Identify the lower class limits, upper class limits, class width, class midpoint
ID: 3272664 • Letter: I
Question
Identify the lower class limits, upper class limits, class width, class midpoints, and class boundaries for the given frequency distribution. Also identify the number of individuals included in the summary. Age(yr) when award was won Frequency 15-24 27 25-34 36 35-44 15 45-54 4 55-64 5 65-74 2 75-84 1 Identify the class width. 10 (Type an integer or a decimal. Do not round.) Identify the class midpoints. 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5 (Type integers or decimals. Do not round. Use ascending order.) Identify the class boundaries. (Type integers or decimals. Do not round. Use ascending order.) Identify the number of individuals included in the summary. (Type an integer or a decimal. Do not round.)Explanation / Answer
Solution :
We are given following frequency distribution :
We have to identify the class width
class width = lower limit of second class - lower limit of first class = 25-15=10
Identify class mid points
class mid point = ( Lower limit + Upper limit ) / 2
Class mid point for first class = (15+24) / 2 = 39 /2 = 19.5
Class mid point for second class = ( 25+34) / 2 = 59 / 2 = 29.5
and so on 39.5 , 49.5 , 59.5 , 69.5 , 79.5
Your answers are correct.
Identify the class boundaries.
To find class boundaries , we find gap between two classes
Gap = Lower limit of second class - Upper limit of first class
Gap = 25 - 24 = 1
Find Gap / 2 = 1 / 2 = 0.5
Minus 0.5 from lower limit of each class and add 0.5 in each upper limit to get class boundaries .
Thus answer for class boundaries are :
( 14.5 , 24.5 , 34.5 , 44.5 , 54.5 , 64.5 , 74.5 , 84.5 )
Identify number of individuals :
Number of individuals is the total of frequency = 27+36+15+4+5+2+1=90 .
Age when award was won Frequency 15-24 27 25-34 36 35-44 15 45-54 4 55-64 5 65-74 2 75-84 1