The boundaries are found using these formulas:
#B_U=L_U+1/2(M)# and
#B_L=L_L-1/2(M)#,
where #B_U# is the upper boundary, #B_L# is the lower boundary, #L_U# is the upper limit of the data, #L_L# is the lower limit of the data, and #M# is the unit of measurement.
Firstly we find the unit of measurement. From 125-131 we seem to be counting in ones, thus our unit of measurement is #M=1#. Next we find #L_U and L_L#. These are simply the highest and lowest numbers, so we get #L_U=131 and L_L=125#. We now plug these numbers into the equation to get:
#B_U=131+1/2(1)" "B_L=125-1/2(1)#
#B_U=131+1/2" "B_L=125-1/2#
#B_U=131.5" "B_L=124.5#
Therefore the boundaries of 125-131 are 124.5 and 131.5.
I invite you too lok at Gamaliel B's answer for more details.
The width (or range) is found by subtracting the highest value from the lowest value, which looks like this:
#131-125=6#, therefore 6 is the width (or range).
The midpoint #p# is found like this:
#p=(L_L+L_U)/2#
where #p# is the midpoint, and #L_L# & #L_U# are the lower and upper limits of the data, respectively. (The midpoint is just the average of the lowest and highest values.) We know that #L_L=125# and #L_U=131.# Plug these into the formula to get
#p=(125+131)/2#
#p=256/2#
#p=128,# the midpoint.
I hope I helped!
