Ganit Prakash - Class-X - Statistics: Median
Let us work out 26.2 Measures of Central Tendency
📘 Exercise 26.2 Solutions (Q1 - Q15)
Chapter 26Explanation:
First, arrange the selling prices in ascending order:
$52, 75, 92, 107, 113, 195, 201$
Here, the number of observations ($n$) is $7$, which is an odd number.
Therefore, the median is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ observation.
\[\begin{array}{l} \text{Median} = \left(\frac{7+1}{2}\right)^{\text{th}} \text{ observation} \\ = \left(\frac{8}{2}\right)^{\text{th}} \text{ observation} \\ = 4^{\text{th}} \text{ observation} \end{array}\]Looking at the arranged data, the $4^{\text{th}}$ observation is $107$.
Answer: The median of the selling prices is ` $107$.
Explanation:
First, arrange the ages in ascending order:
$4, 5, 6, 9, 10, 11, 18, 20$
Here, the number of observations ($n$) is $8$, which is an even number.
Therefore, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.
\[\begin{array}{l} \frac{n}{2} = \frac{8}{2} = 4^{\text{th}} \text{ observation (which is } 9) \\ \left(\frac{n}{2} + 1\right) = 4 + 1 = 5^{\text{th}} \text{ observation (which is } 10) \end{array}\]Now, calculate the average of these two middle values:
\[\begin{array}{l} \text{Median} = \frac{9 + 10}{2} = \frac{19}{2} = 9.5 \end{array}\]Answer: The median of the ages is $9.5$ years.
Explanation:
First, arrange the marks in ascending order:
$42, 45, 45, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 64$
Here, the number of observations ($n$) is $14$, which is an even number.
Therefore, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.
\[\begin{array}{l} \frac{n}{2} = \frac{14}{2} = 7^{\text{th}} \text{ observation (which is } 54) \\ \left(\frac{n}{2} + 1\right) = 7 + 1 = 8^{\text{th}} \text{ observation (which is } 55) \end{array}\]Now, calculate the average of these two middle values:
\[\begin{array}{l} \text{Median} = \frac{54 + 55}{2} = \frac{109}{2} = 54.5 \end{array}\]Answer: The median of the marks is $54.5$.
Today the scores of our cricket match of our locality are
$7 \quad 9 \quad 10 \quad 11 \quad 11 \quad 8 \quad 7 \quad 7 \quad 10 \quad 6 \quad 9$
$7 \quad 9 \quad 9 \quad 6 \quad 6 \quad 8 \quad 8 \quad 9 \quad 8 \quad 7 \quad 8$
Let us find the Median of scores in our cricket match.
Explanation:
First, let's list all the scores and count them. There are $22$ scores in total.
Now, arrange the scores in ascending order:
$6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 11, 11$
Here, the number of observations ($n$) is $22$, which is an even number.
Therefore, the median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ observations.
\[\begin{array}{l} \frac{n}{2} = \frac{22}{2} = 11^{\text{th}} \text{ observation} \\ \left(\frac{n}{2} + 1\right) = 11 + 1 = 12^{\text{th}} \text{ observation} \end{array}\]Looking at the arranged data, the $11^{\text{th}}$ observation is $8$ and the $12^{\text{th}}$ observation is $8$.
Now, calculate the average of these two middle values:
\[\begin{array}{l} \text{Median} = \frac{8 + 8}{2} = \frac{16}{2} = 8 \end{array}\]Answer: The median score is $8$.
Let us find the median of the weights from the following frequency distribution table of 70 students.
| Weight (kg) ($x_i$) | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
|---|---|---|---|---|---|---|---|---|
| Number of students ($f_i$) | 4 | 6 | 8 | 14 | 12 | 10 | 11 | 5 |
Explanation:
We need to construct a cumulative frequency (less than type) table.
| Weight (kg) ($x_i$) | Number of students ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 43 | 4 | 4 |
| 44 | 6 | $4 + 6 = 10$ |
| 45 | 8 | $10 + 8 = 18$ |
| 46 | 14 | $18 + 14 = 32$ |
| 47 | 12 | $32 + 12 = 44$ |
| 48 | 10 | $44 + 10 = 54$ |
| 49 | 11 | $54 + 11 = 65$ |
| 50 | 5 | $65 + 5 = 70 = n$ |
Here, $n = 70$ (which is even). So, we need to find the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ values.
\[\begin{array}{l} \frac{n}{2} = \frac{70}{2} = 35^{\text{th}} \text{ value} \\ \left(\frac{n}{2} + 1\right) = 36^{\text{th}} \text{ value} \end{array}\]From the cumulative frequency column, we can see that values from the $33^{\text{rd}}$ to the $44^{\text{th}}$ position fall in the category of '$47$ kg'.
