Question

The arithmetic mean of the following frequency distribution is 25. Determine the value of p.
Class 0 - 10 10 - 20 20 - 30 30 - 40 40-50
Frequency 5 18 15 р
16
​

Answer 1

Correct Question:

The arithmetic mean of the following distribution is 25. Determine the value of p.

$\begin{array}{|c|c|}\cline{1-2}\bf\:Class & \bf\:Frequency\\\cline{1-2}\sf\:0 - 10 & \sf\:5\\\cline{1-2}\sf\:10 - 20 & \sf\:18\\\cline{1-2}\sf\:20 - 30 & \sf\:15\\\cline{1-2}\sf\:30 - 40 & \sf\:p\\\cline{1-2}\sf\:40 - 50 & \bf\:6\\\cline{1-2}\end{array}$

Answer:

The value of p is 16.

Step-by-step-explanation:

We have given a data.

The arithmetic mean of the given frequency distribution is 25.

We have to find the value of p.

$\begin{array}{|c|c|c|c|}\cline{1-4}\bf\:Class & \bf\:Class\:Mark\:(\:x_i\:) & \bf\:Frequency\:(\:f_i\:) & \bf\:x_i\:\times\:f_i\\\cline{1-4}\sf\:0 - 10 & \sf\:5 & \sf\:5 & \sf\:25\\\cline{1-4}\sf\:10 - 20 & \sf\:15 & \sf\:18 & \sf\:270\\\cline{1-4}\sf\:20 - 30 & \sf\:25 & \sf\:15 & \sf\:375\\\cline{1-4}\sf\:30 - 40 & \sf\:35 & \sf\:p & \sf\:35p\\\cline{1-4}\sf\:40 - 50 & \sf\:45 & \sf\:6 & \sf\:720\\\cline{1-4}& & \sf\:\sum\:f_i\:=\:44\:+\:p & \sf\:\sum\:f_i\:.\:x_i\:=\:940\:+\:35p\\\cline{1-4}\end{array}$

Now,

$\pink{\sf\:Mean\:\bar{X}\:=\:\dfrac{\sum\:f_i\:x_i}{\sum\:f_i}}\\\\\\\implies\sf\:25\:=\:\dfrac{940\:+\:35p}{44\:+\:p}\\\\\\\implies\sf\:940\:+\:35p\:=\:25\:\times\:(\:44\:+\:p\:)\\\\\\\implies\sf\:940\:+\:35p\:=\:1100\:+\:25p\\\\\\\implies\sf\:35p\:-\:25p\:=\:1100\:-\:940\\\\\\\implies\sf\:10p\:=\:160\\\\\\\implies\sf\:p\:=\:\cancel{\dfrac{160}{10}}\\\\\\\implies\boxed{\red{\sf\:p\:=\:16}}$

Hence,

The value of p is 16.

Answer 2

Answer:

Correct Question:

The arithmetic mean of the following distribution is 25. Determine the value of p.

$$\begin{gathered}\begin{array}{|c|c|}\cline{1-2}\bf\:Class & \bf\:Frequency\\\cline{1-2}\sf\:0 - 10 & \sf\:5\\\cline{1-2}\sf\:10 - 20 & \sf\:18\\\cline{1-2}\sf\:20 - 30 & \sf\:15\\\cline{1-2}\sf\:30 - 40 & \sf\:p\\\cline{1-2}\sf\:40 - 50 & \bf\:6\\\cline{1-2}\end{array}\end{gathered}$$

Answer:

The value of p is 16.

Step-by-step-explanation:

We have given a data.

The arithmetic mean of the given frequency distribution is 25.

We have to find the value of p.

$$\begin{gathered}\begin{array}{|c|c|c|c|}\cline{1-4}\bf\:Class & \bf\:Class\:Mark\:(\:x_i\:) & \bf\:Frequency\:(\:f_i\:) & \bf\:x_i\:\times\:f_i\\\cline{1-4}\sf\:0 - 10 & \sf\:5 & \sf\:5 & \sf\:25\\\cline{1-4}\sf\:10 - 20 & \sf\:15 & \sf\:18 & \sf\:270\\\cline{1-4}\sf\:20 - 30 & \sf\:25 & \sf\:15 & \sf\:375\\\cline{1-4}\sf\:30 - 40 & \sf\:35 & \sf\:p & \sf\:35p\\\cline{1-4}\sf\:40 - 50 & \sf\:45 & \sf\:6 & \sf\:720\\\cline{1-4}& & \sf\:\sum\:f_i\:=\:44\:+\:p & \sf\:\sum\:f_i\:.\:x_i\:=\:940\:+\:35p\\\cline{1-4}\end{array}\end{gathered}$$

Now,

$$\begin{gathered}\pink{\sf\:Mean\:\bar{X}\:=\:\dfrac{\sum\:f_i\:x_i}{\sum\:f_i}}\\\\\\\implies\sf\:25\:=\:\dfrac{940\:+\:35p}{44\:+\:p}\\\\\\\implies\sf\:940\:+\:35p\:=\:25\:\times\:(\:44\:+\:p\:)\\\\\\\implies\sf\:940\:+\:35p\:=\:1100\:+\:25p\\\\\\\implies\sf\:35p\:-\:25p\:=\:1100\:-\:940\\\\\\\implies\sf\:10p\:=\:160\\\\\\\implies\sf\:p\:=\:\cancel{\dfrac{160}{10}}\\\\\\\implies\boxed{\red{\sf\:p\:=\:16}}\end{gathered}$$

Hence,

The value of p is 16.