OP Malhotra Class 9 Maths Solutions Chapter 15 Mean Median and Frequency Polygon Chapter

Question 1. The distance (in km) of 40 female engineers from their residence to their place of work were found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 - 5 (5 not included). What main features do you observe from this tabular representation?

Answer: To make a grouped frequency distribution table, we first need to count how many times each distance appears in the given data for specific groups (class sizes). For the class size 0-5 (where 5 is not included), we count all distances from 0 up to 4. Then we make a tally mark for each count and write the total frequency. This helps organize the raw data into a clear summary.

Main features we can observe from this table:

• Most female engineers, 22 of them, live within 5 to 15 km from their work. This includes distances in the 5-10 km and 10-15 km ranges.
• Only a small number, 4 female engineers, live very far, specifically within 20 to 35 km from their workplace. This covers the 20-25 km, 25-30 km, and 30-35 km ranges.

In simple words: We grouped the distances into ranges of 5 km each and counted how many engineers fall into each range. The table shows that most engineers live between 5 km and 15 km from work, and very few live further than 20 km. This kind of table helps us understand where most of the data points are.

🎯 Exam Tip: When constructing grouped frequency distribution tables, always clearly define your class intervals and ensure they are mutually exclusive (no overlap) and exhaustive (cover all data points). Double-check your tally marks and frequency counts to avoid errors.

 

Question 2. The table shows a frequency distribution of the life times of 400 radio tubes tested at a company.
With reference to this table determine the
(ii) lower limit of the eighth class.
(iii) class mark of the seventh class.
(iv) class boundaries of the last class.
(v) class interval size.
(vi) frequency of the fourth class.

Answer: From the given table, we can determine the following values:
(i) The upper limit of the fifth class (700-799) is 799.
(ii) The lower limit of the eighth class (1000-1099) is 1000.
(iii) The class mark of the seventh class is found by averaging its upper and lower limits: \( \frac{900+999}{2}=\frac{1899}{2} = 949.5 \). The class mark represents the midpoint of a class interval.
(iv) The class boundaries of the last class (1100-1199) are simply 1100 and 1199.
(v) The class interval size is the difference between the upper and lower limits plus one (for discrete data), or between consecutive lower limits. In this case, it is \( 399 - 300 + 1 = 100 \) hours (or \( 400-300 = 100 \)).
(vi) The frequency of the fourth class (600-699) is 76.
In simple words: We looked at the table to find specific values like the highest number in a class, the lowest number, the middle point of a class, and how wide each group is. For example, for the group from 900 to 999 hours, the middle value is 949.5 hours.

🎯 Exam Tip: Always clearly identify the class intervals and their corresponding frequencies. Remember that for continuous data, the class interval size is simply the difference between the upper and lower boundaries, and the class mark is the midpoint of the interval.

 

Question 3. What is the median of the values 11, 7, 6, 9, 12, 15, 19
(a) 9
(b) 11
(c) 12
(d) 15
Answer: (b) 11
To find the median, we first arrange the given values in ascending order (from smallest to largest).
The numbers are: 6, 7, 9, 11, 12, 15, 19.
Next, we count how many values there are. Here, \( N = 7 \) (an odd number).
The median is the middle value. For an odd number of observations, the median is the \( \frac{N+1}{2} \)-th value.
So, the median is the \( \frac{7+1}{2} = \frac{8}{2} = 4 \)-th value.
Looking at our ordered list, the 4th value is 11.
In simple words: First, put all the numbers in order from smallest to largest. Since there are 7 numbers, the middle number is the 4th one. The 4th number in our sorted list is 11, which is the median.

🎯 Exam Tip: Always remember to arrange the data in ascending or descending order before finding the median. If you forget this step, your answer will be incorrect. The median helps find the central value of a dataset.

 

Question 4. The following observations have been arranged in ascending order. If the median of the data is 63, the values of x is
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
(a) 63
(b) 62.5
(c) 62
(d) 70
Answer: (c) 62
The given observations are already arranged in ascending order: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.
Let's count the total number of observations, \( N \). There are 10 observations, so \( N = 10 \) (an even number).
For an even number of observations, the median is the average of the two middle values, which are the \( \frac{N}{2} \)-th term and the \( (\frac{N}{2}+1) \)-th term.
Here, the middle terms are the \( \frac{10}{2} = 5 \)-th term and the \( (\frac{10}{2}+1) = 6 \)-th term.
The 5th term is \( x \) and the 6th term is \( x + 2 \).
So, the median is \( \frac{1}{2} (5\text{th term} + 6\text{th term}) \).
Median \( = \frac{1}{2} (x + (x + 2)) \)
Median \( = \frac{1}{2} (2x + 2) \)
Median \( = x + 1 \)
We are given that the median is 63.
So, \( x + 1 = 63 \)
To find \( x \), we subtract 1 from both sides: \( x = 63 - 1 \)
\( \implies x = 62 \).
In simple words: The numbers are already in order. There are 10 numbers, so the median is the average of the 5th and 6th numbers. These are 'x' and 'x+2'. When we average them, we get 'x+1'. Since we know the median is 63, we can find 'x' by solving the simple equation \( x+1=63 \), which means \( x \) is 62.

