Appropriate Question:

Find the mean of the following distribution by direct method.

[tex]\qquad\begin{gathered}\boxed{\begin{array}{c|c} \sf Class\:interval & \sf Frequency \: \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 - 10 & \sf 7\\ \\ \sf 10 - 20 & \sf 10 \\ \\ \sf 20 - 30 & \sf 15 \\ \\ \sf 30 - 40 & \sf 8 \\ \\ \sf 40 - 50 & \sf 10 \end{array}} \\ \end{gathered} \\ \\ [/tex]

Answer:

[tex]\qquad\qquad\qquad\boxed { \sf \: Mean = 25.8 \: }\\ \\ [/tex]

Step-by-step explanation:

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \sf Class\:interval & \sf Frequency \: f_i & \sf \: Midvalue \: x_i& \bf \: f_ix_i\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{}& \frac{\qquad }{}& \frac{\qquad }{} \\ \sf 0 - 10 & \sf 7& \sf 5 & \sf 35\\ \\ \sf 10 - 20 & \sf 10 & \sf 15 & \sf 150 \\ \\ \sf 20 - 30 & \sf 15& \sf 25 & \sf 375 \\ \\ \sf 30 - 40 & \sf 8& \sf 35 & \sf 280 \\ \\ \sf 40 - 50 & \sf 10 & \sf 45 & \sf 450 \end{array}} \\ \end{gathered} \\ \\ [/tex]

So, from above calculations, we concluded

[tex]\sf \: \sum \: f_i = 50 \\ \\ [/tex]

[tex]\sf \: \sum \: f_i x_1= 1290 \\ \\ [/tex]

Now,

[tex]\sf \: Mean = \dfrac{ \sum \: f_i x_1}{ \sum \: f_i } \\ \\ [/tex]

[tex]\sf \: Mean = \dfrac{ 1290}{ 50 } \\ \\ [/tex]

[tex]\sf\implies \sf \: Mean = 25.8\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

1. Mean using Direct Method

[tex]\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ [/tex]

2. Mean using Short Cut Method

[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ [/tex]

3. Mean using Step Deviation Method

[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ [/tex]

4. Mode of the continuous series is given by

[tex] {{ \boxed{\sf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}} \\ \\ [/tex]

where,

l is lower limit of modal class.

[tex] \sf{f_1} [/tex] is frequency of modal class

[tex] \sf{f_0} [/tex] is frequency of class preceding modal class

[tex] \sf{f_2} [/tex] is frequency of class succeeding modal class

h is class height.

Solution :

Find the mean of the following distribution by direct method.

[tex]\begin{gathered}\qquad\begin{gathered}\boxed{\begin{array}{c|c} \sf Class\:interval & \sf Frequency \: \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 - 10 & \sf 7\\ \\ \sf 10 - 20 & \sf 10 \\ \\ \sf 20 - 30 & \sf 15 \\ \\ \sf 30 - 40 & \sf 8 \\ \\ \sf 40 - 50 & \sf 10 \end{array}} \\ \end{gathered} \\ \\ \end{gathered} [/tex]

Mean By Direct Method could be find using :

[tex]\begin{gathered}\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ \end{gathered} [/tex]

To use this Formula we must find the values we need ;

[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c|c} \sf Class\:interval & \sf Frequency \: f_i & \sf \: Midvalue \: x_i& \bf \: f_ix_i\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{}& \frac{\qquad }{}& \frac{\qquad }{} \\ \sf 0 - 10 & \sf 7& \sf 5 & \sf 35\\ \\ \sf 10 - 20 & \sf 10 & \sf 15 & \sf 150 \\ \\ \sf 20 - 30 & \sf 15& \sf 25 & \sf 375 \\ \\ \sf 30 - 40 & \sf 8& \sf 35 & \sf 280 \\ \\ \sf 40 - 50 & \sf 10 & \sf 45 & \sf 450 \end{array}} \\ \end{gathered} \\ \\ \end{gathered}[/tex]

From this table we can conclude these quantities;

[tex]\begin{gathered}\sf \: \sum \: f_i = 50 \\ \\ \end{gathered} [/tex]

[tex]\begin{gathered}\sf \: \sum \: f_i x_1= 1290 \\ \\ \end{gathered}[/tex]

Now, By substituting the values we get in the formula of Direct method of calculating mean ;

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}{ \rm{ \:Mean = \cancel{\dfrac{ 1290}{ 50}} \: }} \\ \\ \end{gathered} [/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\sf{Mean = 25.8}}[/tex]

Therefore , Mean of given distribution is equals to 25.8