Question

solve these two questions!!!!!



Dont answer if you dont know the answer!!​

Answer 1

Answer

$\boxed{\begin{array}{|c|c|} Mode & Frequency \\ \cline{1-2} 0-10 & 4 \\ \cline{1-2} 10-20 & 6 \\ \cline{1-2} 20-30 & 7 \\ \cline{1-2} 30-40 & 12 \\ \cline{1-2} 40-50 & 5 \\ \cline{1-2} 50-60 & 6 \end{array}}}$

Solution :-

Firstly we will look for the highest frequency .

$\boxed{\begin{array}{|c|c|} Mode & Frequency \\ \cline{1-2} 0-10 & 4 \\ \cline{1-2} 10-20 & 6 \\ \cline{1-2} 20-30 & 7\ f_0 \\ \cline{1-2} 30-40 & 12\ f_1 \\ \cline{1-2} 40-50 & 5\ f_2 \\ \cline{1-2} 50-60 & 6 \end{array}}}$

A glance at the series reveals that 30-40 is modal class because it has maximum frequency i.e 12 . We will use the following formula :-

$\sf{\implies Z\ = l_1 + \dfrac{f_1-f_0}{2f_1 - f_0 - f_2}\times i }$

Here lowest class is 20 , interval is 10  so using the formula :-

$\sf{\implies Z\ = 20 + \dfrac{12-7}{2[20] - 7 - 5}\times 10 }$

$\sf{\implies Z\ = 20 + \dfrac{5}{40 - 12 }\times 10 }$

$\sf{\implies Z\ = 20 + \dfrac{50}{28} }$

$\sf{\implies Z\ = 20 + 1.78 \rightarrow 21.78}$

So the mode is 21.78

$\boxed{\begin{array}{|c|c|} Age & no.\ of\ person \\ \cline{1-2} 0-10 & 5 \\ \cline{1-2} 10-20 & 15 \\ \cline{1-2} 20-30 & 20 \\ \cline{1-2} 30-40 & 25 \\ \cline{1-2} 40-50 & 15 \\ \cline{1-2} 50-60 & 11 \\ \cline{1-2}60-70 & 9 \end{array}}}$

To find median convert series in less than o-give

Less Than Ogive :-

$\boxed{\begin{array}{|c|c|} Age & no.\ of\ person \\ \cline{1-2} less\ than\ 10 & 5 \\ \cline{1-2} less\ than\ 20 & 20 \\ \cline{1-2} less\ than\ 30 & 40 \\ \cline{1-2}less\ than\ 40 & 65 \\ \cline{1-2}less\ than\ 50 & 80 \\ \cline{1-2}less\ than\ 60 & 91 \\ \cline{1-2}less\ than\ 70 & 100\end{array}}}$

2. Find out the value of n/2

$\sf{\dfrac{n}{2} = \dfrac{100}{2} \rightarrow 50 }$

Draw a perpendicular from 50 and note where it cut the o-give

The point corresponding to x- axis will be median

So, Median is 34 .