Statistics Exercise -13.1

Hi Friends and Champs!

In the next exercise, we will talk about the mean or average for given data. You already learn about the mean for the ungrouped data in the previous class: Example for the data as given below:   

Observations        Frequency of occurrence

        x₁                                      f₁

        x₂                                      f₂

        x₃                                      f₃

       ---                                      ---

       ---                                      ---

       x_n                                       f_n

Mean for the above ungrouped data is:

$$x̅=\frac{f1x1+f2x2+f3x3+............+fnxn}{f1+f2+f3+............+fn}$$ $$\begin{gathered} \therefore x̅=\frac{\Sigma fixi}{\Sigma fi} \\ -------------(1) \end{gathered}$$

This is called Direct Method to find the mean of the Grouped data , which we can also use for ungrouped data.

For ungrouped data we take class marks "xi" such that

$$x̅i=\frac{Upperclasslimit+Lowerclasslimit}{2}$$

Let "a" is an assumed mean value of the class marks xi, such that
di=xi-a, where "d" is the deviation of the class marks xi from the assumed mean "a" then:

$$d̅=\frac{f1d1+f2d2+f3d3+............+fndn}{f1+f2+f3+............+fn}$$ $$\therefore d̅=\frac{\Sigma fidi}{\Sigma fi}--------(2)$$ $$\therefore d̅=\frac{\Sigma fi(xi-a)}{\Sigma fi}$$ $$\therefore d̅=\frac{\Sigma fixi}{\Sigma fi}-a$$ $$\therefore d̅=x̅-a$$ $$\therefore x̅=d̅+a$$

From equation (2) put value of d̅

$$x̅=a+\frac{\Sigma fidi}{\Sigma fi}----------(3)$$

This is called Assumed Mean Method.
Now to simplify the above formula more , divide the deviation "di" by class interval "h", let's denote it by ui, then:

$$ui=\frac{xi-a}{h}$$

So, the mean of observations with step deviation "ui"

$$u̅=\frac{f1u1+f2u2+f3u3+............+fnun}{f1+f2+f3+............+fn}$$ $$\therefore u̅=\frac{\Sigma fiui}{\Sigma fi}--------(4)$$ $$\therefore u̅=\frac{\Sigma fi\frac{xi-a}{h}}{\Sigma fi}$$ $$\therefore u̅=\frac{1}{h}\frac{\Sigma fixi}{\Sigma fi}-\frac{a}{h}$$ $$\therefore u̅=\frac{1}{h}x̅-\frac{a}{h}$$ $$\therefore x̅=a+h\frac{\Sigma fiui}{\Sigma fi}---------(5)$$

This is called Step Deviation Method for mean.

Exercise 13.1 

 1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Solution:

Since the values are numerically small, so we use direct method to calculate the mean.

$$\because x̅=\frac{\Sigma fixi}{\Sigma fi}$$ $$\therefore x̅=\frac{162}{20}$$ $$\therefore x̅=8.1Plants.$$ Ans

2. Consider the following distribution of daily wages of 50 workers of a factory.

 Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

By step deviation method

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=550+20\frac{-12}{50}$$ $$\therefore x̅=550-4.8$$ $$\therefore x̅=545.2$$  Ans

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Solution:

By step deviation method

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=18+2\frac{f-20}{44}$$ $$\therefore 18=18+\frac{2f-40}{44}$$ $$\therefore 2f=40$$$$\therefore f=20$$

So the missing frequency is 20  Ans

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution:

$$\because x̅=a+\frac{\Sigma fidi}{\Sigma fi}$$ $$\therefore x̅=75.5+\frac{12}{30}$$ $$\therefore x̅=75.5+0.4$$ $$\therefore x̅=75.9$$ Ans.

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

 Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

$$\because x̅=a+\frac{\Sigma fidi}{\Sigma fi}$$ $$\therefore x̅=57+\frac{75}{400}$$ $$\therefore x̅=57+0.19$$ $$\therefore x̅=57.19$$ Ans.

6. The table below shows the daily expenditure on food of 25 households in a locality

 Find the mean daily expenditure on food by a suitable method.
Solution:

By step deviation method

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=225+50\frac{-7}{25}$$ $$\therefore x̅=225-14$$ $$\therefore x̅=211$$ Ans.

7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

 Find the mean concentration of SO2 in the air.

Solution:

$$\because x̅=a+\frac{\Sigma fidi}{\Sigma fi}$$ $$\therefore x̅=0.1+\frac{-0.04}{30}$$ $$\therefore x̅=0.1-0.0013$$ $$\therefore x̅=0.099ppm$$Ans.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

$$\because x̅=a+\frac{\Sigma fidi}{\Sigma fi}$$ $$\therefore x̅=17+\frac{-181}{40}$$ $$\therefore x̅=17-4.525$$ $$\therefore x̅=12.475Days$$Ans.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

By step deviation method

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=70+10\frac{-2}{35}$$ $$\therefore x̅=70-0.0571$$ $$\therefore x̅=69.4285percent$$ Ans.

I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