OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Exercise 28 (C)

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(c)

 

Question 1. Find the mode of the following data :
(i) 3, 4, 7, 11, 4, 3, 4, 5, 6, 4, 1, 2, 4, 4
(ii) Size of shoes : 4, 4.5, 5, 4.5, 5.5, 5, 6, 4.5, 4, 4.5
(iii) Wages (Rs) : 100, 120, 100, 120, 130, 120, 120, 130, 120, 100
(iv) Runs in an innings : 18, 32, 0, 40, 60, 69, 33, 69, 35, 11, 20
Answer:
(i) To find the mode, we first arrange the given data in ascending order: 1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 11. Here, the number 4 appears most often (7 times). Thus, the mode is 4.
(ii) Arranging the shoe sizes in ascending order: 4, 4, 4.5, 4.5, 4.5, 4.5, 5, 5, 5.5, 6. The size 4.5 is repeated the most (4 times). So, the mode is 4.5.
(iii) Arranging the wages in ascending order: 100, 100, 100, 120, 120, 120, 120, 120, 130, 130. The wage Rs 120 appears most frequently (5 times). Therefore, the mode is 120.
(iv) Arranging the runs in ascending order: 0, 11, 18, 20, 32, 33, 35, 40, 60, 69, 69. The number 69 appears twice, which is more than any other number. So, the mode is 69.
In simple words: The mode is the number that shows up most frequently in a list of numbers. To find it easily, you can arrange the numbers from smallest to largest and then count which number appears the most.

🎯 Exam Tip: Always list the data in ascending or descending order first to clearly see the frequency of each value before identifying the mode.

 

Question 2. Find the mode from the following data in question 2 to 7.

Answer: To find the mode from the given data, we look for the mark that has the highest frequency (number of students). In this table, the maximum frequency is 20, which corresponds to the mark of 25. Thus, the required mode is 25.
In simple words: The mode is the value that appears most often. Here, the mark of 25 was achieved by the highest number of students (20 students), making 25 the mode.

🎯 Exam Tip: For simple frequency distributions, the mode is the observation with the highest frequency. Make sure to identify the *observation*, not the frequency itself, as the mode.

 

Question 3.

Answer: We need to calculate the mean, median, and mode from the given data. First, we create a table to calculate \(fx\) (size multiplied by frequency) and cumulative frequency (c.f.).The calculation for the table in the source seems to have a few errors, for example, for x=18, f=8, fx=144, c.f=32. Using the initial table, this is corrected above.
Mean (\( \overline{x} \)) is calculated as \( \frac{\Sigma fx}{\Sigma f} = \frac{616}{39} \approx 15.79 \).
For Median (Md), we first find \( \frac{N+1}{2} = \frac{39+1}{2} = \frac{40}{2} = 20 \). Looking at the cumulative frequency, the 20th item falls in the size category of 16 (since 20 is greater than 16 but less than 24). So, the Median (Md) is 16.
The Mode is the value with the highest frequency. From the original data, the size 16 has the highest frequency of 8. Thus, the Mode is 16.
Using the empirical formula: Mode = 3 Median - 2 Mean
\( \implies \) Mode = \( 3 \times 16 - 2 \times 15.79 \)
\( \implies \) Mode = \( 48 - 31.58 \)
\( \implies \) Mode = \( 16.42 \)
In this case, the direct mode from frequency and empirical mode are close.
In simple words: The mean is the average value. The median is the middle value when all numbers are lined up. The mode is the number that shows up most often. These help us understand the central point of the data.

🎯 Exam Tip: When calculating mean, median, and mode for discrete data, double-check your \(fx\) and cumulative frequency columns carefully, as small errors can lead to incorrect final answers. For mode, remember to also identify the observation with the highest frequency, not just the frequency itself.

 

Question 4.

