RBSE Solutions Class 9 Maths Chapter 15 Statistics More Ques

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 15 Statistics Miscellaneous Exercise

Multiple Choice Questions

Question 1. The frequency of the class interval 3-5 in the following distribution is:
5, 5, 6, 4, 9, 5, 3, 2, 7, 6, 3, 8, 4
(A) 3
(B) 4
(C) 6
(D) 7
Answer: (B) 4
In simple words: To find the frequency, count how many numbers in the list are between 3 and 5, including 3 and 5 themselves. In this list, the numbers 3, 3, 4, and 4 fall into this interval. So there are 4 such numbers.

🎯 Exam Tip: Remember that a class interval like '3-5' usually includes both the lower limit (3) and the upper limit (5) when counting frequencies, unless specified otherwise.

Question 2. The range of the following frequency distribution will be:
3.2, 2.8, 3.1, 2.1, 3.2, 2.4, 2.1, 2.8, 2.7, 2.7
(A) 2.7
(B) 3.1
(C) 2.4
(D) 1.1
Answer: (D) 1.1
In simple words: The range is the difference between the biggest number and the smallest number in a group of data. For this list, the largest number is 3.2 and the smallest is 2.1, so the range is \( 3.2 - 2.1 = 1.1 \).

🎯 Exam Tip: To find the range, first identify the maximum value and the minimum value from the given data set, then subtract the minimum from the maximum.

Question 3. In the following frequency distribution, the number of students whose age is less than 25 years is:

🎯 Exam Tip: For "less than" questions in a frequency distribution table, sum the frequencies of all class intervals that are below the specified upper limit. Do not include the interval that starts at or goes beyond the limit.

Question 5. The comparative study of the result of any class of a school can be done:
(A) by ogive curve
(B) by histogram
(C) by linear curve
(D) All of the options
Answer: (B) by histogram
In simple words: A histogram is a bar graph that shows how often something happens over different periods or groups. It is very useful for comparing results because you can easily see the heights of the bars. This makes it simple to compare how well a class performed.

🎯 Exam Tip: Histograms are excellent for visualizing frequency distributions of continuous data and are commonly used for comparison, as they clearly show the shape, spread, and central tendency of data sets.

Question 6. Range of the distribution 6, 1, 2, 3, 9, 8, 3, 4, 8, 2, 3 is:
(A) 4
(B) 8
(C) 7
(D) 6
Answer: (B) 8
In simple words: The range of a data set is found by subtracting the smallest number from the largest number. In the given set (1, 2, 2, 3, 3, 3, 4, 6, 8, 8, 9), the maximum value is 9 and the minimum value is 1, so the range is \( 9 - 1 = 8 \).

🎯 Exam Tip: Always arrange the data in ascending or descending order first to easily identify the minimum and maximum values before calculating the range.

Question 7. If variate of the distribution are 5, 5, 2, 3, 6, 5, 4, then frequency of variate 5 will be:
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (C) 3
In simple words: Frequency means how many times a particular number appears in a list. In this list of numbers, the number 5 appears three times. So, its frequency is 3.

🎯 Exam Tip: To find the frequency of a variate, simply count how many times that specific value occurs within the given data set.

Question 8. The median of 11, 2, 7, 8, 9, 3, 5 is:
(A) 7
(B) 9
(C) 5
(D) 10
Answer: (A) 7
In simple words: To find the median, first put all the numbers in order from smallest to largest. Then, pick the middle number. If there are two middle numbers, find their average. In this case, there are 7 numbers, so the middle number is the 4th one after sorting.

🎯 Exam Tip: Always arrange the data in ascending order before finding the median. For an odd number of observations, the median is the \(\frac{n+1}{2}\)th term.

Question 9. The mean of 15, 0, 10, 5 will be:
(A) 15
(B) 10
(C) 5
(D) 4
Answer: (D) 4
In simple words: The mean is the average of all the numbers. To find it, add up all the numbers and then divide by how many numbers there are. Here, the sum is \( 15 + 0 + 10 + 5 = 30 \), and there are 4 numbers, so the mean is \( 30 \div 4 = 7.5 \).

🎯 Exam Tip: To calculate the mean, sum all the data points and divide by the total count of data points. Be careful with calculations, especially when dealing with zero.

Very Short Answer Type Questions

Question 11. Write the frequency of the class interval 0-5 from the distribution 3, 2, 0, 10, 8, 5, 13, 5, 6, 6, 0, 14.
Answer: To find the frequency of the class interval 0-5, we count how many numbers in the given distribution fall between 0 and 5 (including 0 and 5). The numbers are 3, 2, 0, 5, 5, 0. There are 6 such numbers in the list.

🎯 Exam Tip: Always make sure to count both the lower and upper limits of the class interval if the interval is inclusive. Tally marks help in systematically counting frequencies without missing any data point.

Question 12. If the mean marks of 5, 8, 4, x, 6, 9, is 7, then find the value of x.
Answer: The mean is calculated by summing all the values and dividing by the count of values. Given marks are 5, 8, 4, x, 6, 9. The total number of marks is 6. The mean is given as 7. We use the formula for the mean: Sum of marks = Mean \( \times \) Number of marks \( 5 + 8 + 4 + x + 6 + 9 = 7 \times 6 \) \( 32 + x = 42 \) Now, to find \( x \), we subtract 32 from 42: \( x = 42 - 32 \) \( x = 10 \)
In simple words: We know the average (mean) of the marks is 7. We add up all the marks we know and then add 'x' to them. This total sum must be equal to 7 multiplied by the number of marks (which is 6). From this, we can figure out what 'x' must be.

