A continuous random variable x has pdf fx where fx frac4kx2...

A continuous random variable X has PDF f(x) where $f(x) = \frac{4k}{x^2}$ for $x \ge 1$ $= 0$ otherwise. (i) Find the value of k. (ii) The lifetime (X in years) of an electrical component has this distribution. Find the time after which 90% of the components will have failed.

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...

Question

A continuous random variable X has PDF f(x) where $f(x) = \frac{4k}{x^2}$ for $x \ge 1$ $= 0$ otherwise. (i) Find the value of k. (ii) The lifetime (X in years) of an electrical component has this distribution. Find the time after which 90% of the components will have failed.

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