Answer: We need to find the mode from the given grouped frequency data.
First, identify the modal class, which is the class interval with the highest frequency. Here, the maximum frequency is 45, which belongs to the class interval 10-20. So, the modal class is 10-20.
Next, identify the values for the mode formula:
Lower limit of modal class (\(I\)) = 10
Frequency of modal class (\(f_m\)) = 45
Frequency of class preceding the modal class (\(f_{m-1}\)) = 10
Frequency of class succeeding the modal class (\(f_{m+1}\)) = 12
Class size (\(i\)) = 20 - 10 = 10
Now, apply the mode formula for grouped data:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies Mode = 10 + \frac{45 - 10}{2(45) - 10 - 12} \times 10 \)
\( \implies Mode = 10 + \frac{35}{90 - 10 - 12} \times 10 \)
\( \implies Mode = 10 + \frac{35}{68} \times 10 \)
\( \implies Mode = 10 + \frac{350}{68} \)
\( \implies Mode = 10 + 5.147058... \)
\( \implies Mode \approx 15.147 \)
In simple words: To find the mode for data grouped into classes, we first find the class with the most items. Then we use a special formula that considers the frequencies of that class and the classes next to it, along with the class size, to get a more exact mode value.

🎯 Exam Tip: Remember to clearly identify \(f_m\), \(f_{m-1}\), \(f_{m+1}\), \(I\), and \(i\) correctly from the modal class and its adjacent classes. A common mistake is using the wrong frequencies in the formula.

 

Question 5.

Answer: We need to find the mode from the given grouped frequency data.
First, identify the modal class, which is the class interval with the highest frequency. Here, the maximum frequency is 60, which belongs to the class interval 30-40. So, the modal class is 30-40.
Next, identify the values for the mode formula:
Lower limit of modal class (\(I\)) = 30
Frequency of modal class (\(f_m\)) = 60
Frequency of class preceding the modal class (\(f_{m-1}\)) = 26
Frequency of class succeeding the modal class (\(f_{m+1}\)) = 38
Class size (\(i\)) = 40 - 30 = 10
Now, apply the mode formula for grouped data:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies Mode = 30 + \frac{60 - 26}{2(60) - 26 - 38} \times 10 \)
\( \implies Mode = 30 + \frac{34}{120 - 26 - 38} \times 10 \)
\( \implies Mode = 30 + \frac{34}{56} \times 10 \)
\( \implies Mode = 30 + \frac{340}{56} \)
\( \implies Mode = 30 + 6.071428... \)
\( \implies Mode \approx 36.071 \)
In simple words: Just like before, we find the class with the most students (modal class). Then, we use the formula for grouped data to calculate the exact mode, which gives us the most common mark students achieved.

🎯 Exam Tip: Double-check your arithmetic, especially in the denominator of the mode formula. Small calculation errors are easy to make and can lead to incorrect results.

 

Question 6.

Answer: The given class intervals are discontinuous (e.g., 1-5, then 6-10). To find the mode, we must first make them continuous by applying an adjustment factor.
The adjustment factor is calculated as \( \frac{\text{lower limit of second class} - \text{upper limit of first class}}{2} \).
Adjustment factor = \( \frac{6 - 5}{2} = \frac{1}{2} = 0.5 \).
We subtract 0.5 from each lower limit and add 0.5 to each upper limit to get continuous class intervals.Now, identify the modal class from the continuous data. The maximum frequency is 32, which belongs to the class interval 15.5-20.5. So, the modal class is 15.5-20.5.
Identify the values for the mode formula:
Lower limit of modal class (\(I\)) = 15.5
Frequency of modal class (\(f_m\)) = 32
Frequency of class preceding the modal class (\(f_{m-1}\)) = 16
Frequency of class succeeding the modal class (\(f_{m+1}\)) = 24
Class size (\(i\)) = 20.5 - 15.5 = 5
Now, apply the mode formula for grouped data:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies Mode = 15.5 + \frac{32 - 16}{2(32) - 16 - 24} \times 5 \)
\( \implies Mode = 15.5 + \frac{16}{64 - 16 - 24} \times 5 \)
\( \implies Mode = 15.5 + \frac{16}{24} \times 5 \)
\( \implies Mode = 15.5 + \frac{80}{24} \)
\( \implies Mode = 15.5 + 3.3333... \)
\( \implies Mode \approx 18.83 \)
In simple words: When the class ranges don't connect perfectly (like 1-5 then 6-10), we first adjust them so they flow smoothly (like 0.5-5.5 then 5.5-10.5). After that, we find the class with the most items and use the usual mode formula for grouped data to get the answer.

🎯 Exam Tip: Always remember to convert discontinuous class intervals into continuous ones before applying the mode formula for grouped data. Forgetting this step is a common error.

 

Question 7.

