
If the median of the distribution given below is 28.5, find the values of x and y.
Class interval Frequency 0 – 10 5 10 – 20 x 20 – 30 20 30 – 40 15 40 – 50 y 50 – 60 5 Total 60
| Class interval | Frequency |
| 0 – 10 | 5 |
| 10 – 20 | x |
| 20 – 30 | 20 |
| 30 – 40 | 15 |
| 40 – 50 | y |
| 50 – 60 | 5 |
| Total | 60 |
Answer
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Hint: In this question first, we will understand Median: Median is that rate of the given observation on which divides it into exactly two parts. After this, we find the cumulative frequency (cf) of all the classes and $\dfrac{n}{2}$, where n = number of observations.
Now, the final class whose cumulative frequency is greater than and nearest to $\dfrac{n}{2}$ and this class is called median class, then use the following formula calculating the median.
$ = L + \left[ {\dfrac{{(\dfrac{n}{2} - cf)}}{f}} \right] \times h$
Where,
L = lower limit of the median class
n = number of observations
c.f = cumulative frequency of class interval preceding the median class
f = frequency of median class
h = class size
Complete step-by-step answer:
We have cumulative frequency table:
It is given that n = 60
i.e. 45 + x + y = 60
x + y = 60 – 40 = 15
$ \Rightarrow $x + y = 15 ………… equation (1)
It is given that the median is 28.5. So, the median lies in the group of 30 – 40.
Therefore, Median class = 20 – 30
L = 20 (lower class), $\dfrac{n}{2}$= 30, c.f = 5 + x, h = 10 and f = 20
Now, substituting these values in the formula of median we get,
Median = $1\, + \,\left[ {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right] \times h$
$
= 28.5 = \,20 + \left[ {\dfrac{{30 - (5 + x)}}{{20}}} \right] \times 10 \\
= 28.5 = \,20\left[ {\dfrac{{25 - x}}{2} \times 10} \right] \\
= 28.5 = \,20 + \dfrac{{25 - x}}{2} \\
= 28.5 = \,\dfrac{{40 + 25 - x}}{2} \\
= 28.5 \times 2 = 40 + 25 - x \\
= 57 = \,65 - x \\
= x = \,65 - 57 \\
= x = \,8 \\
$
Putting the value of x in equation 1, we get;
x + y = 15
$ \Rightarrow $8 + y = 15
$ \Rightarrow $y = 15 – 8 = 7
Hence, the value of x and y is 8 and 7 respectively.
Note: The cumulative frequency of a certain point is formed by adding the frequency at the present point to the cumulative frequency of the previous point. The cumulative frequency of the for the first data point is the same as its frequency since there is no cumulative frequency before it.
Now, the final class whose cumulative frequency is greater than and nearest to $\dfrac{n}{2}$ and this class is called median class, then use the following formula calculating the median.
$ = L + \left[ {\dfrac{{(\dfrac{n}{2} - cf)}}{f}} \right] \times h$
Where,
L = lower limit of the median class
n = number of observations
c.f = cumulative frequency of class interval preceding the median class
f = frequency of median class
h = class size
Complete step-by-step answer:
We have cumulative frequency table:
| Class interval | Frequency | C.F |
| 0 – 10 | 5 | 5 |
| 10 – 20 | x | 5 + x |
| 20 – 30 | 20 | 25 + x |
| 30 – 40 | 15 | 40 + x |
| 40 – 50 | y | 40 + x +y |
| 50 – 60 | 5 | 45 + x + y |
| Total | 60 |
It is given that n = 60
i.e. 45 + x + y = 60
x + y = 60 – 40 = 15
$ \Rightarrow $x + y = 15 ………… equation (1)
It is given that the median is 28.5. So, the median lies in the group of 30 – 40.
Therefore, Median class = 20 – 30
L = 20 (lower class), $\dfrac{n}{2}$= 30, c.f = 5 + x, h = 10 and f = 20
Now, substituting these values in the formula of median we get,
Median = $1\, + \,\left[ {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right] \times h$
$
= 28.5 = \,20 + \left[ {\dfrac{{30 - (5 + x)}}{{20}}} \right] \times 10 \\
= 28.5 = \,20\left[ {\dfrac{{25 - x}}{2} \times 10} \right] \\
= 28.5 = \,20 + \dfrac{{25 - x}}{2} \\
= 28.5 = \,\dfrac{{40 + 25 - x}}{2} \\
= 28.5 \times 2 = 40 + 25 - x \\
= 57 = \,65 - x \\
= x = \,65 - 57 \\
= x = \,8 \\
$
Putting the value of x in equation 1, we get;
x + y = 15
$ \Rightarrow $8 + y = 15
$ \Rightarrow $y = 15 – 8 = 7
Hence, the value of x and y is 8 and 7 respectively.
Note: The cumulative frequency of a certain point is formed by adding the frequency at the present point to the cumulative frequency of the previous point. The cumulative frequency of the for the first data point is the same as its frequency since there is no cumulative frequency before it.
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