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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 $$\frac {n+1} {2} th\ observation$$ otherwise ( if total is even ) the average of the $$\frac {n} {2} + \left< \frac {n} {2}+1 \right>\ th \ observations$$ Example for the data as given below:   

Observations        Frequency of occurrence

        x1                                      f1

        x2                                      f2

        x3                                      f3

       ---                                      ---

       ---                                      ---

       xn                                       fn

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

$$\frac {n+1} {2} th\ observation\ if\ n\ is\ odd$$ else the average of $$\frac {n} {2} + \left< \frac {n} {2}+1 \right>\ th \ observations,\ if\ n\ is\ even$$.
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

       ---                                      ---

       ---                                      ---

       xn                                       fn                                            f1+f2+f3+ -----+ fn

__________________________________________________________

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

__________________________________________________________ 

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.

$$The\ median\ class\ is\ "\ whose\ cumulative\ frequency\ is\ greater\ than\ or\ nearby\ to\ \frac {N} {2}"$$

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

$$Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$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).
$$A\ very\ beautiful\ relationship\ among\ all\ three\ central\ tendanacy\ measures\ is$$ $$3\ Median\ =\ Mode\ +\ 2\ Mean$$
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:
$$\because\ cumulative\ frequency\ N=68\ hence\ average\ of\ \frac {N} {2}\ th\ and \frac {N} {2} +1\ th$$ $$i.e.\ average\ of\ 34\ +\ 35\ =\ 69\ is\ \frac {69} {2}=\ 34.5$$ $$which\ near\ by\ class\ is\ 125-145$$ $$hence\ l=\ 125,\ N=68,\ cf\ =\ 22\ (preceding\ c.f.),\ h=20,\ f=20$$ $$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$ $$\therefore\ Median\ =\ 125+\ \frac {\frac {68} {2} - 22} {20} ×\ 20$$ $$\therefore\ Median\ =\ 137$$ For calculating Mean and Mode use the below formulas as we used in previous exercise:
\[\because \ x̅ =\ a+\ h\frac{Σfiui} {Σfi} \] \[\therefore \ x̅ =\ 135+\ 20\frac{7} {68} \] \[\therefore \ x̅ =\ 135+\ 20×{0.1029} \] \[\therefore \ x̅ =\ 137.058\ Mean\ Value \] Ans
and Mode is: \[\because Mode=l + \frac{f1-f0} {2f1-f0-f2}×h \] \[\therefore Mode=125 + \frac{20-13} {2×20-13-14}×20 \] \[\therefore Mode=135.769\ Mode\ Value \] 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)

$$\because Total\ of\ all\ the\ frequencies\ is\ =\ 60$$
$$\therefore 60=45+x+y$$
$$\therefore x+y=15\ --------\ (1)$$
$$\because median\ is\ 28.5\ which\ will\ exist\ in\ class\ 20-30$$
$$\therefore l=20,\ h=10,\ cf=5+x,\ f=20,\ \frac {N} {2} =30$$
$$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$
$$\therefore\ 28.5=20+\ \frac {30-(5+x)} {20} ×10$$
$$\therefore\ 57=40+\ 30-(5+x)$$
$$\therefore\ x=8\ Ans.$$
$$from\ equation\ (1)\ by\ putting\ value\ of\ x$$
$$y=7\ Ans.$$
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:
$$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$
$$\therefore\ Median=35+\ \frac {50-45} {33} ×5$$
$$\therefore\ Median=35+\ 0.757$$
$$\therefore\ Median=35.757\ Age\ Ans.$$
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.
$$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$
$$\therefore\ Median=144.5+\ \frac {20-17} {12} ×9$$
$$\therefore\ Median=144.5+\ 0.25$$
$$\therefore\ Median=144.75\ mm\ Ans.$$
5. The following table gives the distribution of the life time of 400 neon lamps :

Find the median life time of a lamp.
Solution:
$$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$
$$\therefore\ Median=3000+\ \frac {200-130} {86} ×500$$
$$\therefore\ Median=3000+\ 406.976$$
$$\therefore\ Median=3406.976\ Hours\ Ans.$$
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:
$$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$
$$\therefore\ Median=7+\ \frac {50-36} {40} ×3$$
$$\therefore\ Median=7+\ 1.05$$
$$\therefore\ Median=8.05\ \ Ans.$$
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Solution:
$$\because\ Median\ =\ l+\ \frac {\frac {N} {2} - cf} {f} ×\ h$$
$$\therefore\ Median=55+\ \frac {15-13} {6} ×5$$
$$\therefore\ Median=55+\ 1.67$$
$$\therefore\ Median=56.67\ kg\ Ans.$$

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







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