approaches the upper limit of its support.

Transcript

00:01 Welcome to this lesson.

00:02 In this lesson we have the lifetime in years of some electronic component that is a continuous random variable with density of x equals to k on x for x that is greater than or equals to 1 and 0 else y.

00:18 So in the first part we will find k.

00:20 Now to find k we use the fact that the integral from 1 to infinity of f of x dx should be equals to 1.

00:30 In other words the summation of the total for the continuous random variable would be possible so here we have integral from 1 to infinity of k on x to the power 4 okay the x is equal to 1 so the few adjustment that will make first of 4 let the k come out then we can have the limit as b approaches infinity of integral from 1 to b and this would be 1 on x to the power 4 ds or simply put we can make it x to the power negative four right this should be cosine one so we'll have k and let me bring the limits as b approaches infinity now if we take the integral of this we have extra power negative four plus one all over negative four plus one all right evaluating it from 1 to b this should be cosine 1 so that we can find k now this is k on negative 3 so make it negative k on 3 all right so have the limit as b approaches infinity of 1 on x to the power 3 evaluating it from 1 to b is equals to one okay so let's put in the limit so we'll have one on b to the path rate all right minus one on one to the path rate this should be equals to one but immediately we'll put the limit in there so i can bring k negative k on trade there now if we put the limit here if we as b approaches infinity the whole of this goes to zero right so have zero minus one right that should be close now let's multiply through by three so we have negative k minus 1 equals 3 and that's actually half k equals 3 so the value of k is 3 so now that we know the value of k equals 3 f of x equals u the piecewise function so 3 on x to the power 4 for x greater than or cost 1 and 0 elsewhere all right then the next part will be nice to find the cumulative distribution function so cumulative or the cdf alright so here we have the cdf capital f of x that is equals to integral from 1 to x of the f of t dt right so here we have the cumulative distribution function that's supposed to go from 1 to x of 3 t to the power 4 and that's like you can solve this you can bring that three out we have t to the power negative 4 t so big big f of x this equals you we have this three out then t the power negative 3 or negative 3 right evaluating it from 1 to x so here we'll have negative that is negative then we have t to the power negative 3 right from 1 to x so this is equals to the negative out we have one on x the pathway minus one on one to the pathway so the cumulative distribution function this will have as one minus one on x to the pathway right then the last part that is the pathway will find probability that the lifespan exceeds two years.

05:59 So say that the probability that the lifetime...

06:16 Alright, let's continue.

06:18 So probability for the lifespan to exceed two years.

06:23 This is equal to you 1 minus probability that x is less than or equal to 2...

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