Ganit Prakash - Class-X - Statistics: Mode
Let us work out 26.4 Measures of Central Tendency
📘 Exercise 26.4 Solutions (Q1 - Q10)
Chapter 26Paid amounts of a day of our 16 friends for going to school and other expenditures are
15, 16, 17, 18, 17, 19, 17, 15, 15, 10, 17, 16, 15, 16, 18, 11
Let us find the mode of paid amounts of a day of our friends.
Explanation:
To find the mode, we count the frequency of each value to see which one appears the most.
| Paid Amount | Frequency (Number of friends) |
|---|---|
| 10 | 1 |
| 11 | 1 |
| 15 | 4 |
| 16 | 3 |
| 17 | 4 |
| 18 | 2 |
| 19 | 1 |
The highest frequency is 4. However, this frequency occurs for two different values: 15 and 17.
Since there are two values with the highest frequency, this data is bimodal.
Answer: The modes of the paid amounts are 15 and 17.
Heights (cm) of some students of our class are given below.
131, 130, 130, 132, 131, 133, 131, 134, 131, 132, 132, 131, 133,
130, 132, 130, 133, 135, 131, 135, 131, 135, 130, 132, 135, 134, 133
Let us find the mode of heights of students.
Explanation:
We create a frequency distribution table to find the mode.
| Height (cm) | Frequency |
|---|---|
| 130 | 5 |
| 131 | 7 |
| 132 | 5 |
| 133 | 4 |
| 134 | 2 |
| 135 | 4 |
From the table, we can see that the height with the maximum frequency is 131 cm (it occurs 7 times).
Answer: The mode of the heights of the students is 131 cm.
Let us find the mode of data given below:
(i) 8, 5, 4, 6, 7, 4, 4, 3, 5, 4, 5, 4, 4, 5, 5, 4, 3,
3, 5, 4, 6, 5, 4, 5, 4, 5, 4, 2, 3, 4
(ii) 15, 11, 10, 8, 15, 18, 17, 15, 10, 19, 10, 11,
10, 8, 19, 15, 10, 18, 15, 3, 16, 14, 17, 2
Explanation for (i):
Let's count the frequency of each number:
- 2 appears 1 time
- 3 appears 4 times
- 4 appears 12 times
- 5 appears 9 times
- 6 appears 2 times
- 7 appears 1 time
- 8 appears 1 time
The maximum frequency is 12, which corresponds to the value 4.
Answer (i): The mode is 4.
Explanation for (ii):
Let's count the frequency of each number:
- 2 appears 1 time
- 3 appears 1 time
- 8 appears 2 times
- 10 appears 5 times
- 11 appears 2 times
- 14 appears 1 time
- 15 appears 5 times
- 16 appears 1 time
- 17 appears 2 times
- 18 appears 2 times
- 19 appears 2 times
The maximum frequency is 5. Both values 10 and 15 appear 5 times.
Answer (ii): The modes are 10 and 15.
The frequency distribution table shows the selling prices of shoes of a special company of shoe shop of our village.
| Size ($x_i$) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|
| Frequency ($f_i$) | 3 | 4 | 5 | 3 | 5 | 4 | 3 | 2 |
Let us find the mode of the above frequency distribution table.
Explanation:
We look at the frequency row to find the highest value.
The maximum frequency is 5.
Looking at the table, the frequency 5 corresponds to two different shoe sizes: size 4 and size 6.
Answer: Since two sizes have the maximum frequency, the modes are 4 and 6.
Let us find the mode from the following frequency distribution table of ages of examinees of an entrance examination.
| Age (year) | 16-18 | 18-20 | 20-22 | 22-24 | 24-26 |
|---|---|---|---|---|---|
| No. of examinees | 45 | 75 | 38 | 22 | 20 |
Explanation:
From the table, the maximum frequency is 75.
Therefore, the modal class is 18-20.
Now we use the formula for Mode for grouped data:
\[\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\]Where:
- $l = 18$ (lower limit of the modal class)
- $f_1 = 75$ (frequency of the modal class)
- $f_0 = 45$ (frequency of the class preceding the modal class)
- $f_2 = 38$ (frequency of the class succeeding the modal class)
- $h = 2$ (size of the class interval)
Substituting the values:
\[\begin{array}{l} \text{Mode} = 18 + \left( \frac{75 - 45}{2 \times 75 - 45 - 38} \right) \times 2 \\ = 18 + \left( \frac{30}{150 - 83} \right) \times 2 \\ = 18 + \left( \frac{30}{67} \right) \times 2 \\ = 18 + \frac{60}{67} \\ \approx 18 + 0.895 \\ = 18.895 \end{array}\]Answer: The mode of the ages is approximately 18.9 years.
