CBSE Class 9 Mathematics Statistics Worksheet Set 02

Question. Form a cumulative frequency table with class intervals of length 50 for the following data:
370, 290, 318, 175, 170, 410, 378, 405, 380, 375, 315, 305, 325, 275, 241, 288, 261, 355, 402, 380, 178, 253, 428, 240, 210, 175, 154, 405, 380, 370, 306, 460, 328, 440, 425.
Answer:
To group the given data with class intervals of length 50, we first identify the minimum value (154) and maximum value (460). We can start our intervals at 150. Below is the cumulative frequency table:

In simple words: A cumulative frequency table helps us see the running total of our data counts. To find it, keep adding each class's count to the sum of the classes before it.
Exam Tip: Always make sure the final cumulative frequency matches the total number of data points given in the question (in this case, 35).

Question. List the different ways for representing data graphically.
Answer:
Data can be represented graphically in several ways:
1. Bar Graphs
2. Histograms (for continuous grouped data)
3. Frequency Polygons
4. Pie Charts (or Circle Graphs)
5. Line Graphs
In simple words: Graphs are visual ways to show numbers. Some common types are bar graphs, histograms, pie charts, and line graphs.
Exam Tip: In examinations, keep your list clear and define each briefly if asked for extra detail.

Question. What is a histogram?
Answer:
A histogram is a two-dimensional graphical representation of continuous grouped data. It is constructed using a series of adjacent rectangles where the class intervals are placed on the horizontal axis (x-axis) and the frequencies are placed on the vertical axis (y-axis). The area of each rectangle is proportional to the corresponding frequency.
In simple words: A histogram is like a bar graph, but the bars touch each other because they show continuous intervals of numbers on the bottom axis.
Exam Tip: Remember that unlike standard bar graphs, there are no gaps between the bars of a histogram because the intervals are continuous.

Question. From the graph given below, read the temperature at 11 a.m. and 4 p.m.
Answer:
By analyzing the coordinates plotted on the graph:
- The point for 11 a.m. is plotted at \( (11, 33) \). Therefore, the temperature at 11 a.m. is 33°C.
- For 4 p.m., we look along the line segment connecting 3 p.m. \( (3, 35) \) and 5 p.m. \( (5, 34) \). At the midpoint of this segment (representing 4 p.m.), the temperature is halfway between 35°C and 34°C, which is 34.5°C.
In simple words: Looking at the points on the graph, the temperature at 11 a.m. is 33°C. At 4 p.m., which is exactly halfway between 3 p.m. and 5 p.m., the temperature is 34.5°C.
Exam Tip: When a point is not explicitly labeled, use a ruler to read the values on the axes or find the mid-point of the connecting line segment.

Question. The following table gives the number of vehicles passing through a busy crossing in Noida in different time intervals on a particular day. Represent the above data by a bar graph.
Answer:
The given data represents categorical time intervals with distinct counts of vehicles. Here is the bar graph:

In simple words: This bar graph shows the number of vehicles passing by during different hours of the day using vertical bars of different heights.
Exam Tip: Remember to label both axes clearly and use a consistent scale for the heights of the bars.

Question. The population of a state in different census year is as given below: Represent the above information with the help of a histogram.
Answer:
Because the years represent consecutive continuous periods, we can plot them touching each other on a histogram:

In simple words: We draw consecutive bars touching each other because census years represent adjacent periods, making a histogram the right choice.
Exam Tip: Histograms require bars of uniform class intervals to touch. Make sure you use a clean, linear scale for the vertical axis.

Question. Number of children in seven different classes are given below: Represent the data with the help of a bar graph.
Answer:
The data represents distinct classes which are separate categories, so a spaced bar graph is most appropriate here:

In simple words: This bar graph represents the number of children in each distinct class from Class VI to Class XII with separated vertical bars.
Exam Tip: Since classes are categorical data, leave equal spaces between the bars to distinguish it from a histogram.

Question. Read the bar graph shown in the figure and answer the following questions :
(a) What is the information given by the bar graph ?
(b) What is the order of the change of number of students over several years?
(c) In which year is the increase of students maximum?
(d) State whether true or false :
The enrolment during 1996 - 97 is twice that of 1995 - 96.
Answer:
(a) The bar graph shows the academic enrollment (number of students) over several academic years from 1995-1996 to 1999-2000.
(b) The number of students increases steadily in an ascending order by exactly 50 students each year (from 150 to 350 over 5 years).
(c) The year-on-year increase is constant at 50 students per year across all consecutive years (e.g., 200 - 150 = 50, 250 - 200 = 50, etc.). Thus, the increase is uniform throughout the period.
(d) **False**. The enrolment in 1996-1997 is 200 students, while in 1995-1996 it is 150 students. Since 200 is not twice of 150 (which would be 300), the statement is incorrect.
In simple words: The graph shows how many students enrolled over 5 years. It goes up by 50 students every single year, meaning the growth is steady. The statement in (d) is false because 200 is not twice of 150.
Exam Tip: For true-or-false graph questions, calculate the actual numbers from the graph to verify before giving your final answer.

