All questions of Statistics for Class 9 Exam

Frequency polygons are graphical representations of frequency distributions. They are constructed by plotting the frequency of each class interval on the y-axis and the mid value of each class interval on the x-axis. The correct answer is option 'C', which states that the frequency polygon is constructed by plotting the mid value of the class.

Explanation:

Frequency polygons are used to represent the distribution of a set of data values. They are useful for visualizing the shape and pattern of the data. To construct a frequency polygon, follow these steps:

1. Determine the class intervals: Start by determining the class intervals for the data. Class intervals are ranges of values that divide the data into groups or intervals. For example, if you have data on the heights of students in a class, the class intervals could be 140-150 cm, 150-160 cm, 160-170 cm, and so on.

2. Calculate the frequency: Count the number of data values that fall within each class interval. This is known as the frequency. For example, if there are 5 students with heights between 140-150 cm, the frequency for that class interval would be 5.

3. Find the mid value of each class interval: To construct the frequency polygon, you need to plot the mid value of each class interval on the x-axis. The mid value is calculated by adding the lower limit and the upper limit of each class interval and dividing by 2. For example, if the class interval is 140-150 cm, the mid value would be (140 + 150) / 2 = 145 cm.

4. Plot the points: Plot the frequency of each class interval on the y-axis and the mid value of each class interval on the x-axis. For example, if the frequency for the class interval 140-150 cm is 5, you would plot the point (145, 5) on the graph.

5. Connect the points: Once all the points are plotted, connect them using straight lines. This will form the frequency polygon.

In conclusion, the correct answer is option 'C', which states that the frequency polygon is constructed by plotting the mid value of the class. The mid value provides a representative point for each class interval and helps in visualizing the distribution of the data.

Understanding the Median
The median is the middle value of a data set when it is arranged in ascending order. If there is an even number of observations, the median is the average of the two middle values.
Original Data Set
The original data set is:
- 7, 9, 16, 25, 31, 36, 37, 39, 40, 42, 43
Finding the Median of Original Data
1. Count the Numbers:
- There are 11 numbers (odd count).
2. Median Position:
- The median is at the position (11 + 1) / 2 = 6th position.
3. Value at Median Position:
- The 6th number in the sorted list is 36.
Replacing 37 with 5
Now, we replace 37 with 5 and look at the new data set:
- 5, 7, 9, 16, 25, 31, 36, 39, 40, 42, 43
Finding the Median of Modified Data
1. Count the Numbers:
- The new set still contains 11 numbers (odd count).
2. Median Position:
- The median remains at the 6th position.
3. Value at Median Position:
- The 6th number in the new sorted list is 31.
Conclusion
Thus, the new median after replacing 37 with 5 is 31, which corresponds to option 'C'.

Explanation:

To find the lower limit of the first class interval, we need to determine the smallest value in the given data set.

Given class marks distribution: 25, 26, 27, 31, 36, 41, 46, 51, 57, 59

Step 1: Sort the data in ascending order.
25, 26, 27, 31, 36, 41, 46, 51, 57, 59

Step 2: Identify the smallest value in the data set.
The smallest value in the data set is 25.

Step 3: Define the class intervals.
Since the question does not provide the width of the class intervals, we cannot determine the exact class intervals. However, we can infer that the class intervals are of equal width since the data set is presented in a tabular form.

Step 4: Determine the lower limit of the first class interval.
To find the lower limit of the first class interval, we subtract half of the class interval width from the smallest value in the data set.

In this case, since we do not know the exact class interval width, we cannot calculate the exact lower limit of the first class interval. However, we can approximate the lower limit based on the given answer options.

The options provided are:
a) 40
b) 42
c) 23.5
d) 24.5

Step 5: Analyze the answer options.
Given that the lowest value in the data set is 25, options a) and b) can be eliminated since they are greater than 25. Therefore, the lower limit of the first class interval must be less than 25.

Option c) 23.5 is also incorrect because it is less than the smallest value in the data set.

Therefore, the correct option is d) 24.5, which is slightly less than the smallest value in the data set.

Conclusion:
The lower limit of the first class interval is 24.5.

Given Information:
- Mid value of a class interval is 14
- Class size is 2

To Find:
- Lower limit of the class

Solution:
The class interval is given by the mid value ± half of the class size.

Let's consider the mid value of the class interval as x and the class size as y.

x = 14
y = 2

Calculating the Lower Limit:
The lower limit of the class interval can be found by subtracting half of the class size from the mid value.

Lower Limit = Mid Value - Half of Class Size

Lower Limit = x - (y/2)

Substituting the given values,

Lower Limit = 14 - (2/2)
Lower Limit = 14 - 1
Lower Limit = 13

Therefore, the lower limit of the class is 13.

Conclusion:
The correct answer is option 'B' (13) as the lower limit of the class is 13.

Given information:
- Mean of 10 numbers is 16.
- When two consecutive numbers are excluded, the new mean is 18.

To find: The sum of the excluded numbers.

Let's solve this problem step by step:

Step 1: Finding the sum of the original 10 numbers
Since the mean is the sum of all the numbers divided by the total count of numbers, we can find the sum of the original 10 numbers by multiplying the mean by the count.
Mean = Sum of numbers / Count of numbers
16 = Sum of numbers / 10
Sum of numbers = 16 * 10
Sum of numbers = 160

Step 2: Finding the sum of the remaining 8 numbers
When two consecutive numbers are excluded, it means we are left with 8 numbers. Let's assume the two excluded numbers are x and x+1.

The new mean is given as 18. So, we can find the sum of the remaining 8 numbers using the new mean.
Mean = Sum of remaining numbers / Count of remaining numbers
18 = Sum of remaining numbers / 8
Sum of remaining numbers = 18 * 8
Sum of remaining numbers = 144

Step 3: Finding the sum of the excluded numbers
To find the sum of the excluded numbers, we subtract the sum of the remaining 8 numbers from the sum of the original 10 numbers.
Sum of excluded numbers = Sum of original 10 numbers - Sum of remaining 8 numbers
Sum of excluded numbers = 160 - 144
Sum of excluded numbers = 16

Therefore, the sum of the excluded numbers is 16. Hence, option B is the correct answer.