# CLASS-10MEDIAN-QUARTILES-MODE- PROBLEM & SOLUTION

MEDIAN-QUARTILES-MODE- PROBLEM & SOLUTION -

Example.1) The median of the following observation

11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47

Arranged in ascending order is 24.

Find the value of x and hence, find the mean.

Ans.)  Here, n = 9, which is odd

9 + 1

So, Median = Value of (--------)th observation

2

=  Value of 5th observation = (x + 4)

But, as per given condition, median = 24 (given)

so,  (x + 4)  =  24

=>   x  =  24 – 4

=>   x  =  20

So, the given observations are –

11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47

=>    11, 12, 14, (20 – 2), (20 + 4), (20 + 9), 32, 38, 47

=>    11, 12, 14, 18, 24, 29, 32, 38, 47

1

So, mean = ------- (11 + 12 + 14 + 18 + 24 + 29 + 32 + 38 + 47)

9

225

= --------  =  25

9

Hence, x = 20, and mean = 25           (Ans.)

Example.2)  The hearts of 60 patients were examined through X-ray and the observations obtained are given below –

Find the median diameter in mm.

Ans.) Arranging the given terms in an ascending order and preparing the cumulative frequency table, we have –

So,  n  =  60,

n

=> ------ =  30

2

n

And,  ------- +  1  = 30 + 1 =  31

2

Clearly  n is even number

1          n                          n

So, median = ------ { (-----)th observation + (------ + 1)th

2          2                          2

observation}

1

= ----- (30th observation + 31st observation)

2

But, from the above table, we find that each patient from 16th to 31st has an heart of diameter 122 mm

1

So, Median Diameter =  ------ (22 + 22) mm

2

1

= ------- X  44 mm  =  22 mm       (Ans.)

2

Example.3) The height (in cm) of 50 students of a class are given below –

Find – (i) Median                (ii) First Quartile

(ii) Third Quartile       (iv)  Interquartile Range

Ans.)   Arranging the given heights in an ascending order, we may prepare the cumulative frequency table as shown below –

Where, n = 50, which is even.

1        n                          n

(i) Median = ----- {(-----)th observation + (----- + 1)th observation}

2        2                          2

1        50                         50

= ----- {(------)th observation + (------ + 1)th observation}

2        2                           2

1

= ------ {25th observation + 26th observation}

2

(154 + 155)        309

= ------------- = -------- =  154.5     (Ans.)

2                2

n

(ii) First quartile, Q₁ =  (------)th observation

4

50

=  (------)th observation = 12.5th observation

4

=  13th observation

But, each student from 10th 21st has a height of 153cm.

Hence, Q₁ = 153       (Ans.)

3n

(iii) Third quartile, Q₃ = (------)th observation

4

3 X 50

= (---------)th observation

4

= 37.5th observation = 38th observation

But, each student from 36th to 43rd has a height of 156 cm

Hence, Q₃ = 156      (Ans.)

(iv) Interquartile Range = (Q₃ - Q₁)  = (156 – 153) cm  =  3 cm  (Ans.)

Example.4)  The distribution given below shows the marks obtained by 25 students in aptitude test. Find the mean, median, and mode of the distribution –

Ans.)  We may prepare the cumulative frequency as under-

Marks Obtained(a)   Number of Students(f)   Cumulative Frequency   f X a

5                       3                        3                1

6                       9                       12               54

7                       6                       18               42

8                       4                       22               32

9                       2                       24               18

10                       1                       25               10

∑f = 25 (n)                            ∑fa= 171

∑fa          171

Mean = -------- = -------- = 6084

∑f            25

Here, n = 25, which is odd

25 + 1

Median = value of (---------)th term = value of 13th term = 7

2

Since, the frequency of 6 is maximum, so mode = 6

Hence Mean = 6.84, Median = 7, and Mode = 6          (Ans.)