🎯 Exam Tip: When given an even number of observations, remember that the median is the average of the two middle terms. Set up the equation correctly and solve carefully for the unknown variable.

 

Question 5.
(i) The median of the first 10 prime numbers is
(b) 11
(c) 12.5
(d) 12
(ii) The mean of the factors of 36 is
(a) 11
(b) \( 12\frac { 1 }{ 6 } \)
(c) \( 10\frac { 1 }{ 9 } \)
(d) \( 10\frac { 1 }{ 8 } \)
Answer:
(i) The median of the first 10 prime numbers is 12.
First, list the first 10 prime numbers in ascending order: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
There are \( N = 10 \) numbers. Since \( N \) is an even number, the median is the average of the two middle terms: the \( \frac{N}{2} \)-th term and the \( (\frac{N}{2}+1) \)-th term.
The middle terms are the \( \frac{10}{2} = 5 \)-th term (11) and the \( (\frac{10}{2}+1) = 6 \)-th term (13).
Median \( = \frac{1}{2} (5\text{th term} + 6\text{th term}) \)
Median \( = \frac{1}{2} (11 + 13) \)
Median \( = \frac{1}{2} (24) \)
Median \( = 12 \).
(ii) The mean of the factors of 36 is \( 10\frac { 1 }{ 8 } \).
First, list all the factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
There are 9 factors in total. Let's call this \( n = 9 \).
To find the mean, we sum all the factors and divide by the number of factors.
Sum of factors \( = 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91 \).
Mean \( = \frac{\text{Sum of factors}}{\text{Number of factors}} \)
Mean \( = \frac{91}{9} \)
Mean \( = 10\frac{1}{9} \).
In simple words: For the first part, we listed the first 10 prime numbers and found their middle value, which is 12. For the second part, we listed all numbers that divide 36 evenly, added them up, and then divided by how many factors there were to get the average. Prime numbers are numbers only divisible by 1 and themselves.

🎯 Exam Tip: To find the median, always list numbers in order. For the mean, sum all values and divide by the count. Make sure to identify all factors correctly and list prime numbers accurately, as one mistake can change the entire result.

 

Question 6. A discontinuous distribution of integral data is given below. Draw a frequency polygon.

Answer: To draw a frequency polygon from this discontinuous data, we first need to find the mid-value for each class interval. The mid-value is the average of the lower and upper limits of each class. Once we have the mid-values and their corresponding frequencies, we plot these as points on a graph and connect them with straight lines. To close the polygon, we add two imaginary classes with zero frequency on either side of the given data. Frequency polygons are excellent for visualizing the shape of a distribution and comparing different datasets.

Taking points (15.5, 25), (25.5, 12), (35.5, 10), (45.5, 15) and (55.5, 8) on the graph. We also add two points to close the polygon on the x-axis: (5.5, 0) and (65.5, 0). By joining them in order, we get the frequency polygon as given below:
In simple words: We found the middle point for each age group. Then we plotted these middle points against their counts (frequencies) on a graph. We connected these points with straight lines and added points at the start and end on the bottom line to finish the shape. This shows us how the counts change across different age groups.

🎯 Exam Tip: When drawing frequency polygons, make sure to calculate mid-values correctly. Always close the polygon by connecting the first and last plotted points to the x-axis at the mid-points of the imaginary classes adjacent to the given data, with zero frequency.

 

Question 7.
48, 25, 110, 65, 12, 10, 40, 45, 90, 14, 24, 48, 55, 5, 1, 17, 80, 20, 18, 30, 59, 115, 9, 49, 17.
(i) Construct a classified frequency distribution table with overlapping class intervals, 0 - 20, 20 - 40, etc.
(ii) Draw a frequency polygon to display the distribution.

Answer:
(i) To construct a classified frequency distribution table with overlapping class intervals, we go through the raw data and count how many values fall into each specified range. For overlapping intervals like 0-20, 20-40, etc., any value that falls exactly on a boundary (like 20) is typically included in the *upper* interval (e.g., 20 would be in 20-40). After tallying, we calculate the mid-value for each class.

(ii) To draw the frequency polygon, we plot the mid-points of each class interval against their frequencies. The points to be plotted are (10, 9), (30, 4), (50, 7), (70, 1), (90, 2), and (110, 2). To close the polygon, we connect these points to the mid-points of the imaginary classes at the beginning and end, which would have zero frequency. This helps visualize the overall shape of the data distribution.
In simple words: First, we sorted the numbers into groups, counting how many fell into each range. Then, we found the middle value for each group. We plotted these middle values against their counts on a graph and connected them to form a shape. This picture helps us see where most of the numbers are.

🎯 Exam Tip: When dealing with overlapping class intervals, clearly establish the rule for boundary values (e.g., upper limit exclusive, lower limit inclusive). For drawing frequency polygons, make sure to add imaginary classes with zero frequency on both ends to correctly close the polygon on the x-axis.