Answer: We are given mid-values and frequencies. To find the mode, we first need to determine the class intervals.
The difference between consecutive mid-values is \( 64 - 60 = 4 \). This means the class size (\(i\)) is 4.
The lower limit of a class is Mid-value \( - \frac{i}{2} \), and the upper limit is Mid-value \( + \frac{i}{2} \).
So, for the first mid-value 60, the class interval is \( 60 - \frac{4}{2} \) to \( 60 + \frac{4}{2} \), which is \( 60 - 2 \) to \( 60 + 2 \), or 58-62.
The table with class intervals and frequencies is:Now, identify the modal class. The maximum frequency is 25, which belongs to the class interval 58-62. So, the modal class is 58-62.
Identify the values for the mode formula:
Lower limit of modal class (\(I\)) = 58
Frequency of modal class (\(f_m\)) = 25
Frequency of class preceding the modal class (\(f_{m-1}\)) = 0 (since there is no class before 58-62)
Frequency of class succeeding the modal class (\(f_{m+1}\)) = 8
Class size (\(i\)) = 4
Now, apply the mode formula for grouped data:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies Mode = 58 + \frac{25 - 0}{2(25) - 0 - 8} \times 4 \)
\( \implies Mode = 58 + \frac{25}{50 - 8} \times 4 \)
\( \implies Mode = 58 + \frac{25}{42} \times 4 \)
\( \implies Mode = 58 + \frac{100}{42} \)
\( \implies Mode = 58 + 2.38095... \)
\( \implies Mode \approx 60.38 \)
In simple words: When you have mid-values, first figure out the actual class ranges by looking at the gap between mid-points. Then, find the class with the highest frequency and use the mode formula for grouped data. In this case, the mode is found to be about 60.38.

🎯 Exam Tip: When given mid-values, correctly determine the class intervals and class size (\(i\)) before proceeding with the mode calculation. This is a crucial first step.

 

Question 8.
(i) The mode of the following frequency distribution is 48.6 Find the unknown frequency.
(ii) If the frequency of the class 70-85 is 13 instead of 3, then what difference will it make?
Answer:
(i) We are given the mode (48.6) and a frequency distribution with a missing frequency.
The table is:

Let the missing frequency be \(f\).
Since the mode is 48.6, the modal class must be 40-55 (because 48.6 falls within this range).
Now, identify the values for the mode formula:
Lower limit of modal class (\(I\)) = 40
Frequency of modal class (\(f_m\)) = 44
Frequency of class preceding the modal class (\(f_{m-1}\)) = 20
Frequency of class succeeding the modal class (\(f_{m+1}\)) = \(f\) (the unknown frequency)
Class size (\(i\)) = 55 - 40 = 15
Apply the mode formula and substitute the given mode:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies 48.6 = 40 + \frac{44 - 20}{2(44) - 20 - f} \times 15 \)
\( \implies 48.6 - 40 = \frac{24}{88 - 20 - f} \times 15 \)
\( \implies 8.6 = \frac{24}{68 - f} \times 15 \)
\( \implies 8.6 = \frac{360}{68 - f} \)
\( \implies 8.6 (68 - f) = 360 \)
\( \implies 584.8 - 8.6f = 360 \)
\( \implies 584.8 - 360 = 8.6f \)
\( \implies 224.8 = 8.6f \)
\( \implies f = \frac{224.8}{8.6} \)
\( \implies f \approx 26.139... \)
Since frequency must be a whole number, we can round it to 26.
(ii) If the frequency of the class 70-85 changes from 3 to 13, this value corresponds to \(f_{m+2}\) or a class beyond \(f_{m+1}\) relative to the modal class (40-55). The mode calculation only depends on \(f_m\), \(f_{m-1}\), and \(f_{m+1}\). Since the change is in a class far from the modal class, it will not affect the calculated mode of 48.6. The modal class (40-55) and its immediate surrounding frequencies remain unchanged for the calculation.
In simple words: Part (i) asks us to find a missing number (frequency) when we already know the mode. We use the mode formula, put in all the numbers we know, and solve for the missing one. Part (ii) asks if changing a number far away from the 'most common' group would change the mode. The answer is no, because the mode only cares about the group that has the most items and the groups right next to it.

🎯 Exam Tip: When finding a missing frequency using the mode formula, first correctly identify the modal class based on the given mode value. Also, remember that changes in frequencies far from the modal class usually do not affect the mode.

