Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 22 Statistics

Exercise 22.1

Question 1. Define primary data and secondary data.
Answer: Primary data refers to information gathered directly by the researcher themselves using direct observation according to a clear, pre-planned goal. On the other hand, secondary data is information that has already been gathered by a different person or group in the past, which the current researcher then utilizes for a particular task.
In simple words: Primary data is what you collect yourself from scratch. Secondary data is information that someone else already collected and you are just reusing.
Exam Tip: Clearly highlight who collects the data to gain full marks - "investigator themselves" for primary, and "another person/agency" for secondary.

Question 2. Define the following terms:
(i) Variate
(ii) Class Mark
(iii) True Class limits
(iv) Frequency
Answer:
(i) Variate: Any specific value that a variable can take is referred to as a variate.
(ii) Class Mark: This represents the midpoint of a class interval, calculated as half of the sum of its class limits.
(iii) True Class limits: In grouped data, the minimum and maximum values within an interval define its class limits. For instance, in an exclusive frequency distribution with intervals like 10-20, 20-30, and 30-40, for the group 20-30, 20 serves as the lower boundary and 30 as the upper boundary. These actual, precise boundaries are known as the true class limits.
(iv) Frequency: This refers to the count of how many times a particular value or observation occurs within a dataset.
In simple words: A variate is a single value, the class mark is the middle of a class group, class limits show the start and end of a group, and frequency is how many times a value appears.

Exam Tip: In definitions, always include the mathematical formula for class mark - half the sum of class boundaries - and a brief example for class limits to secure maximum score.

Question 3. Based on the following frequency distribution table, answer the questions that follow:
(i) What is the highest score?
(ii) What is the lowest score?
(iii) Find the range of the scores.
(iv) If the passing mark is 20, find the number of failed students, given that 2 students scored exactly 20.
(v) Find the number of students who scored 40 or more marks, given that 2 students scored exactly 40.

Answer:
(i) The maximum score recorded is 49.
(ii) The minimum score recorded is 12.
(iii) Range = 49 - 12 = 37
(iv) Since 20 is the passing mark, all students in the intervals 11-15 and 16-20 failed, except those who scored exactly 20.
Thus, the number of students who failed = 5 + 7 - 2 = 10
(v) The number of students scoring more than 40 is the sum of the frequencies of the classes 41-45 and 46-50:
3 + 5 = 8
Given that 2 students scored exactly 40, the number of students scoring 40 or above = 8 + 2 = 10.
In simple words: Find the highest and lowest numbers, and subtract them to get the range. To count passing or failing students, carefully look at which groups fall below or above the passing marks.
Exam Tip: When finding the range, write down the formula (Maximum value - Minimum value) first before putting in the numbers to ensure step-marks are awarded.

Question 4. In the given figure, ABCD is a parallelogram and APD is an equilateral triangle of side 8 cm. Find the area of the parallelogram ABCD.
Answer: Since the side of the equilateral triangle \( \Delta\text{APD} \) is \( s = 8\text{ cm} \):
\( \text{ar}(\Delta\text{APD}) = \frac{\sqrt{3}s^2}{4} \)
\( \text{ar}(\Delta\text{APD}) = \frac{\sqrt{3} \times 8^2}{4} = \frac{\sqrt{3} \times 64}{4} = 16\sqrt{3}\text{ cm}^2 \)
We know that when a triangle and a parallelogram lie on the same base and are situated between the same parallel lines, the triangle's area is exactly half of the parallelogram's area.
Therefore, \( \text{ar}(\Delta\text{APD}) = \frac{1}{2} \times \text{ar}(\text{parallelogram ABCD}) \)

\( \implies \text{ar}(\text{parallelogram ABCD}) = 2 \times \text{ar}(\Delta\text{APD}) \)

\( \implies \text{ar}(\text{parallelogram ABCD}) = 2 \times 16\sqrt{3}\text{ cm}^2 \)

\( \implies \text{ar}(\text{parallelogram ABCD}) = 32\sqrt{3}\text{ cm}^2 \)
In simple words: Find the area of the equilateral triangle first using its formula. Since the triangle and the parallelogram share the same base and parallels, the parallelogram's area is simply twice the triangle's area.