 

Question 8. The members of several professional basketball teams were measured for their heights. The results were:

Draw a bar chart and a frequency polygon to illustrate this date.

Answer: To illustrate the data with a bar chart (histogram) and a frequency polygon, we first determine the mid-value for each height range. The histogram uses bars that touch each other to show the frequency of continuous data. The height of each bar corresponds to the frequency. A frequency polygon is then drawn by connecting the midpoints of the tops of these histogram bars. This helps visualize how heights are distributed among the basketball players. Histograms and frequency polygons are great ways to display continuous numerical data.

(i) The Bar chart (Histogram) and frequency polygon are given below:
In simple words: We made a bar graph where each bar shows how many players have heights in a certain range, with the bars touching because height is a continuous measurement. Then, we drew a line graph on top by connecting the middle points of the top of each bar. This double graph helps us see the pattern of heights clearly.

🎯 Exam Tip: Remember that a histogram uses touching bars for continuous data, while a bar chart uses separated bars for discrete data. For the frequency polygon, connect the midpoints of the top of each bar and ensure the polygon is closed on the x-axis by extending to the midpoints of adjacent imaginary classes with zero frequency.

 

Question 9. Consider the following data :

The arithmetic mean of the above distribution is 2.96. What is the missing frequency?
(a) 4
(b) 6
(c) 7
(d) 8
Answer: (b) 6
To find the missing frequency, we first extend the table to calculate \( x \times y \) for each row, and then find the sum of \( y \) (total frequency) and the sum of \( x \times y \). We know the formula for the arithmetic mean in a frequency distribution is \( \text{Mean} = \frac{\sum (x \times y)}{\sum y} \).

We are given that the Mean \( = 2.96 \).
Using the formula for the mean:
\( \text{Mean} = \frac{\sum (x \times y)}{\sum y} \)
\( 2.96 = \frac{50+4x}{19+x} \)
Now, we cross-multiply to solve for \( x \):
\( 2.96 \times (19+x) = 50 + 4x \)
\( 56.24 + 2.96x = 50 + 4x \)
Next, we gather the \( x \) terms on one side and the numbers on the other side:
\( 56.24 - 50 = 4x - 2.96x \)
\( 6.24 = 1.04x \)
To find \( x \), divide both sides by 1.04:
\( x = \frac{6.24}{1.04} \)
\( \implies x = 6 \).
So, the missing frequency is 6.
In simple words: We used the formula for average (mean) which says the mean is the sum of (x times y) divided by the sum of y. We put the known mean (2.96) and the sums from our table into this formula. Then, we solved the equation to find the value of 'x', which is the missing frequency. Finding missing data is a practical application of statistics.

🎯 Exam Tip: For problems with missing frequencies in a mean calculation, always set up your frequency table correctly, calculate \( \sum (x \times y) \) and \( \sum y \) in terms of the unknown, and then solve the resulting algebraic equation carefully.

 

Question 10. Two frequency polygons are shown giving the distribution of the weights of players in two different sports A and B.

Answer:
(i) To find the total number of people who played sport A, we sum up all the frequencies for polygon A. Looking at the graph, the frequencies for Sport A are:
4 (at 55g) + 8 (at 60g) + 14 (at 65g) + 16 (at 70g) + 10 (at 75g) + 6 (at 80g) + 4 (at 85g) = 62.
So, 62 people played sport A.
(ii) We can observe two main differences between the two frequency polygons:

• Sport B generally has players with heavier weights compared to Sport A. The peak of Sport B's polygon is at a higher weight value than Sport A's.
• Sport A has a much smaller range of weights (from 55 kg to 85 kg, effectively 90 kg if considering closure) compared to Sport B, which has players weighing up to 105 kg (effectively 110 kg if considering closure). This means Sport A players have weights that are closer together.

(iii) Based on the frequency polygons:

• Sport A shows a narrower range of weights, with most players having lighter weights. This pattern suggests a sport that does not require heavy players or where a specific lighter build is advantageous. Cricket or badminton could be suitable sports for distribution A. These games often favor agility and specific skills over sheer mass.
• Sport B, on the other hand, shows a broader range and more players in higher weight categories. This distribution is typical for sports that might involve more physical contact, strength, or require heavier builds. Kabbadi, wrestling, or weightlifting are good examples of games where players' weights tend to be higher.

In simple words: First, we counted all the players for Sport A by adding up the numbers on the graph. Then, we compared the two graphs. Sport A players are generally lighter and have a smaller range of weights, while Sport B players are heavier and have a wider range of weights. This suggests that Sport A could be something like cricket or badminton, where being light helps, and Sport B could be wrestling or Kabbadi, which often involves heavier players. Frequency polygons give a quick visual comparison of two datasets.

🎯 Exam Tip: When interpreting frequency polygons, look at the peak (mode) to understand typical values, and the spread (range) to understand variation. For comparative questions, explicitly state at least two distinct differences or similarities you observe between the distributions. Connect observed patterns to real-world scenarios to explain your choices.