 

Question 9. Find the mean, median and mode of the following :
(i) The data 17, 32, 35, 33, 15, 21, 41, 32, 11, 18, 20, 22, 11, 15, 35, 23, 38, 12.
Answer:
(i) We need to find the mean, median, and mode for the given individual data points: 17, 32, 35, 33, 15, 21, 41, 32, 11, 18, 20, 22, 11, 15, 35, 23, 38, 12.
There are \(n = 18\) observations.
**1. Mean:**
Sum of observations \( \Sigma x = 17+32+35+33+15+21+41+32+11+18+20+22+11+15+35+23+38+12 \)
\( \Sigma x = 431 \)
Mean \( \overline{x} = \frac{\Sigma x}{n} = \frac{431}{18} \approx 23.94 \)
**2. Median:**
First, arrange the data in ascending order:
11, 11, 12, 15, 15, 17, 18, 20, 21, 22, 23, 32, 32, 33, 35, 35, 38, 41
Since \(n = 18\) (an even number), the median is the average of the \( \frac{n}{2} \)th and \( (\frac{n}{2} + 1) \)th items.
Median = \( \frac{\text{size of 9th item} + \text{size of 10th item}}{2} \)
The 9th item is 21 and the 10th item is 22.
Median = \( \frac{21 + 22}{2} = \frac{43}{2} = 21.5 \)
**3. Mode:**
Count the frequency of each observation:
11 (appears 2 times)
12 (appears 1 time)
15 (appears 2 times)
17 (appears 1 time)
18 (appears 1 time)
20 (appears 1 time)
21 (appears 1 time)
22 (appears 1 time)
23 (appears 1 time)
32 (appears 2 times)
33 (appears 1 time)
35 (appears 2 times)
38 (appears 1 time)
41 (appears 1 time)
The observations 11, 15, 32, and 35 all repeat twice, which is the maximum frequency. Thus, this data set is multi-modal (it has multiple modes).
The modes are 11, 15, 32, and 35.
In simple words: For a list of numbers, the mean is their average sum divided by how many numbers there are. The median is the middle number after sorting them. If there are two middle numbers, we average them. The mode is the number that shows up most often. Sometimes, there can be more than one mode if several numbers appear the same highest number of times.

🎯 Exam Tip: For ungrouped data, always sort the data first for finding the median and mode. When calculating the median for an even number of observations, remember to average the two middle values.

 

Question 9. (ii) The table of values is given as under:

Answer:
**1. Mean:**
From the table, \( \Sigma fx = 1825 \) and \( \Sigma f = N = 130 \).
Mean \( \overline{x} = \frac{\Sigma fx}{\Sigma f} = \frac{1825}{130} \approx 14.04 \)
**2. Median:**
For median, we find \( \frac{N+1}{2} = \frac{130+1}{2} = \frac{131}{2} = 65.5 \).
Looking at the cumulative frequency (c.f.) column, 65.5 falls within the c.f. of 83, which corresponds to the size of item 14. So, the Median is 14.
**3. Mode:**
The maximum frequency (\(f\)) in the table is 37, which corresponds to the size of item 14. Therefore, the Mode is 14.
In simple words: Using the data in the table, we found the average (mean) by adding up all the \(fx\) values and dividing by the total number of items (\(N\)). The middle value (median) was found by locating the item corresponding to the middle cumulative frequency. The most common item (mode) was the one with the highest frequency.

🎯 Exam Tip: For frequency distribution tables, remember that the median is found by looking at the cumulative frequency and the mode is simply the item with the highest frequency. Ensure your \(fx\) and \(c.f.\) calculations are accurate.

 

Question 10. The mode of the following distribution is 240. Find out the missing frequency :

