Samacheer Kalvi Class 9 Maths Solutions Chapter 8 Statistics More Ques

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions

I. Choose the Best Answer

 

Question 1. The Arithmetic mean of all the factors of 10 is .........
(a) 4.5
(b) 5.5
(c) 10
(d) 5
Answer: (a) 4.5
In simple words: First, find all the numbers that divide 10 evenly (its factors). Then, add these factors together and divide by how many factors there are. For 10, the factors are 1, 2, 5, and 10.

🎯 Exam Tip: Remember to include 1 and the number itself when listing all factors of a number.

 

Question 2. The mean of five numbers is 27, if one number is excluded, then mean is 25. Then the excluded number is .........
(a) 0
(b) 15
(c) 25
(d) 35
Answer: (d) 35
In simple words: To find the excluded number, first calculate the total sum of the five numbers. Then, calculate the sum of the remaining four numbers. The difference between these two sums will give you the number that was removed.

🎯 Exam Tip: The sum of numbers is always equal to the mean multiplied by the count of numbers. Use this formula to solve problems involving changes in means.

 

Question 3. The mean of 8 numbers is 15. If each number is multiplied by 2, then the new mean will be .........
(a) 7.5
(b) 30
(c) 10
(d) 25
Answer: (b) 30
In simple words: When every number in a data set is multiplied by the same value, the mean of that data set also gets multiplied by the same value. So, if the old mean was 15, the new mean will be twice that.

🎯 Exam Tip: Operations like adding, subtracting, multiplying, or dividing a constant to all data points directly affect the mean by the same operation. However, median and mode behave differently with these operations.

 

Question 4. The median of 11, 8, 4, 9, 7, 5, 2, 4, 10 is .........
(a) 1
(b) 8
(c) 7
(d) 11
Answer: (c) 7
In simple words: To find the median, first arrange all the numbers in order from smallest to largest. Then, find the number that is exactly in the middle of this ordered list. If there are an odd number of values, the median is the single middle value.

🎯 Exam Tip: Always sort the data set in ascending or descending order before identifying the median. If there are two middle numbers (even count), the median is their average.

 

Question 5. Median is .........
(a) the most frequent value
(b) the least frequent value
(c) middle most value
(d) mean of first and last values
Answer: (c) middle most value
In simple words: The median is the value that separates the higher half from the lower half of a data sample. It's the number right in the middle after you arrange all the numbers in order.

🎯 Exam Tip: Understand the definitions of mean, median, and mode to avoid confusion. Mean is the average, median is the middle value, and mode is the most frequent value.

 

Question 6. The mode of the distribution is .........

(a) 3
(b) 4
(c) 6
(d) 14
Answer: (b) 4
In simple words: The mode is the number that shows up most often in a set of data. In this table, we look for the highest frequency (f), and the 'x' value next to it is our mode. The number 4 has the highest frequency of 14.

🎯 Exam Tip: For a frequency distribution table, the mode is the value of the variate (x) that corresponds to the highest frequency (f).

 

Question 7. Mode is .........
(a) the middle value
(b) extreme value
(c) minimum value
(d) the most repeated value
Answer: (d) the most repeated value
In simple words: The mode tells us which item or number appears most frequently in a given set of data. It's all about how often something repeats.

🎯 Exam Tip: A dataset can have one mode (unimodal), multiple modes (multimodal), or no mode if all values appear with the same frequency.

 

Question 8. The mode of the data 72, 33, 44, 72, 81, 72, 15 is .........
(a) 72
(b) 33
(c) 81
(d) 15
Answer: (a) 72
In simple words: The mode is the number that shows up most often in the list. Look at the numbers given and count how many times each one appears. The number 72 appears three times, which is more than any other number.

🎯 Exam Tip: Carefully scan the entire data set to identify repetitions. It's helpful to list each unique value and its count to find the highest frequency.

 

Question 9. The Arithmetic mean of 10 number is -7. If 5 is added to every number, then the new arithmetic mean is .........
(a) 17
(b) 12
(c) -2
(d) -7
Answer: (c) -2
In simple words: If you add a certain number to every single item in a list, the average (mean) of that list will also increase by the same certain number. So, if the original average was -7 and you add 5 to everything, the new average will be -7 + 5.

🎯 Exam Tip: Understand how basic arithmetic operations affect measures of central tendency. Adding or subtracting a constant shifts the mean directly by that constant.

