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Course: AP®︎/College Statistics > Unit 8
Lesson 3: Transforming random variablesExample: Transforming a discrete random variable
Example of transforming a discrete random variable.
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Could someone please explain how to read the question? "The table show the probability distribution of X, the number of shots that Anush makes in a set of 2 attempts" what does this sentence suggest.(5 votes)- The game requires you to shoot 2 free throws. X is the random variable which we make equal to the number of free throws she makes. When you take two shots, one of three things can happen. Either you will miss both shots (X=0), make one shot (X=1), or make both shots (X=2). These are the only possible results for X, therefore X is a discrete random variable, because the possible values of the variable are countable. The probability distribution of X just means to show the probability of each value of X occurring. The table shows the probability of each value of X, therefore it shows the probability distribution of X. There is a 16% chance that she makes zero shots, or P(X=0) = .16. There is a 48% chance that she makes one shot, P(X=1) = .48 and there is a 36% probability that she makes two shots P(X=2) = .36. As you can see, these probabilities add up to 1 because these are the only possible values for X.(5 votes)
Could someone help me? Thanks
I understand N = 10x - 15.
But I don't understand why x can be replaced by ux (the mean of x) and get the answer.
I have watched this video for at least 10 times! OMG I am so stupid(3 votes)- We can think of it as first multiplying a particular point from the random variable X by a scaling factor which in this case is the mean of X, and that would leave 10*mean of X, and shifting the random variable will mean it shifts this new mean by the same amount so combining the two effects, mean of N = 10* mean of X -15(1 vote)
- I'm not clear why the mean is 10(1.2) - 15 = 3
If we take the outcomes of the scaled distribution and perform the mean calculation with those values we get:
-15(0.16) + -5(0.45) + 5(0.36) = -2.85
What am I not understanding here?(2 votes)- You have a typo.
The probability of net gain of -5 is not 0.45, but 0.48.
Hence, the mean is the same in both calculation.
-15(0.16) + -5(0.48) + 5(0.36) = -3(2 votes)
- Why are we scaling the standard deviation by 10? Or rather why is the random variable being scaled by 10?(2 votes)
- Because we change the random variable from "X = the number of shots" to "Y = the net gain" and Y = 10 X - 15 where 10 = the gain by shot and 15 = the cost by game (containing 2 attempts).(1 vote)
- What bothers me in such questions is the amount of rounding error that occurs. Calculating from scratch, I have a mean of -2.85, while Sal calculated the new mean from the old mean and came up with -3.
Hope this is pointed out in questions in a way that one knows which answer is required.(1 vote)- Rounding errors can indeed occur, especially when dealing with approximations or calculations involving multiple steps. It's essential to be aware of potential rounding discrepancies and to ensure consistency in rounding throughout the calculation process. In this case, if you calculated a mean of -2.85 while the solution derived a mean of -3, it could be due to differences in rounding methods or intermediate calculations. It's crucial to follow the instructions provided in the question regarding rounding and precision to obtain the desired result.(1 vote)
I'm a little bit confused in this exercise. At1:51, the probabilities are preserved because they come from X. I understand and it makes sense to me.
However, I don't understand the following:
As I understand, the mean and std are for both the random variable and probability distribution. Then, if we transform the mean and std (by scaling or shifting), then the probability will also be affected. However, in this exercise, the probabilities don't change, why? Maybe, my understanding is wrong.(1 vote)
In this problem, the three probabilities that Anush makes 0, 1, or 2 shots, as given by the table, never change. They don't depend on how much each game costs or how much he wins per shot. All we are doing is "mapping" the expected value and standard deviation of successful shots and using that information to calculate the net gain. That is why we are doing transformations.
For instance, since we know that he is expected to make an average of 1.2 shots each game over many trials, we multiply that by $10 since one shot = 10 dollars gained. We then subtract $15 dollars since we needed to pay that much to actually play the game. This transformation results in the average net gain per game we were looking for.(1 vote)
- Hi, the following question was in one of the quizzes fo this section:
A construction company submitted bids for three contracts. The company estimates that it has a P percent chance of winning any given bid. Let X represent the number of bids the company wins. Here is the probability distribution of X along with summary statistics.
The question was to calculate the net gain based on the number of bids won given that the cost for each bid is 2000 $.
