Example: Transforming a discrete random variable | Random variables | AP Statistics | Khan Academy

Anush is playing a carnival game that involves shooting 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 values. Sometimes people say it takes on a countable number of values, but we see he can either make 0 free throws, 1, or 2 of the two. The probability that he makes zero is here, one is here, and two is here. They also give us the mean of ( x ) and the standard deviation of ( x ).

Then they tell us if the game costs Anush fifteen dollars to play and he wins ten dollars per shot he makes, what are the mean and standard deviation of his net gain from playing the game ( n )?

Right, so let's define a new random variable ( n ), which is equal to his net gain. Net gain can be defined in terms of ( x ). What is his net gain going to be? Well, let's see: ( n ) is going to be equal to 10 times however many shots he makes. So it's going to be ( 10 \times x ), and then no matter what, he has to pay 15 dollars to play, minus 15.

In fact, we could set up a little table here for the probability distribution of ( n ). So let me make it right over here. I'll make it look just like this one. ( n ) is equal to net gain, and here we'll have the probability of ( n ). There's three outcomes here.

The outcome that corresponds to him making 0 shots: well, that would be ( 10 \times 0 - 15 ); that would be a net gain of negative 15. It would have the same probability ( 0.16 ).

When he makes one shot, the net gain is going to be ( 10 \times 1 - 15 ), which is negative 5. But it's going to have the same probability; he has a 48% chance of making one shot, and so it's a 48% chance of losing 5.

Last but not least, when ( x ) is 2, his net gain is going to be positive 5, ( +5 ). And so this is a 0.36 chance.

So what they want us to figure out are what the mean and standard deviation of his net gain are.

First, let’s 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 going to be shifted by the same amount.

So the mean of ( n ) is going to be ( 10 \times \text{mean of } x - 15 ), which is equal to ( 10 \times 1.2 - 15 ). This is ( 1.2 ), so it is 12 minus 15, which is equal to negative 3.

Now the standard deviation of ( n ) is going to 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 \text{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, 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 here. So this is going to be ( 10 \times 0.69 ), which is going to be approximately equal 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.