Exam Tip: State the area theorem explicitly as a reason: "Area of a triangle is half that of a parallelogram on the same base and between the same parallels." To earn full marks, ensure that your final calculated unit is clearly written as cm².

Question 5. For the class intervals 55-59, 60-64, 65-69, 70-74, 75-79, 80-84, 85-89, 90-94, and 95-99, find the class boundaries and the class mark for each class.
Answer: For the initial class interval 55-59:
The true lower boundary is calculated as: \( 55 - 0.5 = 54.5 \)
The true upper boundary is calculated as: \( 59 + 0.5 = 59.5 \)
Thus, the actual boundaries for this class are 54.5 and 59.5.
The class mark is determined by taking the average of these boundaries: \( \frac{1}{2}(54.5 + 59.5) = 57 \).
Applying this same method to all the remaining intervals, we obtain the table below:

In simple words: Since the class intervals have gaps (like 55-59 and 60-64), find the true boundaries by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits. The class mark is the middle of each group.
Exam Tip: For discontinuous (inclusive) classes, always convert them to continuous (exclusive) class boundaries before doing further calculations like finding the class mark.

Question 8. A football team (AFC) scored the following number of goals in various matches: [data represented in the table below]. Construct a discrete frequency distribution table and answer the following questions:
(i) Find the range of goals scored.
(ii) How many times did AFC score 3 or more goals?
(iii) Which number of goals has the highest frequency?
Answer: The discrete frequency distribution table is as below:

(i) The highest number of goals made is 6, while the lowest is 0.
So, the range of goals = 6 - 0 = 6
(ii) The total matches where the team scored 3 or more goals is found by summing the frequencies for 3, 4, 5, and 6 goals:
6 + 8 + 4 + 2 = 20 times.
(iii) The goal count that occurred most frequently (with a peak frequency of 8) is 4.
In simple words: Use tally marks to count how many times each goal number occurs to make a table. Use this table to easily answer questions about the goals.
Exam Tip: Verify that the sum of the frequencies in your table matches the total number of matches given in the question to prevent counting mistakes.

Question 9. For the given class intervals: 5-12, 13-20, 21-28, 29-36, 37-44, 45-52, and 53-60:
(i) Find the true class limits for the class 21-28.
(ii) Determine the class size and class mark for the class 45-52.
(iii) Write the true class limits for all the classes in tabular form.
Answer:
(i) For the interval 21-28: the lower boundary is 21 and the upper boundary is 28.
The true lower boundary is \( 21 - 0.5 = 20.5 \).
The true upper boundary is \( 28 + 0.5 = 28.5 \).
Hence, the boundaries are 20.5 and 28.5.
(ii) Regarding the class 45-52:
The true boundaries are \( 45 - 0.5 = 44.5 \) and \( 52 + 0.5 = 52.5 \).
The size of this class is \( 52.5 - 44.5 = 8 \).
The class midpoint (class mark) is \( \frac{1}{2}(44.5 + 52.5) = 48.5 \).
(iii) In an exclusive frequency distribution, the actual boundaries are used. The table of true class limits is given below:

In simple words: To make true boundaries, subtract 0.5 from the bottom limit and add 0.5 to the top limit of each group. The class size is the width of a group, and the class mark is its center.
Exam Tip: Keep in mind that for exclusive class intervals, the true class limits are identical to the given class boundaries.

Question 10. Given the class marks: 15, 25, 35, 45, 55, 65, and 75, find the corresponding class intervals.
Answer: The class marks are spaced evenly.
We can determine the class size by finding the difference between any two successive class marks:
Class size = 25 - 15 = 10
For the first class with a mark of 15:
The lower boundary is \( 15 - \frac{10}{2} = 10 \)
The upper boundary is \( 15 + \frac{10}{2} = 20 \)
This gives us the first interval as 10-20.
Using the same procedure for the remaining class marks, we get the intervals listed in the table below:

In simple words: Find the difference between consecutive class marks to get the group size. Subtract half of this size from each mark for the lower limit, and add half of it for the upper limit.
Exam Tip: Remember the formula: Lower Limit = Class Mark - (Class Size / 2), and Upper Limit = Class Mark + (Class Size / 2).