Question. The results of pass percentage of Class X and XII for 5 years are given below in the table: Represent this data using a double bar graph.
Answer:
A double bar graph is perfect for comparing two groups (Class X and Class XII) side-by-side across several years:

In simple words: This double bar graph helps us compare the pass results of Class X and Class XII next to each other over a five-year period.
Exam Tip: Include a distinct legend (color scheme) to make it clear which bar belongs to Class X and which to Class XII.

Question. What is a bar graph?
Answer:
A bar graph is a diagrammatic representation of data using rectangular bars of uniform width. These bars can be drawn vertically or horizontally with equal spacing between them, and the height or length of each bar is directly proportional to the numerical value it represents.
In simple words: A bar graph is a chart that uses spaced-out bars of different heights to show and compare amounts easily.
Exam Tip: Make sure to emphasize in your definition that the bars are of uniform width and have equal spacing between them.

Question. The bar graph shown in figure represents the circulation of newspaper in 10 languages. Study the bar graph and answer the following questions:
1. What is the total number of newspapers published in Hindi, English, Urdu, Punjabi and Bengali?
2. Name two pairs of languages which publish the same number of newspapers.
3. State the language in which the largest number of news papers is published.
4. State the language in which the number of news papers published is between 2500 and 3600.
Answer:
By reading the values from the bar graph on Page 101:
1. The respective circulation values are:
- Hindi = 3200
- English = 3400
- Urdu = 800
- Punjabi = 200
- Bengali = 1000
\( \text{Total} = 3200 + 3400 + 800 + 200 + 1000 = 8600 \) newspapers.
2. The two pairs with equal circulation are:
- **Pair 1**: Gujarati and Bengali (each has a circulation of 1000).
- **Pair 2**: Marathi and Malayalam (each has a circulation of 1200).
3. The language with the largest circulation is **English** (with 3400 newspapers).
4. The languages having a circulation between 2500 and 3600 are **Hindi** (3200) and **English** (3400).
In simple words: 1. Adding up the newspapers for these five languages gives 8,600. 2. Gujarati/Bengali (1000 each) and Marathi/Malayalam (1200 each) have identical numbers. 3. English has the highest circulation. 4. Hindi and English circulation falls between 2500 and 3600.
Exam Tip: Always look carefully at the scale on the x-axis to ensure you read values like 3200 or 1200 with precision.

Question. The marks scored by 750 students in an examination are given in the form of a frequency distribution table : Represent this data in the form of a histogram and construct a frequency polygon.
Answer:
Below is the joint representation of the histogram and frequency polygon. The midpoints (class marks) are plotted to construct the polygon:

In simple words: The shaded bars make the histogram, while the orange line joining the center tops of the bars forms the frequency polygon.
Exam Tip: Extend your frequency polygon to the base axis on both sides using imaginary adjacent intervals with zero frequency.

Question. Construct a frequency polygon for the following data :
Answer:
By plotting the midpoints (class marks) of each age interval against their frequencies, we construct the following polygon:

In simple words: This line graph connects the center values of our age groups to show how frequency rises and falls.
Exam Tip: Class marks are calculated as \( \frac{\text{Upper Limit} + \text{Lower Limit}}{2} \). Use these values as your x-coordinates.

Question. Construct a frequency polygon for the following data :
Answer:
This table matches the data in Question 27 exactly. Its frequency polygon is identical:

In simple words: This graph is identical to the previous one because the given data set is exactly the same.
Exam Tip: Be sure to double check if a repeated question has identical numbers before drawing.

Question. The following are the scores of two groups of class IV students in a test of reading ability. Construct a frequency polygon for each of these two groups on the same axes.
Answer:
First, we calculate the class marks (midpoints) for each score interval:
- 50 - 52: 51
- 47 - 49: 48
- 44 - 46: 45
- 41 - 43: 42
- 38 - 40: 39
- 35 - 37: 36
- 32 - 34: 33
We add two imaginary intervals on the ends with 0 frequency: 29 - 31 (class mark 30) and 53 - 55 (class mark 54).