Let us see the frequency distribution table of obtaining marks in a periodical examination of 80 students in a class and let us find the mode.
| Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
|---|---|---|---|---|---|---|---|---|
| No. of students | 2 | 6 | 10 | 16 | 22 | 11 | 8 | 5 |
Explanation:
From the table, the maximum frequency is 22.
Therefore, the modal class is 20-25.
Using the formula for Mode:
\[\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\]Where:
- $l = 20$
- $f_1 = 22$
- $f_0 = 16$
- $f_2 = 11$
- $h = 5$
Substituting the values:
\[\begin{array}{l} \text{Mode} = 20 + \left( \frac{22 - 16}{2 \times 22 - 16 - 11} \right) \times 5 \\ = 20 + \left( \frac{6}{44 - 27} \right) \times 5 \\ = 20 + \left( \frac{6}{17} \right) \times 5 \\ = 20 + \frac{30}{17} \\ \approx 20 + 1.76 \\ = 21.76 \end{array}\]Answer: The mode of the marks is approximately 21.76.
Let us find the mode of frequency distribution table given below.
| Class interval | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
|---|---|---|---|---|---|---|---|
| Frequency | 5 | 12 | 18 | 28 | 17 | 12 | 8 |
Explanation:
From the table, the maximum frequency is 28.
Therefore, the modal class is 15-20.
Using the formula for Mode:
\[\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\]Where:
- $l = 15$
- $f_1 = 28$
- $f_0 = 18$
- $f_2 = 17$
- $h = 5$
Substituting the values:
\[\begin{array}{l} \text{Mode} = 15 + \left( \frac{28 - 18}{2 \times 28 - 18 - 17} \right) \times 5 \\ = 15 + \left( \frac{10}{56 - 35} \right) \times 5 \\ = 15 + \left( \frac{10}{21} \right) \times 5 \\ = 15 + \frac{50}{21} \\ \approx 15 + 2.38 \\ = 17.38 \end{array}\]Answer: The mode is approximately 17.38.
Let us find the mode of frequency distribution table given below.
| Class interval | 45-54 | 55-64 | 65-74 | 75-84 | 85-94 | 95-104 |
|---|---|---|---|---|---|---|
| Frequency | 8 | 13 | 19 | 32 | 12 | 6 |
[Hints : In the case of finding mode, we take lower limit of modal class, so we first correct the class limit in inclusive form of class limit.]
Explanation:
First, we need to convert the discontinuous class intervals into continuous class boundaries by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits.
| Class Boundaries | Frequency ($f$) |
|---|---|
| 44.5 - 54.5 | 8 |
| 54.5 - 64.5 | 13 |
| 64.5 - 74.5 | 19 |
| 74.5 - 84.5 | 32 |
| 84.5 - 94.5 | 12 |
| 94.5 - 104.5 | 6 |
The maximum frequency is 32.
Therefore, the modal class is 74.5 - 84.5.
Using the formula for Mode:
\[\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\]Where:
- $l = 74.5$
- $f_1 = 32$
- $f_0 = 19$
- $f_2 = 12$
- $h = 10$ (since $84.5 - 74.5 = 10$)
Substituting the values:
\[\begin{array}{l} \text{Mode} = 74.5 + \left( \frac{32 - 19}{2 \times 32 - 19 - 12} \right) \times 10 \\ = 74.5 + \left( \frac{13}{64 - 31} \right) \times 10 \\ = 74.5 + \left( \frac{13}{33} \right) \times 10 \\ = 74.5 + \frac{130}{33} \\ \approx 74.5 + 3.94 \\ = 78.44 \end{array}\]Answer: The mode is approximately 78.44.
📘 9. Very short answer type question
(A) M.C.Q.Explanation:
An Ogive (cumulative frequency curve) is specifically used in statistics to graphically represent cumulative frequencies. By plotting a "less than" or "more than" ogive (or both), we can determine the median of the grouped data by finding the value on the x-axis corresponding to $N/2$ on the y-axis.
Correct Option: (d) Ogive
Explanation:
The given numbers are: $6, 7, x, 8, y, 14$.
Total number of items ($n$) = 6
Mean of these numbers = 9
We know that, $\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$
\[\begin{array}{l} 9 = \frac{6 + 7 + x + 8 + y + 14}{6} \\ 9 \times 6 = 35 + x + y \\ 54 = 35 + x + y \\ x + y = 54 - 35 \\ x + y = 19 \end{array}\]Correct Option: (b) $x+y=19$
Explanation:
Step 1: Find the original median.
