# In a frequency distribution, the mid-value of a class is 20 and the width of the class is 8.then what is the lower limit ?

Mid value (a) = 20

Width (h) = 8

Lower limit = a - h/2

= 20 - 8/2

= 20 - 4

= 16

Upper limit = a + h/2

= 20 + 8/2

= 20 + 4

= 24

Thus, the class limit = 16 - 24.

Also, a - h/2 and a + h/2 is the formula for finding out the lower and upper limit when class mark i.e mid value and class interval or width is given where a = mid value and h = width..

Hope u get it !!

:) :)

• 62
Mid value (a) = 20

Width (h) = 8

Lower limit = a - h/2

= 20 - 8/2

= 20 - 4

= 16

Upper limit = a + h/2

= 20 + 8/2

= 20 + 4

= 24

Thus, the class limit = 16 - 24.

Also, a - h/2 and a + h/2 is the formula for finding out the lower and upper limit when class mark i.e mid value and class interval or width is given where a = mid value and h = width.
• 5
mid value = 20
width = 8
lower limit = 20 - 8 / 2
= 20 - 4
= 16
upper limit = 20 + 8 / 2
= 20 + 4
= 24
CLASS LIMIT = 16 - 24
• 0
ans is lower limit 20
• 0
no so sorry it is 16
• -2
is a -h /2 a formula
• -4

• 3

• 5
Mid value is 20 and width is 8.
The formula to find lower limit= a - h/2
= 20-8/2
=20-4
=16
• 7
Write the polynomial whose zeroes are-5 and 4
• -4
The mean of 15 observation is 10 .If 2 is subtracted from each observation ,find the new mean
• 0
Mid value (a) = 20

Width (h) = 8

Lower limit = a - h/2

= 20 - 8/2

= 20 - 4

= 16

Upper limit = a + h/2

= 20 + 8/2

= 20 + 4

= 24

Thus, the class limit = 16 - 24.

Also, a - h/2 and a + h/2 is the formula for finding out the lower and upper limit when class mark i.e mid value and class interval or width is given where a = mid value and h = width..

Hope u get it !!

:) :)
• -1
Yy
• 7
The lower limit is 16
• 1
Punnu mere punnu
• 0
Punnu
• 2
Lower limit=a-h/2.If we consider a=20 and h=8,then 20-8/2=20-4=16.
• 0
Ans.12
• 0
yes you are correct

• 0
Lower limit of the class is 16
• 0
uyuuyuguuy

• 0
16 is the lower limit
• 0
• 0

• 0
how many students are in the class
• 0