 

Question 10. The Arithmetic mean of integers from -5 to 5 is ........
(a) 25
(b) -5
(c) 3
(d) 0
Answer: (d) 0
In simple words: The integers from -5 to 5 are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. When you add all these numbers together, the negative numbers cancel out the positive numbers. For example, -1 and 1 add to 0, -2 and 2 add to 0, and so on. The only number left is 0. So, the total sum is 0, and the average of 0 divided by any count (here, 11 numbers) is still 0.

🎯 Exam Tip: For a set of consecutive integers symmetric around zero (like -n to n), the sum is always zero, and thus the arithmetic mean is zero. Remember to include zero in the count of integers.

II. Answer the Following Questions

 

Question 1. Find the Arithmetic mean for the following data.

Answer:
To find the arithmetic mean, we first need to calculate the mid-value (x) for each class interval and then multiply it by its frequency (f) to get fx.The formula for Arithmetic mean \( (\bar{X}) \) is:
\( \bar{X} = \frac{\Sigma fx}{\Sigma f} \)
\( \implies \bar{X} = \frac{5540}{96} \)
\( \implies \bar{X} = 57.7083 \)
So, the Arithmetic mean is approximately \( 57.7 \).
In simple words: We first find the middle value for each group of numbers. Then, we multiply this middle value by how many times it appears. Finally, we add up all these multiplied values and divide by the total count of numbers to get the average.

🎯 Exam Tip: When dealing with grouped data, ensure accurate calculation of mid-values for each class interval. Rounding at intermediate steps should be avoided until the final answer.

 

Question 2. Calculate the Arithmetic mean of the following data using step deviation method

Answer:
We will use the step deviation method. First, we assume a mean (A), calculate deviations (d), and then step deviations (u).
Let the assumed mean \( A = 35 \). The class width \( c = 10 \).The formula for Arithmetic mean \( (\bar{X}) \) using the step deviation method is:
\( \bar{X} = A + \frac{\Sigma fd}{\Sigma f} \times c \)
\( \implies \bar{X} = 35 + \frac{(-20)}{100} \times 10 \)
\( \implies \bar{X} = 35 + (-0.2 \times 10) \)
\( \implies \bar{X} = 35 - 2 \)
\( \implies \bar{X} = 33 \)
So, the Arithmetic mean is \( 33 \).
In simple words: This method helps find the average faster for large data sets. We pick a guessed average, then calculate how far each group's middle value is from our guess. We adjust for this difference and scale it by the group width, then combine it with our guessed average to get the true average.

🎯 Exam Tip: The step deviation method simplifies calculations for grouped data with equal class intervals, especially when numbers are large. Choose an assumed mean (A) that is a mid-value near the center of the data for easier calculations.

 

Question 3. Find the median for the following data.

Answer:
The given class intervals are of inclusive type. We need to convert them into exclusive type by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.
Total frequency \( N = 70 \).
\( \frac{N}{2} = \frac{70}{2} = 35 \)
The cumulative frequency just greater than 35 is 56, which corresponds to the class interval 25.5 - 30.5.
This means the median class is 25.5 - 30.5.
Here,
Lower limit of median class \( (L) = 25.5 \)
Frequency of median class \( (f) = 26 \)
Cumulative frequency of the class preceding the median class \( (cf) = 30 \)
Class width \( (h) = 5 \)
The formula for Median is:
\( \text{Median} = L + \frac{\frac{N}{2} - cf}{f} \times h \)
\( \implies \text{Median} = 25.5 + \frac{35 - 30}{26} \times 5 \)
\( \implies \text{Median} = 25.5 + \frac{5}{26} \times 5 \)
\( \implies \text{Median} = 25.5 + \frac{25}{26} \)
\( \implies \text{Median} = 25.5 + 0.9615 \)
\( \implies \text{Median} = 26.4615 \)
So, the Median is approximately \( 26.46 \).
In simple words: First, we adjust the number groups so they flow smoothly. Then, we find the middle position in the total count of items. We locate the group where this middle item falls and use a special formula that considers the group's starting point, how many items are before it, how many items are in it, and the group's size, to pinpoint the exact median.

🎯 Exam Tip: Always convert inclusive class intervals to exclusive ones before calculating the median. Carefully identify the median class, its lower limit, frequency, and the cumulative frequency of the preceding class for accurate results.

 

Question 4. Calculate the mode of the following data.

