124 students, 5 grills, sum of grills
a. Findprobabilitydistributionforx.
b.
What is the probability that x exceeds 10?
x1= 85
x2= 124
x3= 40
x4= 12
x5= 77
Sum of all = 338
Thus the probability for each is
x1= 85 / 338
x2= 124 / 338
x3= 40 / 338
x4= 12 / 338
x5= 77 / 338
This wouldnt this show the most likely choice is 1-2-5?
and b) is 80% because only x4 doesnt have the probability to be atleast .10
well, as far as I can understand, x is the sum of the grill numbers. So x1=the sum of grill numbers is 1+2+3=6.
and x2=sum of the grill numbers is (1+2+4=)7, x3= the sum of grill numbers is(1+2+5=)8, and so on...
the sum of grill numbers exceeds 10 only in the last combination(2-4-5)
Thats seems a bit to easy doesnt it? In that fashion you would disregard the size of the grill that the student selected?
Because if there is 124 students and grill 1 (x1) gets chosen 85 times within all combinations picked. Grill 2 gets picked in all combinations, etc. Wouldnt that mean 338 grills are selected in total after all combinations are exhausted?