Using the information given, it is found that the class width for this frequency distribution table is of 1.
In this problem, these following classes are given:
0 – 1 14
2 – 3 1
4 – 5 8
6 – 7 12
8 – 9 12
The classes not given, which are 1 - 2, 3 - 4 and 5 - 6, have values of 0.
The <u>difference between the bounds of the classes is of 1</u>, thus, the class width is of 1.
A similar problem is given at brainly.com/question/24701109
Answer:
47.4 ;
50
Step-by-step explanation:
Given the data :
X ($) : 85 139 161 175 85 133 149 145 136 131 290 235 132 149 322 214 105 90 162 229 121 113 149 126139 118 156 214 172 87 172 230 195 126 128 142 118 139
The smallest class interval :
Range / number of classes
Number of classes to use = 5
Range = Maximum - Minimum = (322 - 85) =237
Hence, smallest class interval :
237 / 5 = 47.4
A better class interval would be, one without decimal, rounded to the nearest 10; this will be easier and make more statistical sense
Hence, smallest class interval rounded to the nearest 10 :
47.4 = 50 (nearest 10)
Sounds like "formulas"! y = ax^2 + bx + c involves the variable x and the constants {a, b, c}. The "2" indicates "squaring function."