Answer:
First, we identify the class with the highest frequency, which is called the modal class.
From the table, the highest frequency is 30, and it corresponds to the class interval 25-30.
So, the modal class is 25-30.
Here,
Lower limit of modal class \( (L) = 25 \)
Frequency of the modal class \( (f_1) = 30 \)
Frequency of the class preceding the modal class \( (f_0) = 18 \)
Frequency of the class succeeding the modal class \( (f_2) = 20 \)
Class width \( (h) = 5 \)
The formula for Mode is:
\( \text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( \implies \text{Mode} = 25 + \frac{30 - 18}{2(30) - 18 - 20} \times 5 \)
\( \implies \text{Mode} = 25 + \frac{12}{60 - 18 - 20} \times 5 \)
\( \implies \text{Mode} = 25 + \frac{12}{22} \times 5 \)
\( \implies \text{Mode} = 25 + \frac{60}{22} \)
\( \implies \text{Mode} = 25 + 2.7272... \)
\( \implies \text{Mode} = 27.7272... \)
So, the Mode is approximately \( 27.73 \).
In simple words: We look for the group of numbers that appears most often (the modal class). Then, we use a special formula that takes into account the beginning of this group, its frequency, and the frequencies of the groups just before and after it, along with the group's size, to find the exact mode.

🎯 Exam Tip: Ensure correct identification of \(f_0\), \(f_1\), and \(f_2\) when applying the mode formula. \(f_1\) is the frequency of the modal class, \(f_0\) is the one before it, and \(f_2\) is the one after it.

 

Question 5. Find the mean, median and mode of marks obtained by 20 students in an examination. The marks are given below.

Answer:
We will calculate Mean, Median, and Mode step by step.
**Arithmetic Mean:**
The total sum of \( fx = 560 \) and total frequency \( \Sigma f = 20 \).
The formula for Arithmetic mean \( (\bar{X}) \) is:
\( \bar{X} = \frac{\Sigma fx}{\Sigma f} \)
\( \implies \bar{X} = \frac{560}{20} \)
\( \implies \bar{X} = 28 \)

**Median:**
Total frequency \( N = 20 \).
\( \frac{N}{2} = \frac{20}{2} = 10 \)
The cumulative frequency just greater than or equal to 10 is 10, which corresponds to the class interval 20-30.
So, the median class is 20-30.
Here,
Lower limit of median class \( (L) = 20 \)
Frequency of median class \( (f) = 5 \)
Cumulative frequency of the class preceding the median class \( (cf) = 5 \) (for class 10-20)
Class width \( (h) = 10 \)
The formula for Median is:
\( \text{Median} = L + \frac{\frac{N}{2} - cf}{f} \times h \)
\( \implies \text{Median} = 20 + \frac{10 - 5}{5} \times 10 \)
\( \implies \text{Median} = 20 + \frac{5}{5} \times 10 \)
\( \implies \text{Median} = 20 + 1 \times 10 \)
\( \implies \text{Median} = 20 + 10 \)
\( \implies \text{Median} = 30 \)

**Mode:**
The highest frequency in the table is 8, which corresponds to the class interval 30-40.
So, the modal class is 30-40.
Here,
Lower limit of modal class \( (L) = 30 \)
Frequency of the modal class \( (f_1) = 8 \)
Frequency of the class preceding the modal class \( (f_0) = 5 \) (for class 20-30)
Frequency of the class succeeding the modal class \( (f_2) = 2 \) (for class 40-50)
Class width \( (h) = 10 \)
The formula for Mode is:
\( \text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( \implies \text{Mode} = 30 + \frac{8 - 5}{2(8) - 5 - 2} \times 10 \)
\( \implies \text{Mode} = 30 + \frac{3}{16 - 5 - 2} \times 10 \)
\( \implies \text{Mode} = 30 + \frac{3}{9} \times 10 \)
\( \implies \text{Mode} = 30 + \frac{30}{9} \)
\( \implies \text{Mode} = 30 + 3.333... \)
\( \implies \text{Mode} = 33.33 \)

Therefore, for the given data:
Mean \( = 28 \)
Median \( = 30 \)
Mode \( = 33.33 \)
In simple words: To find the mean, median, and mode for these groups, we first calculate the midpoint for each group. For the mean, we average these midpoints based on how many students are in each group. For the median, we find the middle value of all students, considering their grouped marks. For the mode, we find the mark range where most students fall and then calculate the specific mode value within that range.

🎯 Exam Tip: When a question asks for all three measures (mean, median, mode), it's crucial to organize your calculations clearly. Make sure to use the correct formulas and values for each specific measure from the same data set.