Gina surveyed students about their lunch time preference. Out of 50 students, 20 prefer eating lunch at 11:00am, 15 prefer eating at 11:15a.m., 10 prefer eating at 11:30a.m. , and 5 have no preference. Based on the information, what is the probability that a student chosen at random has no lunch time preference?
To find a specific percent of the number, first find the smallest scale division of the designated percent, or 1%. To find 1% of 13,530, divide it by 100, which yields 135.30. Then, multiply this 135.30 by 6 to find 6%. This comes out to 811.8.