The data is already in ascending order: $30, 34, 35, 36, 37, 38, 39, 40$.
Number of terms ($n$) = 8 (even).
Median is the average of the $\left(\frac{8}{2}\right)^{\text{th}}$ and $\left(\frac{8}{2} + 1\right)^{\text{th}}$ terms, which are the 4th and 5th terms.
\[\text{Original Median} = \frac{36 + 37}{2} = \frac{73}{2} = 36.5\]Step 2: Find the new median after removing 35.
The new data is: $30, 34, 36, 37, 38, 39, 40$.
New number of terms ($n$) = 7 (odd).
Median is the $\left(\frac{7+1}{2}\right)^{\text{th}} = 4^{\text{th}}$ term.
\[\text{New Median} = 37\]Step 3: Calculate the increase.
\[\text{Increase} = \text{New Median} - \text{Original Median} = 37 - 36.5 = 0.5\]Correct Option: (d) 0.5
Explanation:
Let's count the frequency of each number in the given data (excluding x for a moment):
- 14 appears 1 time
- 15 appears 2 times
- 16 appears 2 times
- 17 appears 2 times
- 19 appears 1 time
Currently, the numbers 15, 16, and 17 all have the highest frequency (2). The mode is defined as the value that appears most frequently in a data set.
We are given that the mode of the entire data set is 15. For 15 to be the unique mode, its frequency must be greater than the frequency of any other number.
Therefore, the unknown value '$x$' must be 15, which would make the frequency of 15 equal to 3 (the highest).
Correct Option: (a) 15
Explanation:
The given data in ascending order is: $8, 9, 12, 17, x+2, x+4, 30, 31, 34, 39$.
Total number of terms ($n$) = 10 (even).
The median will be the average of the $\left(\frac{10}{2}\right)^{\text{th}}$ and $\left(\frac{10}{2} + 1\right)^{\text{th}}$ terms, which are the 5th and 6th terms.
- 5th term = $x + 2$
- 6th term = $x + 4$
Correct Option: (b) 12
📘 9. Very short answer type question
(B) True or FalseValue of mode of data $2, 3, 9, 10, 9, 3, 9$ is $10$
Explanation:
Let's count the frequency of each number in the given data set:
- $2$ appears $1$ time
- $3$ appears $2$ times
- $9$ appears $3$ times
- $10$ appears $1$ time
The number $9$ has the maximum frequency ($3$). Therefore, the mode of the data is $9$, not $10$.
Answer: False
Median of data $3, 14, 18, 20, 5$ is $18$
Explanation:
To find the median, we first arrange the given data in ascending order:
$3, 5, 14, 18, 20$
The number of terms ($n$) is $5$, which is an odd number. The median is the $\left(\frac{n+1}{2}\right)^{\text{th}}$ term.
\[\begin{array}{l} \text{Median} = \left(\frac{5+1}{2}\right)^{\text{th}} \text{ term} \\ = \left(\frac{6}{2}\right)^{\text{th}} \text{ term} \\ = 3^{\text{rd}} \text{ term} \end{array}\]Looking at the arranged data, the $3^{\text{rd}}$ term is $14$. The statement claims it is $18$.
Answer: False
📘 9. Very short answer type question
(C) Fill in the blanksExplanation:
Mean, median, and mode are statistical tools used to find the central or typical value of a dataset. They describe the center of a data distribution.
Answer: central tendency
Explanation:
We are given that $\frac{x_1 + x_2 + \dots + x_n}{n} = \bar{x}$.
We need to find the mean of $ax_1, ax_2, \dots ax_n$:
\[\begin{array}{l} \text{New Mean} = \frac{ax_1 + ax_2 + \dots + ax_n}{n} \\ = \frac{a(x_1 + x_2 + \dots + x_n)}{n} \\ = a \times \left( \frac{x_1 + x_2 + \dots + x_n}{n} \right) \\ = a\bar{x} \end{array}\]If each observation is multiplied by a constant '$a$', the new mean is also multiplied by '$a$'.
Answer: $a\bar{x}$
Explanation:
When calculating the arithmetic mean using the direct method ($\frac{\sum f_i x_i}{\sum f_i}$) or the assumed mean method, the class sizes (lengths) do not strictly need to be equal. The formula works correctly for both equal and unequal class intervals as long as we can calculate the class mark ($x_i$).
Answer: equal or unequal
📘 10. Short answer type question :
Solutions| Class | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |
|---|---|---|---|---|---|---|---|
| Frequency | 4 | 15 | 3 | 20 | 14 | 7 | 14 |
Let us find the difference between upper class limit in median class and lower class limit of modal class of the above frequency distribution table.
