Question

On annual day of a school 400 students participated in different programmes. frequency distribution showing their ages is as shown in the following, find the mean and median of the following data.​

Answer 1

The mean of the data will be 10.17 years and the median of the data will be 9.625 years.

Step-by-step explanation:

Given:

N= 400,

Age in years = 5-7  7-9  9-11  11-13  13-15  15-17  17-19

No. of students ($f_{i}$)= 70  120  32  100  45   28  5

To find:

mean and median of the data

Solution:

First, we calculate the classmarks ($x_{i}$) for the classes given

Classmark ($x_{i}$)= Upper limit of class+ Lower limit of class

                                                  2

Classmark for the class 5-7 = 7+5/2 = 6

Classmark for the class 7-9 = 9+7/2 = 8

Classmark for the class 9-11 = 11+9/2 = 10

Classmark for the class 11-13 = 13+11/2 = 12

Classmark for the class 13-15= 15+13/2 = 14

Classmark for the class 15-17 = 17+15/2 = 16

Classmark for the class 17-19 = 19+17/2 = 18

We multiply Classmark ($x_{i}$) & No. of students ($f_{i}$) of each class

∴$x_{i}f_{i}$ for first interval= 6x70= 420

$x_{i}f_{i}$ for second interval= 8x120= 960

$x_{i}f_{i}$ for third interval=10x32= 320

$x_{i}f_{i}$ for fourth interval= 12x100= 1200

$x_{i}f_{i}$ for fifth interval=14x45= 630

$x_{i}f_{i}$ for sixth interval=16x28= 448

$x_{i}f_{i}$ for seventh interval=18x5= 90

Now, ∑$x_{i}f_{i}$= 420+960+320+1200+630+448+90

∑$x_{i}f_{i}$= 4068

∑$f_{i}$=N= 400

Mean(${\bar x}$)= $\frac{\sum f_{i} x_{i} }{\sum f_{i} }$

Mean(${\bar x}$)= $\frac{4068}{400}$

∴ Mean(${\bar x}$)= $10.17$ years

For median, we take out the cumulative frequencies using the frequencies given

c.f. for the class 5-7 will be the same as the frequency=70

c.f. for the class 7-9=70+120=190

c.f. for the class 9-11= 190+32= 222

c.f. for the class 11-13=222+100= 322

c.f. for the class 13-15=322+45= 367

c.f. for the class 15-17= 367+28= 395

c.f. for the class 17-19= 395+5= 400=N

$\frac{N}{2}$= $\frac{400}{2}$ = $200$th term

The closest and greater than c.f. to 200 is 222

∴ The corresponding class 9-11 is the median class

Median= $l+\frac{\frac{N}{2} -c.f.}{f} (h)$

where $l$= lower limit of the median class = 9

$\frac{N}{2}$=200

$c.f.$= cumulative frequency of the class preceding the median class

   = 190

$f$= frequency of the median class= 32

$h$= difference between the upper limit and lower limit of the median class

 = 11-9 = 2

Median= $l+\frac{\frac{N}{2} -c.f.}{f} (h)$

Median= $9+\frac{200 -190}{32} (2)$

Median= $9+\frac{10}{32} (2)$

Median= $9+\frac{20}{32}$

Median= $9+0.625$

∴ Median= $9.625$ years

Hence, the mean and median of the data will be 10.17 years and 9.625 years respectively.

Answer 2

Answer:

The mean of the given students is 10.17 years and their median is 9.625 years.

Step-by-step explanation:

Given: Total number of students = 400,

             Age (class intervals) = 5-7  7-9  9-11  11-13  13-15  15-17  17-19

             No. of students (frequency)= 70  120  32  100  45   28  5

To find: mean and median

Solution:

First, we calculate the class mark ($x_i$) -

Classmark ($x_i$)= $\frac{Upper\;class\;limit+lower\;class\;limit}{2}$

Thus,
$x_1=\frac{7+5}{2} = 6$

$x_2=\frac{9+7}{2} = 8$

$x_3=\frac{9+11}{2} = 10$

$x_4=\frac{13+11}{2} = 12$

$x_5=\frac{15+13}{2} = 14$

$x_6=\frac{17+15}{2} = 16$

$x_7=\frac{19+17}{2} = 18$

Next, we multiply the class mark of an interval ($x_i$) by their respective frequency ($f_i$)

$x_1f_1=6\times 70=420\\\\x_2f_2=8\times 120=960\\\\x_3f_3=10\times 32=320\\\\x_4f_4=12\times 100=1200\\\\x_5f_5=14\times 45=630\\\\x_6f_6=16\times 28=448\\\\x_7f_7=18\times 5=90$

Now, we find $\sum(x_if_i)$

$\sum(x_if_i)= 420+960+320+1200+630+448+90\\\\\sum(x_if_i) = 4068$

∑=N= 400

Mean($\bar x$)= $\frac{\sum(x_if_i)}{(f_i)}$

Mean($\bar x$) = $\frac{4068}{400} = 10.17$
Thus, the mean of the given students is 10.17 years.

Now, to calculate the median, we find the cumulative frequencies as follows -
$cf_1 = 70\\\\cf_2 = 70 + 120 = 190\\\\cf_3 = 190 + 32 = 222\\\\cf_4 = 222 + 100 = 322\\\\cf_5 = 322 + 45 = 367\\\\cf_6 = 367 + 28 = 395\\\\cf_7 = 395 + 5 = 400 = \sum f_i$

Now, $\frac{N}{2} = \frac{400}{2} = 200$
The closest (and greater than) to 200 is 222.
Thus, the corresponding class 9-11 is the median class

Median= $l + \frac{\frac{N}{2} - cf }{f} \times h$
Median=  $9 + \frac{\frac{400}{2} - 190 }{32} \times 2$

Median= $9 + \frac{200 - 190 }{32} \times 2$

Median= $9 + \frac{20}{32}$

Median= $9.625$

Thus, the median of the students is 9.625 years.