Statistics Exercise -13.3 (Median)

 Hi Friends and Champs! 

We will discuss about the median for grouped given data in the upcoming exercise. You were already familiar with finding the median for an ungrouped data from the prior class. To determine the median for ungrouped data, follow these steps:

Step-1 Rearrange the given frequency (Data ) in the ascending order.

Step-2  If the total of the given frequency (Data ) is odd then the median is $$\frac{n+1}{2}thobservation$$ otherwise ( if total is even ) the average of the $$\frac{n}{2}+⟨\frac{n}{2}+1⟩thobservations$$ Example for the data as given below:   

Observations        Frequency of occurrence

        x₁                                      f₁

        x₂                                      f₂

        x₃                                      f₃

       ---                                      ---

       ---                                      ---

       x_n                                       f_n

Median for the above ungrouped data when f1,f2,......fn are in ascending order is:

$$\frac{n+1}{2}thobservationifnisodd$$ else the average of $$\frac{n}{2}+⟨\frac{n}{2}+1⟩thobservations,ifniseven$$.
Now for the grouped data , this may be difficult to find the median in the usual way , because the median value will be somewhere within the class interval. So to find the median for grouped data , we first find the median class , with the help of c.f. ( cumulative frequency) method. Cumulative meaning is successive addition i.e. we add the successive frequency with the previous one, for more clearance, please check the following table:

Observations        Frequency of occurrence            Cumulative Frequency (c.f.)

        x₁                                      f₁                                                    f₁                                                

        x₂                                      f₂                                                f₁+f₂f

        x₃                                      f₃                                               f₁+f₂+f₃

       ---                                      ---

       ---                                      ---

       x_n                                       f_n                                            f₁+f₂+f₃+ -----+ f_n

__________________________________________________________

Total                                                                 N = f₁+f₂+f₃+ -----+ f_n

__________________________________________________________ 

Let N is the total of c.f. then, we use the formula as given above for ungrouped data to find the median class, i.e.

$$Themedianclassis"whosecumulativefrequencyisgreaterthanornearbyto\frac{N}{2}"$$

After that, we use the following formula to find the median:

$$Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$where l = lower limit of median class,

N = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming all class size is equal).
$$Averybeautifulrelationshipamongallthreecentraltendanacymeasuresis$$ $$3Median=Mode+2Mean$$
Now let's do the next exercise.$$Exercise-13.3$$
1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them:

Solution:

$$\because cumulativefrequencyN=68henceaverageof\frac{N}{2}thand\frac{N}{2}+1th$$ $$i.e.averageof34+35=69is\frac{69}{2}=34.5$$ $$whichnearbyclassis125-145$$ $$hencel=125,N=68,cf=22(precedingc.f.),h=20,f=20$$ $$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$ $$\therefore Median=125+\frac{\frac{68}{2}-22}{20}\times 20$$ $$\therefore Median=137$$ For calculating Mean and Mode use the below formulas as we used in previous exercise:

$$\because x̅=a+h\frac{\Sigma fiui}{\Sigma fi}$$ $$\therefore x̅=135+20\frac{7}{68}$$ $$\therefore x̅=135+20\times 0.1029$$ $$\therefore x̅=137.058MeanValue$$ Ans
and Mode is: $$\because Mode=l+\frac{f1-f0}{2f1-f0-f2}\times h$$ $$\therefore Mode=125+\frac{20-13}{2\times 20-13-14}\times 20$$ $$\therefore Mode=135.769ModeValue$$ Ans.

So all the three measures are approximately same.

2. If the median of the distribution given below is 28.5, find the values of x and y.

Solution:

Median =28.5 (given)

$$\because Totalofallthefrequenciesis=60$$

$$\therefore 60=45+x+y$$

$$\therefore x+y=15--------(1)$$

$$\because medianis28.5whichwillexistinclass20-30$$

$$\therefore l=20,h=10,cf=5+x,f=20,\frac{N}{2}=30$$

$$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$

$$\therefore 28.5=20+\frac{30-(5+x)}{20}\times 10$$

$$\therefore 57=40+30-(5+x)$$

$$\therefore x=8Ans.$$

$$fromequation(1)byputtingvalueofx$$

$$y=7Ans.$$

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Solution:

$$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$

$$\therefore Median=35+\frac{50-45}{33}\times 5$$

$$\therefore Median=35+0.757$$

$$\therefore Median=35.757AgeAns.$$

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Solution: We first convert the given data in continuous classes form, because we have the formula for continuous classes, then will find the c.f. class.

$$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$

$$\therefore Median=144.5+\frac{20-17}{12}\times 9$$

$$\therefore Median=144.5+0.25$$

$$\therefore Median=144.75mmAns.$$

5. The following table gives the distribution of the life time of 400 neon lamps :

Find the median life time of a lamp.

Solution:

$$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$

$$\therefore Median=3000+\frac{200-130}{86}\times 500$$

$$\therefore Median=3000+406.976$$

$$\therefore Median=3406.976HoursAns.$$

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution:

$$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$

$$\therefore Median=7+\frac{50-36}{40}\times 3$$

$$\therefore Median=7+1.05$$

$$\therefore Median=8.05Ans.$$

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

$$\because Median=l+\frac{\frac{N}{2}-cf}{f}\times h$$

$$\therefore Median=55+\frac{15-13}{6}\times 5$$

$$\therefore Median=55+1.67$$

$$\therefore Median=56.67kgAns.$$

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