Bob sells bagels for $0.80 each or $9.00 for a dozen. If I is the amount of money Bob collects from a randomly selected customer, find and interpret the expected value of c. NOTE: In the real world, a probability distribution for a bagel shop would consist of many more possible random variables. Remember, a probability distribution should include all possible outcomes. We have created a shorter version of a probability distribution to a make practicing finding the expected value and standard deviation by hand less tedious and time consuming. Solution Our current probability distribution shows the number of bagels sold and the probabilities. We are asked to find the expected value of the money Bob collects. We need to add a column to our probability distribution and determining , the amount of money collected by Bob for each of the given variables, number of bagels. Each bagel costs $0.80 cents. Customers get a discount when they buy 12 bagels, purchasing them for $9 Number of Bagels I = Money collected P(z) = Probability 1 0.80(1)=0.80 0.5 0.80(2) = 1.60 0.1 0.80(3) = 0.1 0.80(6) = 0.1 12 0.2 The expected value of \\, the amount of money Bob can expect to make per customer, is E(X) = EXP(X) = $ So in the long run, Bob can expect to make an average of $3.08 per customer. What is the standard deviation? Number of Bagels (X - () (Round to nearest hundredth) P(I) (x - ) . P(I) 1 (0.80 - 3.08) 2 - 5.20 0.5 2.6 2 (1.60 - 3.08)~ 0.1 0.219 3 (2.40 - 3.08)2 0.46 0.1 6 (4.80 - 3.08)2 0.296 12 (9 - 3.08)2 ~ 35.05 0.2 0 = VI-HYPE) V10.171 ~ Round to nearest hundredth.)