F) Frequency Use the given frequency distribution to find the (a) class width. (
ID: 2908516 • Letter: F
Question
F) Frequency Use the given frequency distribution to find the (a) class width. (b) class midpoints. (c) class boundaries 32-36 37-41 42-46 47-51 52-56 57-61 62-66 (a) What is the class width? Type an integer or a decimal.) (b) What are the class midpoints? Complete the table below. (Type integers or decimals.) 32-36 37-41 42-46 47-51 52-56 57-61 62-66 (c) What are the class boundaries? Complete the table below. (Type integers or decimals.) Class boundaries 32-36 37-41 42-46 47-51 52-56 57-61 62-66Explanation / Answer
Solution:
we are given the following frequency distribution:
Part a) Class width:
Class width = Difference between two successive lower limits or difference between two successive upper limits
Thus
Class width = 37 - 32
Class width = 5
Part b) Class Midpoints:
Class midpoint = ( Upper limit + Lower Limit ) / 2
For First class:
Class midpoint = ( 32 + 36) / 2
Class Midpoint = ( 68 ) / 2
Class Midpoint = 34
For second class:
Class midpoint = ( 37 +41) / 2
Class Midpoint = ( 78 ) / 2
Class Midpoint = 39
Similarly we calculate next class midpoints:
Part c) Class Boundaries:
To find Class Boundaries, first, find the gap between two consecutive upper and lower limits.
Gap = Upper limit of a class - Lower limit of its next class
Gap = 37- 36
Gap =1
Then find Gap / 2 = 1 / 2 = 0.5
To get Class boundaries, Minus 0.5 from each lower limit and add 0.5 in each upper limit:
Calculations:
Class Class boundaries
32 - 36 32-0.5 = 31.5 , 36+0.5=36.5
37-41 37-0.5= 36.5 , 41+0.5=41.5
42-46 42-0.5= 41.5 , 46+0.5=46.5
47-51 47-0.5= 46.5 , 51+0.5=51.5
52-56 52-0.5= 51.5 , 56+0.5=56.5
57-61 57-0.5= 56.5 , 61+0.5=61.5
62-66 62-0.5= 61.5 , 66+0.5=66.5
Thus we get :
Temperature Frequency 32-36 1 37-41 3 42-46 5 47-51 11 52-56 7 57-61 7 62-66 1Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.