🎯 Exam Tip: Remember the mean formula: \( \text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}} \). Use this to form an equation and solve for the unknown variable.

Question 13. What is range?
Answer: The range in statistics is a simple measure of how spread out a set of numbers is. It is defined as the difference between the maximum (highest) and minimum (lowest) values in a given data set. A larger range indicates greater variability in the data. Range = Maximum value - Minimum value
In simple words: The range tells us how much the numbers in a group spread out. You find it by taking the biggest number and subtracting the smallest number from it.

🎯 Exam Tip: The range is the simplest measure of dispersion and provides a quick idea of the spread, but it's sensitive to outliers (extreme values).

Question 14. What is histogram?
Answer: A histogram is a special type of bar graph that represents the frequency distribution of continuous data. In a histogram, the data is grouped into ranges (called class intervals), and each bar's width shows the class interval, while its height shows the frequency of values within that interval. The bars in a histogram always touch each other, unlike regular bar graphs.
In simple words: A histogram is a type of graph with bars that are used to show how often different numbers appear in a group of data. The bars touch each other because the numbers are continuous.

🎯 Exam Tip: Key features of a histogram are adjacent bars (no gaps between bars) and that the x-axis represents class intervals (continuous data), while the y-axis represents frequency.

Question 15. Prepare frequency distribution table from the following data:
9, 7, 9, 8, 3, 9, 8, 3, 5, 7, 5, 3
Answer: To prepare a frequency distribution table, we first list all the unique data points (variates) in ascending order. Then, we count how many times each data point appears in the given data set using tally marks and write down its frequency. Given data: 9, 7, 9, 8, 3, 9, 8, 3, 5, 7, 5, 3 Unique variates are: 3, 5, 7, 8, 9. Let's count their frequencies: - For 3: It appears 3 times (3, 3, 3) - For 5: It appears 2 times (5, 5) - For 7: It appears 2 times (7, 7) - For 8: It appears 2 times (8, 8) - For 9: It appears 3 times (9, 9, 9)

🎯 Exam Tip: Always double-check your tally counts against the total number of data points to ensure accuracy and avoid missing any values. The sum of frequencies should equal the total number of observations.

Question 16. Mean of any frequency distribution is 15 and \( \sum f = 20 \), then find the value of \( \sum fx \).
Answer: We know the formula for the mean (\( \bar{x} \)) of a frequency distribution: \( \bar{x} = \frac { \sum fx }{ \sum f } \) Here, we are given: Mean (\( \bar{x} \)) = 15 Sum of frequencies (\( \sum f \)) = 20 We need to find the sum of (frequency \( \times \) variate) (\( \sum fx \)). Substitute the given values into the formula: \( 15 = \frac { \sum fx }{ 20 } \)
\( \implies \) To find \( \sum fx \), we multiply the mean by the sum of frequencies: \( \sum fx = 15 \times 20 \) \( \sum fx = 300 \)
In simple words: The average of a group of numbers is found by dividing the total sum of "fx" (which is frequency times each value) by the total number of frequencies. If we already know the average and the total frequency, we can just multiply them to find the total sum of "fx".

🎯 Exam Tip: Understand that \( \sum fx \) represents the sum of all observations when individual values are weighted by their frequencies. This formula is fundamental for calculating the mean of grouped data.

Question 18. Find the median of the following distribution 12, 1, 6, 4, 10, 8, 1, 4.
Answer: To find the median, we first need to arrange the given numbers in ascending order (from smallest to largest). The given distribution is: 12, 1, 6, 4, 10, 8, 1, 4. Arranging them in ascending order: 1, 1, 4, 4, 6, 8, 10, 12. Now, count the total number of observations (\( n \)). Here, \( n = 8 \). Since \( n \) is an even number, the median will be the average of the two middle terms. The two middle terms are the \( \left(\frac {n}{2}\right) \)th term and the \( \left(\frac {n}{2} + 1\right) \)th term. \( \frac {n}{2} = \frac {8}{2} = 4 \) So, the middle terms are the 4th term and the \( (4+1) \)th = 5th term. From the ordered list (1, 1, 4, 4, 6, 8, 10, 12): The 4th term is 4. The 5th term is 6. Now, calculate the median by taking the average of these two terms: Median = \( \frac { \text{4th term} + \text{5th term} }{ 2 } \) Median = \( \frac { 4 + 6 }{2} \) Median = \( \frac { 10 }{2} \) Median = 5
In simple words: First, put all the numbers in order. Since there's an even count of numbers, the median is the average of the two numbers exactly in the middle of the sorted list. This helps give a central value for the data.

🎯 Exam Tip: For an even number of observations, the median is the average of the two middle terms. Always re-check your ascending order to prevent calculation errors.

Question 19. Find the mode of the distribution 4, 3, 4, 1, 2, 4, 7, 5, 3.
Answer: The mode of a distribution is the value that appears most frequently in the data set. To find the mode, we simply count how many times each number appears. Given distribution: 4, 3, 4, 1, 2, 4, 7, 5, 3 Let's list the frequencies of each number: - 1 appears 1 time. - 2 appears 1 time. - 3 appears 2 times. - 4 appears 3 times. - 5 appears 1 time. - 7 appears 1 time. The variate with the highest frequency is 4, which appears 3 times. Hence, the mode of the distribution is 4.
In simple words: The mode is just the number that shows up most often in a list. We look at all the numbers and count how many times each one appears. The one that appears the most is the mode.

🎯 Exam Tip: The mode can be unique, or there can be multiple modes (bimodal or multimodal) if several values share the highest frequency. If all values have the same frequency, there is no mode.