Explanation:
1. Finding the Median Class:
We need to create a cumulative frequency table.
| Class | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 65 - 85 | 4 | 4 |
| 85 - 105 | 15 | $4+15 = 19$ |
| 105 - 125 | 3 | $19+3 = 22$ |
| 125 - 145 | 20 | $22+20 = 42$ |
| 145 - 165 | 14 | $42+14 = 56$ |
| 165 - 185 | 7 | $56+7 = 63$ |
| 185 - 205 | 14 | $63+14 = 77$ |
Total frequency $N = 77$. So, $\frac{N}{2} = 38.5$.
The cumulative frequency just greater than $38.5$ is $42$, which corresponds to the class $125 - 145$. Thus, this is the median class.
Upper class limit of the median class = $145$.
2. Finding the Modal Class:
The modal class is the class with the highest frequency. From the table, the highest frequency is $20$, which corresponds to the class $125 - 145$.
Lower class limit of the modal class = $125$.
3. Calculating the difference:
Difference = Upper limit of median class - Lower limit of modal class
Difference = $145 - 125 = \mathbf{20}$
The following frequency distribution shows the time taken to complete 100 metre hurdle race of 150 athletics :
| Time (seconds) | 13.8-14 | 14-14.2 | 14.2-14.4 | 14.4-14.6 | 14.6-14.8 | 14.8-15 |
|---|---|---|---|---|---|---|
| No. of athletics | 2 | 4 | 5 | 71 | 48 | 20 |
Let us find the difference between the upper class limit of modal class and lower class limit of modal class.
Explanation:
The modal class is the class with the highest frequency.
Looking at the table, the highest frequency is $71$, which corresponds to the class $14.4 - 14.6$.
This is the modal class.
- Upper class limit of the modal class = $14.6$
- Lower class limit of the modal class = $14.4$
The difference between them is simply the class size (or class length).
Difference = Upper limit - Lower limit
Difference = $14.6 - 14.4 = \mathbf{0.2}$
Explanation:
We are given:
- Mean ($\bar{x}$) = $8.1$
- $\sum f_i x_i = 132 + 5k$
- $\sum f_i = 20$
We know the formula for the direct method of calculating the mean is:
\[\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\]Substitute the given values into the formula:
\[\begin{array}{l} 8.1 = \frac{132 + 5k}{20} \\ 8.1 \times 20 = 132 + 5k \\ 162 = 132 + 5k \\ 5k = 162 - 132 \\ 5k = 30 \\ k = \frac{30}{5} \\ k = \mathbf{6} \end{array}\]Explanation:
This problem uses the step-deviation method for finding the mean.
The standard substitution is $u_i = \frac{x_i - a}{h}$, where '$a$' is the assumed mean and '$h$' is the class size.
Comparing $u_i = \frac{x_i - 25}{10}$ with the standard form, we can identify:
- Assumed Mean ($a$) = $25$
- Class Size ($h$) = $10$
We are also given:
- $\sum f_i u_i = 20$
- $\sum f_i = 100$
The formula for the mean using the step-deviation method is:
\[\bar{x} = a + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right)\]Substitute the values:
\[\begin{array}{l} \bar{x} = 25 + 10 \times \left( \frac{20}{100} \right) \\ \bar{x} = 25 + 10 \times 0.2 \\ \bar{x} = 25 + 2 \\ \bar{x} = \mathbf{27} \end{array}\]| Marks | less than 10 | less than 20 | less than 30 | less than 40 | less than 50 | less than 60 |
|---|---|---|---|---|---|---|
| No. of students | 3 | 12 | 27 | 57 | 75 | 80 |
Let us write the modal class from the above frequency distribution table.
Explanation:
The given table represents a cumulative frequency distribution (less than type). To find the modal class, we must first convert it into a standard frequency distribution table to find the individual frequencies for each class interval.
| Class Interval (Marks) | Cumulative Frequency ($cf$) | Frequency ($f$) |
|---|---|---|
| 0 - 10 | 3 | 3 |
| 10 - 20 | 12 | $12 - 3 = 9$ |
| 20 - 30 | 27 | $27 - 12 = 15$ |
| 30 - 40 | 57 | $57 - 27 = \mathbf{30}$ |
| 40 - 50 | 75 | $75 - 57 = 18$ |
| 50 - 60 | 80 | $80 - 75 = 5$ |
The modal class is the class interval with the highest frequency.
From our derived table, the highest frequency is $30$, which belongs to the class interval $30 - 40$.
Answer: The modal class is $30 - 40$.
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