Question

The following table shows the marks scored by \[140\] students in an examination of a certain paper:

Marks:	\[0 - 10\]	\[10 - 20\]	\[20 - 30\]	\[30 - 40\]	\[40 - 50\]
Number of students:	\[20\]	\[24\]	\[40\]	\[36\]	\[20\]

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Answer 1

Hint: Here, we have to find the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method. It is given that there is a table with marks of \[140\] students. We have to find the average of the marks using the required method solution. By using the given data into the required method, we will get the final answer of that required method.

Formula used: We know that:
By direct method, the mean \[ = \dfrac{{\sum fx}}{{\sum f}}\]
By assumed mean method, mean\[ = A + \dfrac{{\sum fu}}{{\sum f}}\]
By stop deviation method, mean\[ = A + h \times \dfrac{{\sum fu}}{{\sum f}}\]

Complete step-by-step solution:
It is given that;

Marks:	\[0 - 10\]	\[10 - 20\]	\[20 - 30\]	\[30 - 40\]	\[40 - 50\]
Number of students:	\[20\]	\[24\]	\[40\]	\[36\]	\[20\]

The following table shows the marks scored by \[140\] students in an examination of a certain paper:
We have to find the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Direct method:

Size	\[x\]	\[f\]	\[xf\]
\[0 - 10\]	\[5\]	\[20\]	\[100\]
\[10 - 20\]	\[15\]	\[24\]	\[360\]
\[20 - 30\]	\[25\]	\[40\]	\[1000\]
\[30 - 40\]	\[35\]	\[36\]	\[1260\]
\[40 - 50\]	\[45\]	\[20\]	\[900\]

We know that,
 Mean\[ = \dfrac{{\sum fx}}{{\sum f}}\]
Here, \[\sum fx\] means the sum of \[fx\] and \[\sum f\] means the sum of \[f\].
Substitute the values we get,
Mean\[ = \dfrac{{3620}}{{140}}\]
Solving we get,
Mean\[ = 25.857\]
Assumed mean method:

Size	\[x\]	\[u = x - 25\]	\[f\]	\[uf\]
\[0 - 10\]	\[5\]	\[ - 20\]	\[20\]	\[ - 400\]
\[10 - 20\]	\[15\]	\[ - 10\]	\[24\]	\[ - 240\]
\[20 - 30\]	\[25\]	\[0\]	\[40\]	\[0\]
\[30 - 40\]	\[35\]	\[10\]	\[36\]	\[360\]
\[40 - 50\]	\[45\]	\[20\]	\[20\]	\[400\]

We know that,
 Mean\[ = A + \dfrac{{\sum fu}}{{\sum f}}\]
Here, \[\sum fu\] means the sum of \[fu\] and \[\sum f\] means the sum of \[f\].
Substitute the values we get,
Mean\[ = 25 + \dfrac{{120}}{{140}}\]
Solving we get,
Mean\[ = 25.875\]
Stop deviation method:

Size	\[d = x - 25\]	\[u = \dfrac{{x - 25}}{{10}}\]	\[f\]	\[uf\]
\[0 - 10\]	\[ - 20\]	\[ - 2\]	\[20\]	\[ - 40\]
\[10 - 20\]	\[ - 10\]	\[ - 1\]	\[24\]	\[ - 24\]
\[20 - 30\]	\[0\]	\[0\]	\[40\]	\[0\]
\[30 - 40\]	\[10\]	\[1\]	\[36\]	\[36\]
\[40 - 50\]	\[20\]	\[2\]	\[20\]	\[40\]

We know that,
 Mean\[ = A + h \times \dfrac{{\sum fu}}{{\sum f}}\]
Here, \[\sum fu\] means the sum of \[fu\] and \[\sum f\] means the sum of \[f\].
Substitute the values we get,
Mean\[ = 25 + 10 \times \dfrac{{12}}{{140}}\]
Solving we get,
Mean\[ = 25.875\]
Hence,
By direct method: Mean\[ = 25.857\]
By assumed mean method: Mean\[ = 25.875\]
By stop deviation method: Mean\[ = 25.875\]

Note: Mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations.
By direct method, the mean \[ = \dfrac{{\sum fx}}{{\sum f}}\]
In statistics, the assumed mean method is used for calculating mean or arithmetic mean of a grouped data. If the given data is large, then this method is recommended rather than a direct method for calculating mean. This method helps in reducing the calculations and results in small numerical values.
By assumed mean method, mean\[ = A + \dfrac{{\sum fu}}{{\sum f}}\]
Sometimes, during the application of the short-cut method for finding the mean, the deviations d, are divisible by a common number \[h\]. In this case the \[{d_i} = {x_i} - A\] is reduced to a great extent as di becomes \[\dfrac{{{d_i}}}{h}\].
By stop deviation method, mean\[ = A + h \times \dfrac{{\sum fu}}{{\sum f}}\]