In simple words: Two lines are drawn on the same graph to compare how both groups performed in their reading tests side-by-side.
Exam Tip: Draw the frequency polygons using distinct line styles or colors to avoid confusion on the same axes.

Question. Represent the following data by means of a histogram :
Answer:
Because the class widths are unequal, we must calculate the adjusted frequencies first. The minimum class width is 5.
Adjusted Frequency = \( \frac{\text{Frequency}}{\text{Class Width}} \times \text{Minimum Class Width} \)

Using these adjusted frequencies, we draw the histogram below:

In simple words: Because some wage groups are wider than others, we adjust the height of the bars so that the area of each block accurately shows the real distribution of workers.
Exam Tip: Whenever class intervals have unequal widths, always recalculate the frequencies using the standard formula before drawing the histogram.

Question. Find the arithmetic mean of first 7 numbers whole numbers .
Answer:
The first 7 whole numbers are: 0, 1, 2, 3, 4, 5, and 6.
The formula for calculating the mean is:
\( \text{Mean} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} \)
\( \text{Mean} = \frac{0 + 1 + 2 + 3 + 4 + 5 + 6}{7} \)
\( \text{Mean} = \frac{21}{7} = 3 \)
In simple words: The first seven whole numbers are 0 through 6. Adding them up gives 21, and dividing by 7 gives a mean of 3.
Exam Tip: Remember that whole numbers start from 0, not 1. If you start from 1, your calculation will be incorrect.

Question. If the mean of 2, 4, 6 and p is 11, find the value of p.
Answer:
The sum of the four observations divided by 4 is equal to 11:
\( \frac{2 + 4 + 6 + p}{4} = 11 \)
\( \implies \frac{12 + p}{4} = 11 \)
\( \implies 12 + p = 44 \)
\( \implies p = 44 - 12 = 32 \)
In simple words: The average of the four values is 11. By solving the equation, we find that the missing value \( p \) must be 32.
Exam Tip: Cross-multiply the denominator of your mean formula first to simplify the algebraic equation quickly.

Question. Find the median of the following data : 37, 31, 42, 43, 46, 25, 39, 45, 32
Answer:
First, we arrange the observations in ascending order:
25, 31, 32, 37, 39, 42, 43, 45, 46
Here, the number of observations is \( n = 9 \) (which is an odd number).
The median is the \( \left(\frac{n+1}{2}\right)\)-th term:
\( \text{Median} = \left(\frac{9+1}{2}\right)\text{-th term} = 5\text{-th term} \)
Looking at our ordered list, the 5th term is 39.
In simple words: If you write the numbers in order from smallest to largest, the exact middle number is 39.
Exam Tip: Never calculate the median without sorting the numbers in ascending or descending order first.

Question. Find the median of the following data : 25, 34, 31, 23, 22, 26, 35, 28, 20, 32.
Answer:
First, we arrange the observations in ascending order:
20, 22, 23, 25, 26, 28, 31, 32, 34, 35
Here, the number of observations is \( n = 10 \) (which is an even number).
The median is the average of the \( \left(\frac{n}{2}\right)\)-th and the \( \left(\frac{n}{2} + 1\right)\)-th terms:
\( \text{Median} = \frac{5\text{-th term} + 6\text{-th term}}{2} \)
\( \text{Median} = \frac{26 + 28}{2} = \frac{54}{2} = 27 \)
In simple words: Since there is an even number of data points, we take the average of the two middle numbers (26 and 28), which gives us 27.
Exam Tip: When \( n \) is even, the median value does not have to be one of the original numbers in the given list.

Question. Five people were asked about the time in a week they spend in doing social work in their community. They said 10, 7, 13, 20 and 15 hours, respectively. Find the mean (or average) time in a week devoted by them for social work.
Answer:
The given hours are: 10, 7, 13, 20, 15.
\( \text{Mean} = \frac{\text{Sum of hours}}{\text{Number of people}} \)
\( \text{Mean} = \frac{10 + 7 + 13 + 20 + 15}{5} \)
\( \text{Mean} = \frac{65}{5} = 13 \text{ hours} \)
In simple words: If you add all their hours together, they worked for 65 hours total. Spread equally among the 5 people, the average is 13 hours.
Exam Tip: Include the correct unit (hours) with your final answer to get full marks.