Question 11. Find the class intervals for the given class marks: 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, and 77.
Answer: The given class marks are distributed at equal intervals.
The class width is computed by subtracting any class mark from the one following it:
Class width = 32 - 27 = 5
For the first class mark of 27:
The lower boundary is \( 27 - \frac{5}{2} = 24.5 \)
The upper boundary is \( 27 + \frac{5}{2} = 29.5 \)
Therefore, the first class interval spans from 24.5 to 29.5.
By calculating the boundaries for the rest of the class marks in the same manner, we complete the table:

In simple words: Find the class size by subtracting two consecutive class marks. Use half of this size to find the lower and upper limits of each interval.
Exam Tip: Always double check that the average of your calculated lower and upper limits equals the given class mark.

Question 12. From a given set of data containing 60 observations where the minimum value is 14 and the maximum value is 55, construct a frequency distribution table with a class size of 7.
Answer: The minimum value in the data is 14 and the maximum is 55.
So, the range = 55 - 14 = 41
Given a class width of 7:
The number of class groups needed = \( \frac{41}{7} \approx 6 \)
We can construct the frequency table using these classes as follows:

In simple words: Find the total range of the data and divide it by the group size to see how many classes you need. Then, group the data into intervals and count the frequency of each using tallies.
Exam Tip: When using tally marks, make sure to bundle them in groups of five (four vertical lines crossed by a diagonal line) for clear presentation.

Exercise 22.2

Question 1. Given the marks obtained by 60 students in a test, construct a cumulative frequency distribution table:
Answer: Using the frequencies provided, the cumulative frequency distribution is calculated by progressively adding up the number of students in each successive mark range:

In simple words: To find cumulative frequency, keep adding each group's frequency to the sum of all previous frequencies as you go down the table.
Exam Tip: The final cumulative frequency must always equal the total number of observations (the sum of all individual frequencies).

Question 2. Construct a cumulative frequency table for the given distribution of ages of patients admitted to a hospital:
Answer: The cumulative frequency is obtained by adding the number of patients in each age group to the running total from preceding groups:

In simple words: Add the frequencies step-by-step from top to bottom to make a running total for the ages of the patients.
Exam Tip: Present the columns clearly and label them correctly as "Class Interval", "Frequency", and "Cumulative Frequency" to avoid losing presentation marks.

Question 3. Construct a frequency and cumulative frequency table with tally marks for the following data on class intervals:
Answer: By counting the occurrences in each class range using tally marks, we determine the frequency and cumulative frequency for the entire dataset:

In simple words: Group the values into classes, use tally marks to find the frequency of each class, and then add them up step-by-step to get the running total.
Exam Tip: Make sure your tally mark column is placed between the class intervals and the frequency column as per standard textbook layouts.

Question 4. From the given cumulative frequency distribution, find the individual frequency of each class:
Answer: To find the individual frequencies of each class from the cumulative frequencies, we subtract the cumulative frequency of the preceding class from that of the current class:

In simple words: To find the original frequencies from the running total, subtract each previous cumulative frequency from the current one.
Exam Tip: The frequency of the first class is always equal to its cumulative frequency. For subsequent classes, use: Frequency = current c.f. - previous c.f.

Exercise 22.3

Question 1. Find the mean of each of the following datasets:
(i) 5, 7, 8, 4, 6
(ii) 3, 0, 5, 2, 6, 2
Answer: The mean is computed by dividing the sum of all observations by their total number:
Mean = \( \frac{\sum x}{N} \)

(i) For the numbers 5, 7, 8, 4, 6:
Mean = \( \frac{5 + 7 + 8 + 4 + 6}{5} = \frac{30}{5} = 6 \)

(ii) For the numbers 3, 0, 5, 2, 6, 2:
Mean = \( \frac{3 + 0 + 5 + 2 + 6 + 2}{6} = \frac{18}{6} = 3 \)
In simple words: Add all the numbers in the list together, and then divide that total by how many numbers there are in the list.

