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The usage time of a telephone belonging to a specific brand may be looked upon as a random variable having normal distribution with a mean of 20 years and a varianc...

Question

The usage time of a telephone belonging to a specific brand may be looked upon as a random variable having normal distribution with a mean of 20 years and a variance 4 years: What is the probability that a randomly selected telephone will be used between 15 and 18 years?

The usage time of a telephone belonging to a specific brand may be looked upon as a random variable having normal distribution with a mean of 20 years and a variance 4 years: What is the probability that a randomly selected telephone will be used between 15 and 18 years?



Answers

A recent study of the lifetimes of cell phones found the average is 24.3 months. The standard deviation is 2.6 months. If a company provides its 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 months. Assume cell phone life is a normally distributed variable.

So we're given the problem. DNC function is two. It is bar minus 30. Do you really this 02 infinity? We need to find that brought the call. Should last nor last more than five minutes. So we need to find the probability is equal to integral 0 to 5. 50 duty. So this equal do point, name, name, name, name 546

All right. So let's say we are told that the average waiting time between two calls um we're trying to find the average waiting time, but between two calls, given that we wait 20 seconds or longer 50% of the time. So we're gonna use this function here. PFT equals one over r E to the minus t over our which is the waiting time function. And we're trying to find the average time given that p of tea greater than 20 Ah, 0.5 s o. We're gonna say 0.5 is gonna equal the integral from 20 to infinity of one over R yeah, to the minus t over our DT. And we're gonna use this to solve for our which is theat bridge waiting time. So Ah, this is gonna be 0.5 on the left side and over here. Ah, this is just is minus e to the minus t over are from 20 to infinity, which equals negative e to the negative t over our when tea's infinity is gonna be just negative. Infinity, which we'll come back to you in a second minus negative VSo plus e to be minus 20 over our And since so e to the minus infinity is just zero. Anything to the minus. Infinity goes to zero because it just becomes one. Overeat to the infinity and then plus plus Still, we have negative 20 over our equals 0.5. So if we take the ln of both sides, we get negative. 20 over our is gonna equal ln of 0.5 and so are are is gonna equal negative 20 over ln of 0.5, which is approximately equal to 28 point 854 So that's the average waiting time.

Problem 59. The function F of X and Y is a joint probability density function between the random variables X and Y. Both X and Y represent the time to failure in years that I am to figure in years of the tablet Pc and smartphone perspective we want to find the probability that the tablet Pc and smartphone we lost four years or longer. Then we want to find the probability for X To be greater than equals four. And why is greater than equals four. We can get this probability by double integrating F. Of X and Y over the center then equals the double integral for F. Of X and Y. The X devoid for the terrible for X from the smallest value from four to the greatest body. It's the greatest value is infinity. And Why is from 4 to anything? Let's integrate with rooms with two X. Then the integration of E. Sorry It's 4.06.06. The integration of gives E. And we divide by the differentiation of the work we substitute From 4 to infinity. And it was a lot of negative open three wide is considered a constant relative to X equals The integration from 4 to infinity for all .06 multiplied by. We substitute by infinity. First we can substitute by infinity. Then we take them we have minus all going to over to here we take the name it for it is about minus or going to X. When X approaches infinity minus we substitute by X equals four. It's E. It is about of negative appointed Want to blow it by. It was a lot of negative 4.3. Y anyway This element gives zero because it was about negative infant gives you then we have -4.3 but deployed by Minour it was over negative 2.8 deployed by the integration from four to infinity For it was about to open three Y do I? It equals 4.3 multiplied by it was a lot of negative 4.8. It is a negative it is a negative The integration of E. And we divide by the differentiation of the power Which is -4.3. We substitute from 4 to infant equals Buying us. It was a lot of negative a .8. But the blue eyed boy We substitute by infinity. Again we can't substitute for infinity. We take dynamic for it. To the vote of negative 4.3. Why when white approaches infinity minus we supposed to buy Y equals four. Then it was a lot of negative one point. Then the answer equals most of this gives zero and it's positive E. To the bottom negative point it multiplied by E. To the bar of negative 1.2 equals It was a lot of negative two equals 4.135

And this question we're told the length of time that the battery and he pulled out a cellphone, well hold enough charge to operate acceptably is normally distributed With a mean of 25.6 hours And the standard deviation of .32 hours. And we're told that she forgot to charge her phone. So today it's been 26 hours and 18 minutes. Today it's been 26.3 hours. And we want the probability that the phone will operate properly. So the probability that it will work. So the probability that the time It'll hold enough charges more than 26.3. Now, we're going to convert that to call it his see more than 26.3 -25.6 over 0.32 So 26.3 Mine instead over .32. That's basically the probability Z Is more than 2.19. So that's one minus the probability Z is less than 2.19. So we're going to look at our table and we're going to see the probability for Z being Less than 2.19 is .98. Yeah point 9857 Inseparably disease, less than point 2.19. So that's 1 -1857. Quality that it will operate normally is .0143. So very low chance.


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