Question. If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, find the mean of first three observations.
Answer:
First, we use the mean of the five observations to calculate \( x \):
\( \frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} = 11 \)
\( \implies \frac{5x + 20}{5} = 11 \)
\( \implies x + 4 = 11 \)
\( \implies x = 7 \)
Now, the first three observations are \( x \), \( x + 2 \), and \( x + 4 \).
Substituting \( x = 7 \), these observations are 7, 9, and 11.
Their mean is:
\( \text{Mean} = \frac{7 + 9 + 11}{3} = \frac{27}{3} = 9 \)
In simple words: First we solve for \( x \) and find it is 7. The first three numbers are therefore 7, 9, and 11, which have an average of 9.
Exam Tip: For evenly spaced consecutive terms, the mean is always equal to the middle term. Here, the mean of the first three terms is simply the second term, which is \( x + 2 \).

Question. Give one example of a situation in which
(a) The mean is an appropriate measure of central tendency.
(b) The mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
(a) **Example**: The average height of students in a class. Since heights generally follow a normal distribution without extreme values, the mean represents the center point well.
(b) **Example**: The average salary of employees in a small firm with a high-earning executive. If 4 workers earn Rs. 5,000 each and the executive earns Rs. 1,00,000, the mean salary is Rs. 24,000, which is misleading. The median salary (Rs. 5,000) represents a typical worker's income much more realistically because it is not affected by outliers.
In simple words: The mean is perfect for normal things like student heights. But for things with massive extremes, like salaries when a CEO is included, the median is a much fairer middle value.
Exam Tip: Use the term "outlier" or "extreme values" to explain why the median is preferred in skewed distributions.

Question. The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 is arranged in ascending order is 24. Find the value of x.
Answer:
The list has \( n = 10 \) observations (even).
The median is the average of the 5th and 6th observations:
\( \text{Median} = \frac{(x + 2) + (x + 4)}{2} \)
\( 24 = \frac{2x + 6}{2} \)
\( 24 = x + 3 \)
\( \implies x = 21 \)
In simple words: The two middle numbers are \( x + 2 \) and \( x + 4 \), meaning their midpoint is \( x + 3 \). Setting this equal to 24 gives \( x = 21 \).
Exam Tip: Since the observations are already in ascending order, directly pick the two middle values and solve.

Question. Find the mode of the following data : 110, 120, 130, 120, 110, 140, 130, 120, 140, 120.
Answer:
We calculate the frequency of occurrence of each value:
- 110: 2 times
- 120: 4 times
- 130: 2 times
- 140: 2 times
Since 120 has the maximum frequency (it appears 4 times), the mode is 120.
In simple words: The number that appears most often in this list is 120, so 120 is the mode.
Exam Tip: Always make a quick frequency count table to ensure you don't miss any values while looking for the mode.

Question. The mean monthly salary of 10 members of a group is Rs. 1445, one more member whose monthly salary is Rs. 1500 has joined the group. Find the mean monthly salary of 11 members of the group.
Answer:
Total salary of the first 10 members = \( 10 \times \text{Rs. } 1445 = \text{Rs. } 14450 \).
After the new member joins, the new total salary becomes:
\( \text{Rs. } 14450 + \text{Rs. } 1500 = \text{Rs. } 15950 \)
The new mean salary for 11 members is:
\( \text{New Mean} = \frac{\text{Rs. } 15950}{11} = \text{Rs. } 1450 \)
In simple words: The total earnings of the 10 people was Rs. 14,450. Adding the new person's salary brings the total to Rs. 15,950, which averages out to Rs. 1,450 per person.
Exam Tip: Find the sum of original values by multiplying the number of items by their mean first, then perform the addition.

Question. Find the mode for the following series : 7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2
Answer:
We calculate the frequencies of each distinct number:
- 7.2: 4 times
- 7.3: 2 times
- 7.4: 1 time
- 7.5: 2 times
- 7.6: 1 time
- 7.7: 2 times
Since 7.2 has the maximum frequency (it occurs 4 times), the mode is 7.2.
In simple words: The number 7.2 appears 4 times, which is more than any other number in the list. Thus, 7.2 is the mode.
Exam Tip: Count the total number of items after grouping to ensure you did not leave out any decimals.

Question. Following table shows the weights of 12 students: Find the mean weight.
Answer:
We use the formula for the mean of grouped data, \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \):

\( \text{Mean} = \frac{843}{12} = 70.25 \text{ kg} \)
In simple words: Multiplying weights by the number of students gives a total of 843 kg. Divided among the 12 students, the average weight is 70.25 kg.
Exam Tip: Set up a column for \( f_i x_i \) to make your calculations systematic and minimize calculation errors.

