Question

class mark of the 1st class interval is 5 and there are five classes if the class size is 10 then the last class interval is​

Answer 1

Given,

Class Mark$\[ = 5\]$

Class Size$\[ = 10\]$

Solution,

Assume that the upper and lower limit of the class is $\[U\]$and $\[L\]$respectively.

Formula used,$\[{\rm{Class}}\,{\rm{Mark}} = \frac{{{\rm{Lower}}\,{\rm{Limit + Upper}}\,{\rm{Limit}}}}{{\rm{2}}}\]$

Formula used,$\[{\rm{Class}}\,{\rm{Size}} = {\rm{Upper}}\,{\rm{Limit}} - {\rm{Lower}}\,{\rm{Limit}}\]$

Therefore,

$\[\begin{array}{l} \Rightarrow 5 = \frac{{U + L}}{2}\\ \Rightarrow U - L = 10\end{array}\]$

Solve the above equation.

$\[\begin{array}{l} \Rightarrow U = 10\\ \Rightarrow L = 0\end{array}\]$

Hence, the last class interval is $\[40 - 50\].$

Answer 2

The last class interval = 40 - 50

Given :

• Class mark of the 1st class interval is 5 and there are five classes.
• The class size is 10

To find : The last class interval

Solution :

Step 1 of 3 :

Find Upper Class boundary

Here it is given that Class mark of the 1st class interval is 5

There are five classes.

The class size is 10

So Class Mark of five classes are 5 , 15 , 25 , 35 , 45

Thus class mark of last class interval

Upper Class boundary

$\sf{ = 45 + \dfrac{10}{2} }$

$\sf{ = 45 + 5 }$

$\sf{ = 50 }$

Step 2 of 3 :

Find Lower class boundary

Lower class boundary

$\sf{ = 45 - \dfrac{10}{2} }$

$\sf{ = 45 - 5 }$

$\sf{ = 40}$

Step 3 of 3 :

Find the last class interval

Hence the required last class interval is 40 - 50

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