Exam Tip: Write the formula \( \text{Mean} = \frac{\sum x}{N} \) clearly. Even if you make a simple calculation error, you will get step-marks for the correct formula.

Question 2. Calculate the mean for the following frequency distribution:
Answer: To calculate the mean of a frequency distribution, we first multiply each value \( x \) by its corresponding frequency \( f \) to get \( fx \), then find the sum of these products and divide by the total frequency:

Mean = \( \frac{\sum fx}{\sum f} = \frac{98}{14} = 7 \)
In simple words: Multiply each number by how many times it appears, add those results together, and then divide by the total number of items.
Exam Tip: Include a dedicated column for \( fx \) in your table and show the sum of \( f \) and \( fx \) clearly at the bottom.

Question 3. The weights of 7 students are 20 kg, 52 kg, 56 kg, 72 kg, 64 kg, 13 kg, and 80 kg. Find their mean weight.
Answer: The arithmetic mean is calculated using the formula:
Mean = \( \frac{\sum x}{N} \)

Substituting the given weights:
Mean = \( \frac{20 + 52 + 56 + 72 + 64 + 13 + 80}{7} = \frac{351}{7} = 51\text{ kg} \)
In simple words: Add all 7 weights together to get the total weight, then divide by 7 to get the average weight.

Exam Tip: Do not forget to write the unit (kg) in your final answer, as omission of units often results in a deduction of half a mark.

Question 4. The marks obtained by 10 students in a test are 17, 15, 16, 7, 10, 14, 12, 19, 16, and 12. Find the mean marks.
Answer: To find the average of the marks obtained:
Mean = \( \frac{\sum x}{N} \)

Adding the marks and dividing by the total number of students:
Average marks = \( \frac{17 + 15 + 16 + 7 + 10 + 14 + 12 + 19 + 16 + 12}{10} = \frac{138}{10} = 13.8 \)
In simple words: Find the sum of all 10 students' marks and divide by 10 to get the average marks.

Exam Tip: When dividing by 10, simply move the decimal point of the sum one place to the left for a quick and error-free calculation.

Question 5. The scores of 11 matches are 10, 9, 31, 45, 0, 4, 8, 15, 12, 0, and 6. Calculate the average score.
Answer: The average score is calculated by dividing the total score by the number of matches:
Mean = \( \frac{\sum x}{N} \)

Average score = \( \frac{10 + 9 + 31 + 45 + 0 + 4 + 8 + 15 + 12 + 0 + 6}{11} = \frac{140}{11} \approx 12.7 \)
In simple words: To find the average score, sum the scores from all 11 matches - including the matches where the score was 0 - and divide by 11.

Exam Tip: Remember to count zero-scores as valid observations in the denominator \( N \); otherwise, your calculated mean will be incorrect.

Question 7. The mean of the observations 2, 5, 3, 8, 0, 9, x, 6, 1, and 8 is 5. Find the value of x.
Answer: The formula for the mean of the observations is:
Mean = \( \frac{\sum x}{N} \)

Adding the given values together with the unknown \( x \):
\( \text{Mean} = \frac{2 + 5 + 3 + 8 + 0 + 9 + x + 6 + 1 + 8}{10} = \frac{42 + x}{10} \)

Given that the mean is equal to 5, we can set up the equation:
\( 5 = \frac{42 + x}{10} \)

\( \implies 50 = 42 + x \)

\( \implies x = 50 - 42 = 8 \)
In simple words: Write the formula for the average with the unknown letter \( x \). Use the given mean to solve the equation and find \( x \).

Exam Tip: Cross-multiply the denominator of the mean formula to the other side first to simplify the algebraic equation.