Therefore, both the $35^{\text{th}}$ and $36^{\text{th}}$ observations are $47$.
\[\begin{array}{l} \text{Median} = \frac{47 + 47}{2} = 47 \end{array}\]Answer: The median weight is $47$ kg.
Let us find the median of length of diameter from the following frequency distribution table of length of diameter of pipe.
| Length of diameter (mm) ($x_i$) | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |
|---|---|---|---|---|---|---|---|---|
| Frequency ($f_i$) | 3 | 4 | 10 | 15 | 25 | 13 | 6 | 4 |
Explanation:
We construct a cumulative frequency (less than type) table.
| Length of diameter (mm) ($x_i$) | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 18 | 3 | 3 |
| 19 | 4 | $3 + 4 = 7$ |
| 20 | 10 | $7 + 10 = 17$ |
| 21 | 15 | $17 + 15 = 32$ |
| 22 | 25 | $32 + 25 = 57$ |
| 23 | 13 | $57 + 13 = 70$ |
| 24 | 6 | $70 + 6 = 76$ |
| 25 | 4 | $76 + 4 = 80 = n$ |
Here, $n = 80$ (which is even). The median will be the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ values.
\[\begin{array}{l} \frac{n}{2} = \frac{80}{2} = 40^{\text{th}} \text{ value} \\ \left(\frac{n}{2} + 1\right) = 41^{\text{st}} \text{ value} \end{array}\]Looking at the cumulative frequency column, values from the $33^{\text{rd}}$ to the $57^{\text{th}}$ position correspond to a diameter of '$22$ mm'.
Thus, both the $40^{\text{th}}$ and $41^{\text{st}}$ observations are $22$.
\[\begin{array}{l} \text{Median} = \frac{22 + 22}{2} = 22 \end{array}\]Answer: The median length of the diameter is $22$ mm.
Let us find the median.
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| f | 7 | 44 | 35 | 16 | 9 | 4 | 1 |
Explanation:
We construct a cumulative frequency (less than type) table.
| x | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0 | 7 | 7 |
| 1 | 44 | $7 + 44 = 51$ |
| 2 | 35 | $51 + 35 = 86$ |
| 3 | 16 | $86 + 16 = 102$ |
| 4 | 9 | $102 + 9 = 111$ |
| 5 | 4 | $111 + 4 = 115$ |
| 6 | 1 | $115 + 1 = 116 = n$ |
Here, $n = 116$ (which is even). The median is the average of the $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2} + 1\right)^{\text{th}}$ values.
\[\begin{array}{l} \frac{n}{2} = \frac{116}{2} = 58^{\text{th}} \text{ value} \\ \left(\frac{n}{2} + 1\right) = 59^{\text{th}} \text{ value} \end{array}\]From the cumulative frequency table, we can observe that observations from the $52^{\text{nd}}$ to the $86^{\text{th}}$ correspond to $x = 2$.
Therefore, both the $58^{\text{th}}$ and $59^{\text{th}}$ observations are $2$.
\[\begin{array}{l} \text{Median} = \frac{2 + 2}{2} = \mathbf{2} \end{array}\]The frequency distribution table of expenditures of tiffin allowances of 40 students is given below.
| Expenditure for tiffin (in `) | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 |
|---|---|---|---|---|---|---|---|
| Students | 3 | 5 | 6 | 9 | 7 | 8 | 2 |
Let us find the median of tiffin allowance.
Explanation:
Let's construct the cumulative frequency table.
| Expenditure (`) | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 35 - 40 | 3 | 3 |
| 40 - 45 | 5 | $3 + 5 = 8$ |
| 45 - 50 | 6 | $8 + 6 = 14$ |
| 50 - 55 | 9 | $14 + 9 = 23$ |
| 55 - 60 | 7 | $23 + 7 = 30$ |
| 60 - 65 | 8 | $30 + 8 = 38$ |
| 65 - 70 | 2 | $38 + 2 = 40 = n$ |
Here, $n = 40$, so $\frac{n}{2} = 20$.
The cumulative frequency just greater than $20$ is $23$, which corresponds to the class $50 - 55$.
