Hi$,$ In this question, since the width is $11,$ shouldn't the upper limit be $($Lower limit $+$ width$),$ so in this case, wouldn't the class limits be $(45-56),(57-68)(69-80)$ so on and so forth? But here the solution says $(45-55),(56-66)$ etc. As per another question $($Question $6,$ Section $2\mathrm{.}1$ of Understandable statistics, $11$th edition$)$ the first logic I mentioned is followed, Upper limit $=$ Lower limit $+$width. Please let me know what the right approach is$,$ I've attached the solution from the other question as well. Thank you!Here the class width is$,$

Class width $=22$

Then the class intervals are,

The first class interval is$,$

The lowest class is the lowest value in the data set.That is$,$

Lower class $=10$

The upper class is$,$

Upper. class $=$ Lower. class $+$ class. width

$=10+22$

$=32$

So$,$ the first class interval is$,$

First class interval is $10-32$

Then the second class interval is$,$

Lower class $=33$

And,

Upper. class $=$ Lower. class $+$ class. width

$=33+22$

$=55$

So$,$ the second class interval is$,$

Second class interval is $33-55$