Question 8. The heights (in cm) of 7 students are 148, 162, 160, 154, 170, 162, and 152. If the mean height of 8 students is 158 cm, find the height of the eighth student.
Answer: Let the height of the eighth student be represented by \( x \).
The mean height of the 8 students is given by:
Mean = \( \frac{\sum x}{N} \)

Substituting the known values and the mean:
\( \text{Mean} = \frac{148 + 162 + 160 + 154 + 170 + 162 + x + 152}{8} = \frac{1108 + x}{8} \)

Since the mean height is 158 cm:
\( 158 = \frac{1108 + x}{8} \)

\( \implies 1264 = 1108 + x \)

\( \implies x = 1264 - 1108 \)

\( \implies x = 156 \)
Thus, the height of the eighth student is 156 cm.
In simple words: Use the given average height of 8 students to find the total height, and subtract the sum of the first 7 students' heights to find the height of the 8th student.

Exam Tip: Set up the equation using a variable like \( x \) for the unknown height, and clearly mention "Let the height of the 8th student be x cm."

Question 9. The mean of the observations 7, 16, 9, 15, 16, a, 12, 8, b, and 11 is 12. Express a in terms of b.
Answer: The mean of the given 10 observations is calculated as:
Mean = \( \frac{\sum x}{N} \)

Summing the values:
\( \text{Mean} = \frac{7 + 16 + 9 + 15 + 16 + a + 12 + 8 + b + 11}{10} = \frac{94 + a + b}{10} \)

We are given that this mean is equal to 12:
\( 12 = \frac{94 + a + b}{10} \)

\( \implies 120 = 94 + a + b \)

\( \implies a + b = 120 - 94 \)

\( \implies a + b = 26 \)

\( \implies a = 26 - b \)
In simple words: Find the sum of the numbers and write the mean equation with \( a \) and \( b \). Rearrange it to show what \( a \) equals.

Exam Tip: Be careful when simplifying terms; keep the variables on one side and the constant values on the other side.

Question 10. The marks scored by 16 students in a test of 25 marks are: 25, 8, 14, 20, 16, 22, 10, 15, 8, 7, 24, 18, 19, 6, 11, and 17.
(i) Find the mean marks.
(ii) If the marks are scaled out of 50 by doubling them, find the new mean marks.
Answer:
(i) The total score of all 16 students combined is:
\( 25 + 8 + 14 + 20 + 16 + 22 + 10 + 15 + 8 + 7 + 24 + 18 + 19 + 6 + 11 + 17 = 240 \)
The mean marks are calculated by:
Mean = \( \frac{\text{Total Marks}}{\text{Number of Students}} = \frac{240}{16} = 15 \)
(ii) If the maximum marks are adjusted to 50, each student's score is doubled.
Thus, the sum of the new marks will be twice the original sum:
\( 2 \times 240 = 480 \)
The revised mean marks will be:
Mean = \( \frac{480}{16} = 30 \)
In simple words: Add all the marks together and divide by 16 to find the mean. If the marks are doubled, the new mean is simply twice the original mean.

Exam Tip: State the property that multiplying each observation by a constant multiplies the mean by the same constant to save calculation time in part (ii).

Question 11. The marks of 8 students are 14, 16, 18, 14, 16, 14, 12, and 16.
(i) Find the mean marks.
(ii) If each student is awarded 2 extra grace marks, find the revised mean marks.
Answer:
(i) The sum of the marks obtained by the 8 students is:
\( 14 + 16 + 18 + 14 + 16 + 14 + 12 + 16 = 120 \)
The mean score is calculated as:
Mean = \( \frac{120}{8} = 15 \)
(ii) When every student receives 2 additional marks, the total score increases by:
\( 2 \times 8 = 16 \text{ marks} \)
The new sum of the marks becomes:
\( 120 + 16 = 136 \)
Hence, the updated mean marks are:
Revised Mean = \( \frac{136}{8} = 17 \)
In simple words: Find the average of the 8 scores. Adding 2 marks to every student increases the overall average marks by exactly 2.

Exam Tip: Note that adding a constant value to each observation increases the mean of the distribution by that same constant.

