Statistics Exercise -13.2

Hi Friends and Champs!
We will talk about the mode of some supplied grouped data today. Remember that you have already learnt the mode for ungrouped data from the last class. As you are previously aware, the mode for ungrouped data is the frequency that is repeated more frequently. As an illustration:

Example - Run score by a batsman in 8 matches are 50,98,99,98,76,75,99,98.Find the mode ?

Solution : 98 runs repeated 3 times , Hence mode is 98.

It is evident from the example above that figure 98 is the mode because it is repeated three times.Thus, it is easy to find the mode for ungrouped data. However, it is a little more complicated with aggregated data. For ungrouped data, we select the class with the highest frequency and use the formula below to determine the mode.
$$Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ where l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f1 = frequency of the modal class,
f0 = frequency of the class preceding the modal class,
f2 = frequency of the class succeeding the modal class.
Now begine the next exercise.


$$Exercise-13.2$$

1. The following table shows the ages of the patients admitted in a hospital during a year:
Age ( in years)          5-15     15-25     25-35     35-45     45-55     55--65
Number of patients     6          11          21          23          14              5
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:

Age ( in Years	Model Class	Number of patients	Class Marks xi	ui=(xi-40)/10	fiui
5-15		6	10	-3	-18
15-25		11	20	-2	-22
25-35	f0	21	30	-1	-21
35-45	Model Class -f1	23	40	0	0
45-55	f2	14	50	1	14
55-65		5	60	2	10
Total		80			-37

We first calculate the Mean $$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=40+10\frac{-37}{80}$$ $$\therefore x̅=40+10\times -0.4625$$ $$\therefore x̅=35.375$$ Ans
and Mode is: $$\because Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ $$\therefore Mode=35+\frac{23-21}{2\times 23-21-14}\times 10$$ $$\therefore Mode=36.82$$ Ans. 

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Solution:

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=3250+500\frac{-235}{200}$$ $$\therefore x̅=3250+500\times -1.175$$ $$\therefore x̅=2662.5Rs.MeanValue$$ Ans
and Mode is: $$\because Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ $$\therefore Mode=1500+\frac{40-24}{2\times 40-24-33}\times 500$$ $$\therefore Mode=1847.83Rs.ModeValue$$ Ans.

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=32.5+5\frac{-23}{35}$$ $$\therefore x̅=32.5+5\times -0.6571$$ $$\therefore x̅=29.21428StudentsPerTeacher$$ Ans
and Mode is: $$\because Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ $$\therefore Mode=30+\frac{10-9}{2\times 10-9-3}\times 5$$ $$\therefore Mode=30.625StudentsPerTeacher$$ Ans.

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Solution:

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=7500+1000\frac{-87}{49}$$ $$\therefore x̅=7500+1000\times -1.7755$$ $$\therefore x̅=5724.489RunsMeanValue$$ Ans
and Mode is: $$\because Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ $$\therefore Mode=4000+\frac{18-4}{2\times 18-4-9}\times 1000$$ $$\therefore Mode=4608.69RunsModeValue$$ Ans.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Solution:

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=45+10\frac{-43}{100}$$ $$\therefore x̅=45+10\times -0.43$$ $$\therefore x̅=40.7CarsMeanValue$$ Ans
and Mode is: $$\because Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ $$\therefore Mode=40+\frac{20-12}{2\times 20-12-11}\times 10$$ $$\therefore Mode=44.71CarsModeValue$$ Ans.

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