Answer: We need to find the missing frequency in the given grouped data, knowing that the mode is 240.
Let the missing frequency for the class 300-400 be \(f\).
Since the mode is 240, the modal class must be 200-300 (because 240 falls within this range).
Now, identify the values for the mode formula:
Lower limit of modal class (\(I\)) = 200
Frequency of modal class (\(f_m\)) = 270
Frequency of class preceding the modal class (\(f_{m-1}\)) = 230
Frequency of class succeeding the modal class (\(f_{m+1}\)) = \(f\) (the missing frequency for 300-400)
Class size (\(i\)) = 300 - 200 = 100
Apply the mode formula and substitute the given mode:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies 240 = 200 + \frac{270 - 230}{2(270) - 230 - f} \times 100 \)
\( \implies 240 - 200 = \frac{40}{540 - 230 - f} \times 100 \)
\( \implies 40 = \frac{40}{310 - f} \times 100 \)
\( \implies 40 = \frac{4000}{310 - f} \)
\( \implies 40 (310 - f) = 4000 \)
\( \implies 12400 - 40f = 4000 \)
\( \implies 12400 - 4000 = 40f \)
\( \implies 8400 = 40f \)
\( \implies f = \frac{8400}{40} \)
\( \implies f = 210 \)
So, the missing frequency is 210.
In simple words: We know the mode of this grouped data and need to find one missing frequency. Since we know the mode is 240, we know which group is the "modal class." We then use the mode formula for grouped data, plug in the mode and all known frequencies, and solve for the unknown frequency.

🎯 Exam Tip: When solving for a missing frequency, carefully set up the mode formula with the given mode value. Ensure algebraic manipulations are correct to avoid calculation errors.

 

Question 11. Find the median, mode, third quartile, 5th decile and 65th percentile from the following :

Answer: We need to calculate the median, mode, third quartile (Q3), 5th decile (D5), and 65th percentile (P65) for the given data. First, we'll create a cumulative frequency (c.f.) table.**1. Median (Md):**
The position of the median is \( \frac{N+1}{2} = \frac{355+1}{2} = \frac{356}{2} = 178 \).
Looking at the c.f. column, the value corresponding to the 178th item is 5 (since 178 is greater than 138 but less than 184).
So, Median (Md) = 5.
**2. Mode:**
The mode is the value of the item with the highest frequency. In the table, the maximum frequency is 58, which corresponds to the value of the item 6.
So, Mode = 6.
**3. Third Quartile (Q3):**
The position of the third quartile is \( \frac{3(N+1)}{4} = \frac{3 \times 356}{4} = \frac{1068}{4} = 267 \).
Looking at the c.f. column, the value corresponding to the 267th item is 7 (since 267 is greater than 242 but less than 274).
So, Q3 = 7.
**4. 5th Decile (D5):**
The position of the 5th decile is \( \frac{5(N+1)}{10} = \frac{5 \times 356}{10} = \frac{1780}{10} = 178 \).
Looking at the c.f. column, the value corresponding to the 178th item is 5 (since 178 is greater than 138 but less than 184).
So, D5 = 5.
**5. 65th Percentile (P65):**
The position of the 65th percentile is \( \frac{65(N+1)}{100} = \frac{65 \times 356}{100} = \frac{23140}{100} = 231.4 \).
Looking at the c.f. column, the value corresponding to the 231.4th item is 6 (since 231.4 is greater than 184 but less than 242).
So, P65 = 6.
In simple words: We find various central values and positions in the data. The median is the middle value. The mode is the most common value. Quartiles divide the data into four parts, deciles into ten, and percentiles into one hundred. We find where each specified position falls in the sorted data (using cumulative frequency) to identify the corresponding item value.

🎯 Exam Tip: For median, quartiles, deciles, and percentiles for discrete data, always use the \( \frac{kN}{100} \) or \( \frac{k(N+1)}{100} \) type formulas for position and then locate the value using the cumulative frequency. Pay attention to whether N or N+1 is used based on the formula being taught.

 

Question 12. Find median, mean and modal marks and also determine the limits between which 80% of the students have secured marks.

