Question

in a frequency distribution, the mid value of a class is 20 and the width of the class is 12. the upper limit of class interval is

Answer 1

let the lower limit be x

∴upper limit =(x+12) (∵ class size is 12)

mid value of class =(upper limit +lowerlimt)÷2

⇒20 =[(x+12)+x]÷2

⇒40=2x+12

⇒40 - 12 =2x

⇒ 28=2x

=> 14= x

Answer 2

Answer:

26

Step-by-step explanation:

Mid value+ width/2 =upper limit

20+12/2= 26

if we find lower limit

Mid value - width /2