Question

In a frequency distribution ,the mid-value of a class is 15 and the class interval is 4 . The lower limit of the class is

Answer 1

Answer:

The lower limit is 13.

Step-by-step explanation:

Let the lower limit be l and upper limit be u.  

Then, u−l=4...(1)

and  

2

u+l

​  

=15 or u+l=30...(2)

Subtracting (2) and (1),

u+l−u+l=30−4

2l=26

l=13

Thus, lower limit =13

Answer 2

Answer:

The lower limit of the class is $13$.

Step-by-step explanation:

Step 1 of 2

It is given that,

The mid-value of a class = 15

The class interval = 4

To find:- The $lower$ limit of the class.

Let the lower class limit be L and the upper class limit be U.

As we know,

The upper class limit - the lower class limit = The class interval

⇒ U - L = 4 ____ (1)

Also,

(The upper class limit + the lower class limit)/2 = The mid value of a class

⇒ $\frac{U+L}{2}=15$

⇒ U + L = 15 × 2

⇒ U + L = 30 ____ (2)

Step 2 of 2

Find the value of L.

Subtract the equation $(1)$ from $(2)$, we get

(U + L) - (U - L) = 30 - 4

2L = 26

L = 26/2

L = 13

Final answer: The lower limit of the class is $13$.

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