Question

8. 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 of Number Cheeseburgers of Days Sold 70 100 10 6 1 2 ترا 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 WORK FOR EACH ANSWER

Answer 1

Solution:

Given:

Number of Cheeseburgers sold	Number of Days
0	70
1	100
2	10
3	6
4	4
	N= 190

Part a)

Total number of sample points = N = 190

Part b)

To get probabilities we divide each frequency ( Number of days) by total points N = 190

That is:

Probability of 0 Cheeseburgers sold = 70 / 190 = 0.3684

Probability of 1 Cheeseburgers sold = 100 / 190 = 0.5263

and so on.

thus we get:

Number of Cheeseburgers sold	Number of Days	Probability
0	70	0.3684
1	100	0.5263
2	10	0.0526
3	6	0.0316
4	4	0.0211
	N = 190

_{(Round final answer to specified number of decimal places)}

Part c) Find:

P( The Taco stand will not sell any cheeseburgers in a given day) =..........?

P( 0 cheeseburgers in a given day) =.........?

P( 0 cheeseburgers in a given day) = 0.3684

Part d) Find:

P( Selling at least 2 cheeseburgers )= ............?

P( Selling at least 2 cheeseburgers )=P( 2 or more)

P( Selling at least 2 cheeseburgers )= P(2)+P(3)+P(4)

P( Selling at least 2 cheeseburgers )=0.0526+0.0316+0.0211

P( Selling at least 2 cheeseburgers )=0.1053

Part e) Find:

P( Selling 1 or 2 cheeseburgers )=...............?

P( Selling 1 or 2 cheeseburgers )= P(1) + P(2)

P( Selling 1 or 2 cheeseburgers )= 0.5263 + 0.0526

P( Selling 1 or 2 cheeseburgers )= 0.5789

Part f) Find:

P( Selling less than 3 cheeseburgers )=..............?

P( Selling less than 3 cheeseburgers )=P(0) + P(1) +P(2)

P( Selling less than 3 cheeseburgers )= 0.3684 + 0.5263 +0.0526

P( Selling less than 3 cheeseburgers )= 0.9473