When I did the transformation I subtracted the cost for the three bids the company submitted (3*2000) but the solution expresses the net gain as the gross gain for any given bid won minus only 2000 (the cost for that bid). Which does not make sense to me. Whatever number of bids the company wins, the cost will always be 6000.(1 vote)- In the context of the construction company bidding scenario, your approach of subtracting the total cost for all bids submitted makes sense. If the company submits three bids, the total cost would indeed be $6000 (3 * $2000), regardless of how many bids they win. Therefore, it seems logical to subtract $6000 from the gross gain to calculate the net gain. However, if the solution expresses the net gain as the gross gain for any given bid won minus only $2000 (the cost for that bid), it might be an oversight or a simplification made in the context of the question. To accurately represent the net gain, it should indeed reflect the total cost of all bids submitted.(1 vote)
- Would like to know , why would we need ti find the net gain at first if we can just come up with the "equation" N=10X-15 in the first place. If I came up with that eqt in the first place then I can just skip to2:50and just come up with the answers ( as the original mean and SD were given)(1 vote)
You don't need it. Sal does it in the video only for illustration purposes.(1 vote)
- why is the variance of the transformed random variable
not a^2 *var(N)?
Please let's engage.(1 vote)- The variance of a transformed random variable is indeed affected by both scaling and shifting. If a random variable X is transformed into a new random variable N by scaling it by a factor a and shifting it by a constant b, the variance of N is given by a^2 times the variance of X, not a^2 times the standard deviation of X. So, in your context, if N is the net gain (transformed from X representing the number of bids won), the variance of N would be a^2 times the variance of X, where a is the scaling factor.(1 vote)
- i disagree with the editing. I know that the editing did not remove any important information, but too much editing will disrupt the follow of thought, especially when the mouse/pointer jumps from one place to another. Please don't over edit the video. Edit when it can improve the flow or make the concept clearer, but don't edit just for the sake of saving a second or two here and there.(0 votes)
1:51Video transcript
- [Instructor] Anush is
playing a carnival game that involves two free throws. The table below displays the
probability distribution of X, the number of shots that
Anush makes in a set of two attempts, along with
some summary statistics. So here's the random variable X it's a discrete random variable. It only takes on a finite number of value, sometimes you can say
it takes on a countable number of values. We see we can either
make zero free throws, one, or two of the two. And the probability that
he makes zero is here, one is here, two is here. And then they also give us the mean of X and the standard deviation of X. Then they tell us if the
game costs Anush $15 to play and he wins $10 per shot he makes, what are the mean and standard
deviation of his net gain from playing the game N? All right, so let's define
a new random variable N. Which is equal to his net gain. Net gain. We can define N in terms of X. What is his net gain going to be? Well let's see, N, it's
going to be equal to 10 times however many shots he makes. So it's gonna be 10 times X. And then no matter what,
he has to pay $15 to play. Minus 15. In fact, we can set up a little table here for the probability distribution of N. So, let me make it right over here. So I'll make it look just like this one. N is equal to net gain. And here we'll have the probability of N. And there's three outcomes here. The outcome that corresponds
to him making zero shots, well that would be 10 times zero minus 15. That would be a net gain of negative 15. And would have the same probability, 0.16. When he makes one shot,
the net gain is gonna be 10 times one minus 15 which is, negative five. Which is going to have
the same probability. He has a 48% chance of making one shot. And so it's a 48% chance
of losing five dollars. And then last, but not least, when X is two his net gain
is gonna be positive five. Plus five. And so this is a 0.36 chance. So what they want us to figure out are, what are the mean and standard deviation of his net gain? So let's first figure out the mean of N. Well, if you scale a random variable the corresponding mean is going to be scaled by the same amount. And if you shift a random variable the corresponding mean is gonna be shifted by the same amount. So the mean of N is gonna be 10 times the mean of X minus 15. Which is equal to 10 times 1.2 minus 15. 1.2. So there's 12 minus 15 which is equal to negative three. Now the standard deviation of N is gonna be slightly different. For the standard
deviation, scaling matters. If you scale a random
variable by a certain value you would also scale
the standard deviation by the same value. So this is going to be equal to 10 times the standard deviation of X. Now you might say, what
about the shift over here? Well, the shift should
not affect the spread of the random variable. If you're scaling the
random variable, well, your spread should grow by the amount that you're scaling it. But by shifting it, it doesn't affect how much you disperse from the mean. So standard deviation is
only affected by the scaling, but not by the shifting. So this is going to be 10 times 0.69 which is going to. This was an approximation. So I'll say this is approximately to 6.9. So this is our new
distribution for our net gain. This is the mean of our net gain. And this is roughly the standard
deviation of our net gain.