Therefore, the median class is $50 - 55$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
- Lower limit of median class ($l$) = $50$
- Cumulative frequency of preceding class ($cf$) = $14$
- Frequency of median class ($f$) = $9$
- Class size ($h$) = $5$
Let us find the median of heights of students from the table below.
| Height(cm) | 135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 | 165-170 |
|---|---|---|---|---|---|---|---|
| No. of students | 6 | 10 | 19 | 22 | 20 | 16 | 7 |
Explanation:
| Height (cm) | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 135 - 140 | 6 | 6 |
| 140 - 145 | 10 | 16 |
| 145 - 150 | 19 | 35 |
| 150 - 155 | 22 | 57 |
| 155 - 160 | 20 | 77 |
| 160 - 165 | 16 | 93 |
| 165 - 170 | 7 | 100 = n |
Here, $n = 100$, so $\frac{n}{2} = 50$.
The cumulative frequency just greater than $50$ is $57$, which corresponds to the class $150 - 155$.
Therefore, the median class is $150 - 155$.
- Lower limit of median class ($l$) = $150$
- Cumulative frequency of preceding class ($cf$) = $35$
- Frequency of median class ($f$) = $22$
- Class size ($h$) = $5$
Let us find the median of data from the following frequency distribution table.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|---|
| Frequency | 4 | 7 | 10 | 15 | 10 | 8 | 5 |
Explanation:
| Class interval | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0 - 10 | 4 | 4 |
| 10 - 20 | 7 | 11 |
| 20 - 30 | 10 | 21 |
| 30 - 40 | 15 | 36 |
| 40 - 50 | 10 | 46 |
| 50 - 60 | 8 | 54 |
| 60 - 70 | 5 | 59 = n |
Here, $n = 59$, so $\frac{n}{2} = 29.5$.
The cumulative frequency just greater than $29.5$ is $36$, which corresponds to the class $30 - 40$.
- Lower limit of median class ($l$) = $30$
- Cumulative frequency of preceding class ($cf$) = $21$
- Frequency of median class ($f$) = $15$
- Class size ($h$) = $10$
Let us find the median of given data.
| Class interval | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 3 | 2 |
Explanation:
| Class interval | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 5 - 10 | 5 | 5 |
| 10 - 15 | 6 | 11 |
| 15 - 20 | 15 | 26 |
| 20 - 25 | 10 | 36 |
| 25 - 30 | 5 | 41 |
| 30 - 35 | 4 | 45 |
| 35 - 40 | 3 | 48 |
| 40 - 45 | 2 | 50 = n |
Here, $n = 50$, so $\frac{n}{2} = 25$.
The cumulative frequency just greater than $25$ is $26$, which corresponds to the class $15 - 20$.
- Lower limit of median class ($l$) = $15$
- Cumulative frequency of preceding class ($cf$) = $11$
- Frequency of median class ($f$) = $15$
- Class size ($h$) = $5$
Let us find the median of given data.
| Class interval | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 | 31-35 |
|---|---|---|---|---|---|---|---|
| Frequency | 2 | 3 | 6 | 7 | 5 | 4 | 3 |
Explanation:
The given class intervals are discontinuous. To make them continuous, we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of each class.
| Class Boundaries | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0.5 - 5.5 | 2 | 2 |
| 5.5 - 10.5 | 3 | $2 + 3 = 5$ |
| 10.5 - 15.5 | 6 | $5 + 6 = 11$ |
| 15.5 - 20.5 | 7 | $11 + 7 = 18$ |
| 20.5 - 25.5 | 5 | $18 + 5 = 23$ |
| 25.5 - 30.5 | 4 | $23 + 4 = 27$ |
| 30.5 - 35.5 | 3 | $27 + 3 = 30 = n$ |
Here, $n = 30$, so $\frac{n}{2} = 15$.
The cumulative frequency just greater than $15$ is $18$, which corresponds to the median class $15.5 - 20.5$.