Answer: We need to find the mean, median, mode, and the range for 80% of the students. We will use the step-deviation method for the mean, and then calculate median and mode.
Let's choose Assumed Mean (\(A\)) = 45 and Class size (\(i\)) = 10.
The table of values with calculations is given below:**1. Mean (\( \overline{x} \)) by Step-deviation method:**
\( \overline{x} = A + \frac{\Sigma fd'}{N} \times i \)
\( \implies \overline{x} = 45 + \frac{-15}{393} \times 10 \)
\( \implies \overline{x} = 45 - \frac{150}{393} \)
\( \implies \overline{x} = 45 - 0.3816... \)
\( \implies \overline{x} \approx 44.62 \)
**2. Median (Md):**
The position of the median is \( \frac{N}{2} = \frac{393}{2} = 196.5 \).
Looking at the cumulative frequency (c.f.) column, 196.5 lies in the class 40-50 (since 196.5 is greater than 148 but less than 226). This is our median class.
Lower limit of median class (\(I\)) = 40
Frequency of median class (\(f\)) = 78
Cumulative frequency of class preceding median class (\(C\)) = 148
Class size (\(i\)) = 10
Median formula for grouped data:
\( Md = I + \frac{\frac{N}{2} - C}{f} \times i \)
\( \implies Md = 40 + \frac{196.5 - 148}{78} \times 10 \)
\( \implies Md = 40 + \frac{48.5}{78} \times 10 \)
\( \implies Md = 40 + \frac{485}{78} \)
\( \implies Md = 40 + 6.2179... \)
\( \implies Md \approx 46.22 \)
**3. Mode:**
Identify the modal class (class with the highest frequency). The maximum frequency is 80, which belongs to the class 50-60. So, the modal class is 50-60.
Lower limit of modal class (\(I\)) = 50
Frequency of modal class (\(f_m\)) = 80
Frequency of class preceding the modal class (\(f_{m-1}\)) = 78
Frequency of class succeeding the modal class (\(f_{m+1}\)) = 70
Class size (\(i\)) = 10
Mode formula for grouped data:
\( Mode = I + \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \times i \)
\( \implies Mode = 50 + \frac{80 - 78}{2(80) - 78 - 70} \times 10 \)
\( \implies Mode = 50 + \frac{2}{160 - 78 - 70} \times 10 \)
\( \implies Mode = 50 + \frac{2}{12} \times 10 \)
\( \implies Mode = 50 + \frac{20}{12} \)
\( \implies Mode = 50 + 1.666... \)
\( \implies Mode \approx 51.67 \)
**4. Limits for 80% of students:**
This means finding the marks between which the middle 80% of students fall. This is typically between the 10th percentile (P10) and the 90th percentile (P90).
**a. 10th Percentile (P10):**
Position of P10 = \( \frac{10 \times N}{100} = \frac{10 \times 393}{100} = \frac{3930}{100} = 39.3 \).
Looking at the c.f. column, 39.3 lies in the class 10-20 (since 39.3 is greater than 15 but less than 40). This is our P10 class.
Lower limit of P10 class (\(I\)) = 10
Frequency of P10 class (\(f\)) = 25
Cumulative frequency of class preceding P10 class (\(C\)) = 15
Class size (\(i\)) = 10
P10 formula: \( P_{10} = I + \frac{\frac{10N}{100} - C}{f} \times i \)
\( \implies P_{10} = 10 + \frac{39.3 - 15}{25} \times 10 \)
\( \implies P_{10} = 10 + \frac{24.3}{25} \times 10 \)
\( \implies P_{10} = 10 + \frac{243}{25} \)
\( \implies P_{10} = 10 + 9.72 \)
\( \implies P_{10} = 19.72 \)
**b. 90th Percentile (P90):**
Position of P90 = \( \frac{90 \times N}{100} = \frac{90 \times 393}{100} = \frac{35370}{100} = 353.7 \).
Looking at the c.f. column, 353.7 lies in the class 60-70 (since 353.7 is greater than 306 but less than 376). This is our P90 class.
Lower limit of P90 class (\(I\)) = 60
Frequency of P90 class (\(f\)) = 70
Cumulative frequency of class preceding P90 class (\(C\)) = 306
Class size (\(i\)) = 10
P90 formula: \( P_{90} = I + \frac{\frac{90N}{100} - C}{f} \times i \)
\( \implies P_{90} = 60 + \frac{353.7 - 306}{70} \times 10 \)
\( \implies P_{90} = 60 + \frac{47.7}{70} \times 10 \)
\( \implies P_{90} = 60 + \frac{477}{70} \)
\( \implies P_{90} = 60 + 6.814... \)
\( \implies P_{90} \approx 66.81 \)
The required limits between which 80% of students have secured marks are 19.72 and 66.81.
In simple words: We calculated the average marks (mean), the middle mark (median), and the most frequent mark (mode) using special formulas for grouped data. To find the range where 80% of students' marks fall, we looked at the bottom 10% and top 10% marks, and the range between them covers the middle 80%.

🎯 Exam Tip: When calculating percentiles for grouped data, ensure you identify the correct percentile class by checking the cumulative frequency. Pay close attention to the specific value of \( \frac{kN}{100} \) and its corresponding class limits and frequencies.