Question. The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
Answer:
The incorrect sum of the 40 observations is:
\( \text{Incorrect Sum} = 40 \times 160 = 6400 \)
Now, we calculate the correct sum by removing the incorrect value and adding the correct value:
\( \text{Correct Sum} = 6400 - 125 + 165 = 6440 \)
The correct mean is:
\( \text{Correct Mean} = \frac{6440}{40} = 161 \)
In simple words: The original calculation missed 40 units because 165 was misread as 125. Adding those 40 units back raises the total from 6400 to 6440, giving a corrected average of 161.
Exam Tip: A quicker method is to find the difference \( \frac{165 - 125}{40} = 1 \) and add it directly to the old mean.

Question. The mean of 5 numbers is 18. If one numbers is excluded, their mean is 16. Find the excluded number.
Answer:
Sum of the 5 numbers is:
\( 5 \times 18 = 90 \)
Sum of the remaining 4 numbers is:
\( 4 \times 16 = 64 \)
The value of the excluded number is:
\( 90 - 64 = 26 \)
In simple words: The five numbers added up to 90. When one number was taken away, the remaining four added up to 64. The missing number is 26.
Exam Tip: Subtracting the sum of the smaller dataset from the sum of the larger dataset always yields the excluded value directly.

Question. Find the median of the following data : 19, 25, 59, 48, 35, 31, 30, 32, 51. If a student replaces 25 by 52 by mistake, what will be the new median?
Answer:
1. **Original Median**:
Arrange the original 9 observations in ascending order:
19, 25, 30, 31, 32, 35, 48, 51, 59
Since \( n = 9 \) is odd, the median is the 5th term:
\( \text{Original Median} = 32 \)

2. **New Median**:
If 25 is replaced by 52, the new set of observations is:
19, 30, 31, 32, 35, 48, 51, 52, 59
The new median is the 5th term:
\( \text{New Median} = 35 \)
In simple words: Originally, sorting the numbers placed 32 in the middle. When 25 is swapped for 52, the numbers shift and the new middle value becomes 35.
Exam Tip: Replacing a value below the median with a value above the median shifts the median position to the next higher term.

Question. The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean?
Answer:
If a constant \( k \) is subtracted from every observation in a data set, the mean of the new observations also decreases by the same constant \( k \).
Given: original mean = 20, value subtracted \( k = 5 \).
\( \text{New Mean} = 20 - 5 = 15 \)
In simple words: If you take 5 away from every single number in a group, their overall average will also drop by exactly 5.
Exam Tip: You do not need to assume values for the 10 numbers; apply this property directly as a theorem for full marks.

Question. If the mean of the following distribution is 6, find the value of p.
Answer:
We use the grouped mean formula \( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \):

Given that the mean is 6:
\( \frac{2p + 52}{11} = 6 \)
\( \implies 2p + 52 = 66 \)
\( \implies 2p = 14 \)
\( \implies p = 7 \)
In simple words: Setting up the mean calculation with the variable \( p \) gives us an equation. Solving it shows that \( p \) must equal 7.
Exam Tip: Be careful to multiply \( 2 \) by both parts of \( (p + 5) \) to get \( 2p + 10 \) rather than \( 2p + 5 \).

Question. Consider a small unit of a factory where there are 5 employees : A supervisor and four labourers. The labourers draw a salary of Rs 5,000 per month each while the supervisor gets Rs 15,000 per month. Calculate the mean, median and mode of the salaries of this unit of the factory. Interpret the findings.
Answer:
The monthly salaries are: Rs. 5000, Rs. 5000, Rs. 5000, Rs. 5000, and Rs. 15000.
1. **Mean**:
\( \text{Mean} = \frac{5000 + 5000 + 5000 + 5000 + 15000}{5} = \frac{35000}{5} = \text{Rs. } 7000 \)

2. **Median**:
Since the values sorted in ascending order are 5000, 5000, 5000, 5000, 15000, the 3rd (middle) value is:
\( \text{Median} = \text{Rs. } 5000 \)

3. **Mode**:
Since Rs. 5000 occurs most frequently (4 times):
\( \text{Mode} = \text{Rs. } 5000 \)

**Interpretation**:
The median and mode (Rs. 5000) accurately represent the typical salary of a worker at this factory since 80% of the employees earn exactly this amount. The mean salary (Rs. 7000) is skewed upwards because of the supervisor's higher salary (Rs. 15000), making the mean unrepresentative of a typical worker's monthly pay.
In simple words: The average (mean) salary is Rs. 7,000, which is misleading because most workers only earn Rs. 5,000. The median and mode of Rs. 5,000 are much better representations of a typical salary here.
Exam Tip: When interpreting statistics, always point out how extreme values (outliers) affect the mean but leave the median and mode unaffected.