Question

For some reason, Elliot's World O' Tacos sells cheeseburgers - but not many. It turns out he makes bad cheeseburgers and the word has gotten out. The sales records of Elliot's World O' Tacos show the following sales of cheeseburgers over the past 200 days: Number Cheeseburgers Sold: 0 1 2 3 4 Number of Days: 70 100 10 6 4

a. How many sample points are there? b. Assign probabilities to the sample points and show their values. c. What is the probability that the Taco stand will not sell any cheeseburgers in a given day? d. What is the probability of selling at least 2 cheeseburgers? e. What is the probability of selling 1 or 2 cheeseburgers? f. What is the probability of selling less than 3 cheeseburgers?

PLEASE SHOW DETAILED WORK ON EACH PROBLEM

Answer 1

Number Cheeseburgers Sold: Number of Days: 0 1 3 4 Total N 70 100 10 5 4 190

a)   The possible number of values for "Number of Cheeseburgers Sold" = 5                      
   Hence, number of sample points = 5                      
                          
b)   Total number of observations = 190                      
   Probability of each sample point = Number of days for the sample point / Total number of days                      

0 1 2 3 4 Total Number Cheeseburgers Sold: Number of Days: Probability 6 4 70 0.3684 100 0.5263 10 0.0526 190 1 0.0316 0.0211

    Let X be the number of cheeseburgers sold                       

c) To find P(taco stand will not sell cheeseburgers on any given day)                      
   that is to find P(X = 0)                      
   From the probability table in (b), we get                      
   P(X = 0) = 0.3684                      
   Probability that taco stand will not sell cheeseburgers on any given day = 0.3684                    
                          
d) To find P(selling at least 2 cheeseburgers)                      
   that is to find P(X ≥ 2)                      
   P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)                      
                    = 0.0526 + 0.0316 + 0.0211                      
                    = 0.1053                      
   Probability of selling at least 2 cheeseburgers = 0.1053                      
                          
e) To find P(selling 1 or 2 cheeseburgers)                      
   that is to find P(X = 1 OR X = 2)                      
   P(X = 1 OR X = 2) = P(X = 1) + P(X = 2)           …since all the sample points are mutually exclusive           
       = 0.5263 + 0.0526                  
       = 0.5789                  
   Probability of selling 1 or 2 cheeseburgers = 0.5789                      
                          
                          
f) To find P(selling less than 3 cheeseburgers)                      
   that is to find P(X < 3)                      
   P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)                      
                    = 0.3684 + 0.5263 + 0.0526                      
                     = 0.9473                      
   Probability selling less than 3 cheeseburgers = 0.9473         

            Note to student : There could be a mismatch in the last digit of answer (f) due to rounding off.

The answer could also be 0.9474