Question 16. Find the median of the following sets of data:
(i) 15, 8, 14, 20, 13, 12, 16
(ii) 25, 11, 15, 10, 17, 6, 5, 12
Answer:
(i) Let's sort the first dataset in ascending order:
8, 12, 13, 14, 15, 16, 20
Since the number of terms is odd (\( N = 7 \)):
Median = \( \left(\frac{N+1}{2}\right)^{\text{th}} \text{ term} = \left(\frac{7+1}{2}\right) = 4^{\text{th}} \text{ term} \)
The 4th term in the ordered list is 14.
Thus, the median is 14.
(ii) Sorting the second dataset in ascending order:
5, 6, 10, 11, 12, 15, 17, 25
Since the number of terms is even (\( N = 8 \)), we take the average of the two middle terms:
The \( \left(\frac{N}{2}\right)^{\text{th}} \text{ term} \) is the 4th term, which is 11.
The \( \left(\frac{N}{2} + 1\right)^{\text{th}} \text{ term} \) is the 5th term, which is 12.
Median = \( \frac{4^{\text{th}} \text{ term} + 5^{\text{th}} \text{ term}}{2} = \frac{11 + 12}{2} = \frac{23}{2} = 11.5 \)
In simple words: Put the numbers in order from smallest to biggest. If there is an odd number of items, the middle one is the median. If even, average the two middle ones.

Exam Tip: Arranging the data in ascending or descending order is a mandatory first step. Skipping this step is the most common reason students lose all marks on median questions.

Question 17. Find the median of the following numbers: 1, 2, 4, 3, 5, 4, 9, 1, 2, 8, 4, 9, 10, and 6.
Answer: First, arrange the given set of observations in ascending order:
1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10
Since the total number of terms is even (\( N = 14 \)):
The \( \left(\frac{N}{2}\right)^{\text{th}} \text{ term} \) is the 7th term, which is 4.
The \( \left(\frac{N}{2} + 1\right)^{\text{th}} \text{ term} \) is the 8th term, which is 4.
Therefore, the median is the average of these two middle values:
Median = \( \frac{4 + 4}{2} = \frac{8}{2} = 4 \)
In simple words: Sort the 14 numbers in order. Since 14 is even, find the two numbers in the middle (the 7th and 8th numbers) and find their average.

Exam Tip: For even \( N \), clearly show the calculation of both middle terms, \( \frac{N}{2} \) and \( \frac{N}{2}+1 \), before taking their average.

Question 18. The observations 3, 8, 10, x, 14, 16, 18, and 20 are arranged in ascending order. If the median of the data is 13, find the value of x.
Answer: The given data is already arranged in ascending order:
3, 8, 10, x, 14, 16, 18, 20
The number of observations is even (\( N = 8 \)).
The two middle terms are:
The 4th term, which is \( x \).
The 5th term, which is 14.
Since the median of the dataset is given as 13:
\( \text{Median} = \frac{x + 14}{2} = 13 \)

\( \implies x + 14 = 13 \times 2 \)

\( \implies x + 14 = 26 \)

\( \implies x = 26 - 14 = 12 \)
In simple words: The numbers are already in order. Find the two middle numbers, which are \( x \) and 14. Their average must equal the given median of 13. Solve for \( x \).

Exam Tip: Pay attention to the instruction that the data is already arranged in ascending order, so you can directly identify the middle terms.

Question 20. A student's scores in 11 tests are 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, and 16. Find:
(i) The mean marks.
(ii) The median marks.
Answer:
(i) The sum of all marks across the 11 tests is:
\( 15 + 17 + 16 + 7 + 10 + 12 + 14 + 16 + 19 + 12 + 16 = 154 \)
The mean score is:
Mean = \( \frac{154}{11} = 14 \)
(ii) To find the median, we first arrange the scores in ascending order:
7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19
The total number of observations is odd (\( N = 11 \)):
Median = \( \left(\frac{N+1}{2}\right)^{\text{th}} \text{ term} = 6^{\text{th}} \text{ term} \)
The 6th observation in the sorted list is 15.
Thus, the median score is 15.
In simple words: Find the mean by adding all marks and dividing by 11. Find the median by ordering the marks and picking the middle one.

Exam Tip: For the median of an odd number of items, use the formula \( \left(\frac{N+1}{2}\right)^{\text{th}} \text{ term} \) to locate the exact position of the median.