Can anyone answer these very hard maths questions please!?!?!?!?
In order to answer your questions, you have to assume a distribution across the different ranges. Since nothing is given in the question it is reasonable to assume a uniform distribution. The reason this has to be done is because the mean and median will change depending on how they are spread out across the ranges. ie, if they all fell on the first value if the range so 0,2,2,2,4,4,4,4,4,6,6….etc. or if they all fell on the last value of the range so 1.99,3.99,3.99,3.99,….etc or any other combination would yield different mean and median values. So we assume they are uniformly distributed across the range. a) we use the mean formula for a uniform dist. (b+a)/2….so (2+0)/2, (4+2)/2, (6+4)/2….etc we then multiply by their probabilities which we get from their frequency 1st range prob = 1/30, 2nd = 3/30, 3rd = 6/30 etc……30 comes from the total frequencies so mean = (2+0)/2 * (1/30) + (4+2)/2 * (3/30) +… etc it will equal 8.07 b)since their are 30 numbers median will be betw