- Lower limit of median class ($l$) = $15.5$
- Cumulative frequency of preceding class ($cf$) = $11$
- Frequency of median class ($f$) = $7$
- Class size ($h$) = $20.5 - 15.5 = 5$
Let us find the median of given data.
| Class interval | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 | 101-110 |
|---|---|---|---|---|---|---|
| Frequency | 4 | 10 | 15 | 20 | 15 | 4 |
Explanation:
Similar to the previous question, the classes are discontinuous. We adjust them by $0.5$ to make continuous class boundaries.
| Class Boundaries | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 50.5 - 60.5 | 4 | 4 |
| 60.5 - 70.5 | 10 | $4 + 10 = 14$ |
| 70.5 - 80.5 | 15 | $14 + 15 = 29$ |
| 80.5 - 90.5 | 20 | $29 + 20 = 49$ |
| 90.5 - 100.5 | 15 | $49 + 15 = 64$ |
| 100.5 - 110.5 | 4 | $64 + 4 = 68 = n$ |
Here, $n = 68$, so $\frac{n}{2} = 34$.
The cumulative frequency just greater than $34$ is $49$, which corresponds to the median class $80.5 - 90.5$.
- Lower limit of median class ($l$) = $80.5$
- Cumulative frequency of preceding class ($cf$) = $29$
- Frequency of median class ($f$) = $20$
- Class size ($h$) = $90.5 - 80.5 = 10$
Let us find the median of given data.
| Marks | Number of Students |
|---|---|
| less than 10 | 12 |
| less than 20 | 22 |
| less than 30 | 40 |
| less than 40 | 60 |
| less than 50 | 72 |
| less than 60 | 87 |
| less than 70 | 102 |
| less than 80 | 111 |
| less than 90 | 120 |
Explanation:
The given table is a "less than" cumulative frequency distribution. First, we convert it into regular continuous classes to find the frequencies.
| Class Interval | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0 - 10 | 12 | 12 |
| 10 - 20 | $22 - 12 = 10$ | 22 |
| 20 - 30 | $40 - 22 = 18$ | 40 |
| 30 - 40 | $60 - 40 = 20$ | 60 |
| 40 - 50 | $72 - 60 = 12$ | 72 |
| 50 - 60 | $87 - 72 = 15$ | 87 |
| 60 - 70 | $102 - 87 = 15$ | 102 |
| 70 - 80 | $111 - 102 = 9$ | 111 |
| 80 - 90 | $120 - 111 = 9$ | 120 = n |
Here, $n = 120$, so $\frac{n}{2} = 60$.
To find the median class, we look for the cumulative frequency just greater than $\frac{n}{2}$. The cumulative frequency just greater than $60$ is $72$, which corresponds to the class $40 - 50$.
Note: Because the $cf$ value at $30 - 40$ is exactly equal to $60$, the median will lie exactly at the boundary of $40$. Let's prove it with the formula.
- Lower limit of median class ($l$) = $40$
- Cumulative frequency of preceding class ($cf$) = $60$
- Frequency of median class ($f$) = $12$
- Class size ($h$) = $10$
If the median of the following data is 32, let us determine the values of x and y when the sum of the frequencies is 100.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Frequency | 10 | x | 25 | 30 | y | 10 |
Explanation:
First, we create the cumulative frequency table with variables $x$ and $y$.
| Class Interval | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0 - 10 | 10 | 10 |
| 10 - 20 | x | $10 + x$ |
| 20 - 30 | 25 | $35 + x$ |
| 30 - 40 | 30 | $65 + x$ |
| 40 - 50 | y | $65 + x + y$ |
| 50 - 60 | 10 | $75 + x + y$ |
We are given that the sum of frequencies ($n$) is $100$.
\[\begin{array}{l} 75 + x + y = 100 \\ x + y = 100 - 75 \\ x + y = 25 \quad \text{--- (Equation 1)} \end{array}\]We are given that the Median is $32$. Since $32$ lies in the interval $30 - 40$, the median class is $30 - 40$.
- Lower limit of median class ($l$) = $30$
- $\frac{n}{2} = \frac{100}{2} = 50$
- Cumulative freq of preceding class ($cf$) = $35 + x$
- Frequency of median class ($f$) = $30$
- Class size ($h$) = $10$
Using the median formula:
\[\begin{array}{l} \text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \\ 32 = 30 + \left[ \frac{50 - (35 + x)}{30} \right] \times 10 \\ 32 - 30 = \frac{50 - 35 - x}{3} \\ 2 = \frac{15 - x}{3} \\ 2 \times 3 = 15 - x \\ 6 = 15 - x \\ x = 15 - 6 \\ \mathbf{x = 9} \end{array}\]Now, substitute the value of $x$ into Equation 1:
\[\begin{array}{l} x + y = 25 \\ 9 + y = 25 \\ y = 25 - 9 \\ \mathbf{y = 16} \end{array}\]Answer: The values are $x = 9$ and $y = 16$.
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