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Theoumber of beakers that Tallula breaks in random variable X The one day is described by the discrete below: The domain of x is iprobability mass function for X is...

Question

Theoumber of beakers that Tallula breaks in random variable X The one day is described by the discrete below: The domain of x is iprobability mass function for X is flx) and is tabulated [0,7]. The value for f(7) is not given- Ve 0046 10.202 10.296 0.221 10.057 0.049 Jo.o1oCalculate the probability that tomorrow she will break at least 5 beakers. Give your answer as a decimal to three places.

Theoumber of beakers that Tallula breaks in random variable X The one day is described by the discrete below: The domain of x is iprobability mass function for X is flx) and is tabulated [0,7]. The value for f(7) is not given- Ve 0046 10.202 10.296 0.221 10.057 0.049 Jo.o1o Calculate the probability that tomorrow she will break at least 5 beakers. Give your answer as a decimal to three places.



Answers

For a laboratory assignment, if the equipment is working, the density function of the observed outcome, $X,$ is $$ f(x)=\left\{\begin{array}{ll} 2(1-x), & 0<x<1 \\ 0, & \text { otherwise } \end{array}\right. $$ (a) Calculate $P(X<0.5)$. (b) What is the probability that $X$ will exceed $0.4 ?$ (c) Given that $X \geq 0.5,$ what is the probability that $X$ will be less than $0.7 ?$

Yeah. That's probably been given the sample space of a random experience. Experiment with a further random variable being defined. That's false. Now we want to determine the probability mass function from this. Now we know that each outcome is equally likely from A through F. So since we have six options there that means the probability of A. Is equal to the probability of B. Does include all these probabilities Which is 1 6. Since they were equally likely. Now here on a. We want to find the probability the X. Yeah. Is everyone 1.5. Yeah. Now there are two different values on their where access 1.5 and that's at c. and D. So since each of those is equal to 1/6 I mean this probability is 26 which is one third. So the probability of 1.5 is one third mm On dealing with the probability of being between .5 and 2.7. Now count all the different values Where access between 2.4.5 and 2.7. And there are three of them c. d. and E. And so he and since they're equally likely that means this is 36 Which is one half. And so that probability is 1/2 On C. We want the probability access greater than three And there are no values on that table where actually greater than three. And so that's a probability of zero On the we want the probability access between zero and 2. It can be zero. It cannot be too. And at A B. C. And D we all have those values. And so there are four values Each with a probability of 1 6. And so this is 46 Which is 2/3. And then lastly on E We want the probability at zero or that's us too. Mhm. And there are three different values for X0 or two. And so this is 36 which gives us the probability of one half.

So we're given the probability function. 1924. What were they doing? Quite pretty late. We're required to calculate probability that the number lies between 14 men. Good night. The animal selected number. So this given by the integral 14 to 29 if of X BX people 15 by 24 Uh huh. Five by eight is required probability.

All right, So here's our function. Okay, Times x to the fifth to find on this interval zero to six. And we want to find the value of k said that this is a probability distribution function as long as case not negative f will be no negative. And I think we'll find the cable have to be not negative. So we know that if we integrate dysfunction from zero to six k extra dex, that needs to be once or just use in case of this, this mineral has normalized. Okay, so what do we get? We get k over six times X to the sex and I waved from zero to six. It's one. So in other words, uh, see that six to the sixth over. Six Time's kay is one. And if we saw for Kay, we get one over six to the death

Right here We have dysfunction. That is K X cued on to find on the intervals your five. And as long as case not negative on the centre hole we'LL be So you just want to choose K to normalize thie Integral to make this integral of f from zero to five equal to one. So we get K over four x to the fourth, evaluated from zero to five. Needs to be one. Okay, but that's just k times five to the fourth over. Four is one. Or in other words, K is for over twelve, five times in the fourth. Um, well, this let's just leave it inspected the fourth twenty six, twenty five times five. So this is the value of K that will make a probability distribution function.


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