Exam (elaborations) MATH 110 Module 2 Exam.| PORTGE LEARNING

MATH 110 Module 2 Exam.| PORTGE LEARNING Module 2 Exam Exam Page 1 During an hour at a fast food restaurant, the following types of sandwiches are ordered: Cheeseburger Fish Cheeseburger Hamburger Fish Chicken Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Chicken Fish Turkey Fish Hamburger Fish Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken Turkey Fish Hamburger Chicken Fish a) Make a frequency distribution for this data. Sandwiches Frequency Fish 20 Chicken 9 Hamburger 9 Cheeseburger 8 Turkey 7 Total 53 b) Make a relative frequency distribution for this data. Include relative percentages on this table. Copied and pasted from answer above to save on time not having to re-type Sandwiches Calculation Relative Frequency Relative Percentage Fish 20/53 = 0.3773 x 100 = 37.73 = 38% Chicken 9/53 = 0.1698 x 100 = 16.98 = 17% Hamburger 9/53 = 0.1698 x 100 = 16.98= 17% Cheeseburger 8/53 = 0.1509 x 100 = 15.09 = 15% Turkey 7/53 = 0.1320 x 100 = 13.2 = 13% Total 53 1 100% During an hour at a fast food restaurant, the following types of sandwiches are ordered: Cheeseburger Fish Cheeseburger Hamburger Fish Chicken Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Chicken Fish Turkey Fish Hamburger Fish Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken Turkey Fish Hamburger Chicken Fish a) Make a frequency distribution for this data. Major Frequency Hamburger 9 Cheeseburger 8 Fish 20 Turkey 7 Chicken 9 Total 53 b) Make a relative frequency distribution for this data. Include relative percentages on this table. Exam Page 3 Consider the following data: {22, 18, 16, 26, 20, 24} a) Find the sample mean of this data. 22+18+16+26+20+24 = 126 n = 6 xbar = ∑xi / n ∑xi = 126 n = 6 126/6 = 21 sample mean = 21 b) Find the range of this data. range = highest value - lowest value high value = 26 low value = 16 26-16 = 10 Range = 10 c) Find the sample standard deviation of this data. s^2 = variance s = standard deviation s^2 = ∑(xi-xbar)^2 / (n-1) xi = 16, 18, 20, 22, 24, 26 xbar = 21 n = 6 ∑(xi-xbar)^2 / (n-1) = (16-21)^2 = 25 (18-21)^2 = 9 (20-21)^2 = 1 (22-21)^2 = 1 (24-21)^2 = 9 (26 -21)^2 = 25 70 n-1 = 6-1 =5 70/5 = 14 s^2 (variance) = 14 standard deviation (s) = √variance √14 = 3.74 standard deviation of sample = 3.74 d) Find the coefficient of variation. coefficient of variation = (standard deviation / mean) x 100 standard deviation = 3.74 sample mean = 21 (3.74/21) x 100 = 17.8095