4 4 points customers at a restaurant pay with a credit card...

4. [4 points] Customers at a restaurant pay with a credit card (CC), debit card (DC), or cash(C). Assume that successive customers make independent choices, with P(CC)=.5, P(DC)=.2, and P(C)=.3. a). [2 points] Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Please define the random variable you are using to answer this question and write down the distribution your defined variable follow. b). [2 points] Among the next 100 customers, what are the mean and variance of the number who don't pay with cash? Please define the random variable you are using to answer this question and write down the distribution your defined variable follow.

Transcript

00:01 We have three ways to pay at a gas station.

00:03 We could pay with credit card, which we denote by the event a, debit card, which is b, and cash is c.

00:09 So we're given that the customer's choices are independent, and we have some probabilities.

00:15 The probability of paying with credit card for any some customer is 0 .5.

00:20 The probability of a debit card, oops, p of b, is 0 .2, and the probability of a cash is 0 .3.

00:30 So if we have 100 customers, so we'll say n equals 100, because that's our sample size, what are the mean and variance of the number who pay with a credit card? so since we're considering only the credit cards here, we can treat this as a binomial distribution because every customer will either fall into the credit card category or the not credit card category.

00:54 So we can say x, which is, we'll call that the random variable for the number of, of customers that pay with a credit card, it has a binomial distribution with the parameters 100 and 0 .5.

01:12 So this 0 .5 right here is the p for our binomial distribution, and that tells us that q is also 0 equal to 0 .5 because q is always 1 minus p, right? that represents the probability that someone does not pay with a credit card, and that means that they are debit or cash, 0 .3 plus 0 .2, 0 .5.

01:33 So the mean for a binomial distribution is equal to n times p.

01:43 So in this case, since our p is 0 .5, we have 100 times 0 .5, and that comes out to 50.

01:50 We expect half or 150 or 50 of the 100 customers to pay with credit card.

01:58 Now the variance, sigma squared, is given by n times p times q, which would be 100 times 0 .5 times 0 .5, and that's 25.

02:11 So even though there are multiple categories, if we only really care about whether one category is impacted or not that category, we could treat it as a binomial...

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