Samacheer Kalvi Class 9 Maths Solutions Chapter 8 Statistics Exercise 8.2

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.2

 

Question 1. Find the median of the given values: 47, 53, 62, 71, 83, 21, 43, 47, 41
Answer: To find the median, first arrange all the numbers from smallest to largest. The given values sorted in ascending order are: 21, 41, 43, 47, 47, 53, 62, 71, 83. There are 9 numbers in total, which is an odd count. For an odd number of values, the median is the middle value. We find its position by adding 1 to the total number of values and dividing by 2. So, the median is the \( \frac{9+1}{2} = 5^{th} \) variable. The 5th number in the ordered list is 47. The median helps us understand the central point of the data.
In simple words: First, put the numbers in order from smallest to largest. Since there are 9 numbers (an odd count), the median is the middle number. The middle number is the 5th one, which is 47.

🎯 Exam Tip: Always arrange the data in ascending order before finding the median; otherwise, you'll get the wrong middle value.

 

Question 2. Find the median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51
Answer: First, arrange the given numbers from the smallest to the largest. The sorted list is: 31, 35, 36, 37, 44, 44, 51, 60, 86, 86. There are 10 numbers in the list, which is an even count. When the number of values is even, the median is the average of the two middle numbers. The two middle numbers are the \( (\frac{10}{2})^{th} \) and \( (\frac{10}{2}+1)^{th} \) values, which are the 5th and 6th values. These values are 44 and 44. To find the average, we add them and divide by two. This gives us \( \frac{44+44}{2} = \frac{88}{2} = 44 \). The median gives a good sense of the typical value in a dataset.
In simple words: Put the numbers in order from smallest to largest. There are 10 numbers, so it's an even set. The median is the average of the two middle numbers (the 5th and 6th numbers). Both are 44, so the median is 44.

🎯 Exam Tip: For an even number of values, remember to take the average of the two middle numbers, not just one of them.

 

Question 3. The median of observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the values of x.
Answer: The numbers are already in increasing order: 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41. There are 10 numbers in this list, which is an even count. For an even number of values, the median is the average of the two middle terms. In this case, the 5th term is \( x+2 \) and the 6th term is \( x+4 \). We are told that the median is 24. So, we can set up an equation:
\( 24 = \frac{(x+2) + (x+4)}{2} \)
\( \implies \) \( 24 = \frac{2x+6}{2} \)
\( \implies \) \( 24 \times 2 = 2x+6 \)
\( \implies \) \( 48 = 2x+6 \)
\( \implies \) \( 2x = 48-6 \)
\( \implies \) \( 2x = 42 \)
\( \implies \) \( x = \frac{42}{2} \)
\( \implies \) \( x = 21 \). Algebra is essential for solving problems like this.
In simple words: The numbers are already sorted. Since there are 10 numbers (even), the median is the average of the 5th and 6th numbers. The 5th is \( x+2 \) and the 6th is \( x+4 \). We know the median is 24. Set up the equation \( 24 = \frac{x+2+x+4}{2} \), which simplifies to \( 24 = \frac{2x+6}{2} \). Solving this equation gives \( x = 21 \).

🎯 Exam Tip: Always check if the data is already sorted. If not, sort it first. Be careful with algebraic manipulation when solving for unknown variables.

 

Question 4. A researcher studying the behavior of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching its food.
Answer: To find the median time, first arrange all the recorded times from the shortest to the longest. The sorted list is: 27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63. There are 13 different mice, so the number of values is 13, which is an odd number. For an odd number of observations, the median is the middle value. Its position is found by the formula \( \frac{13+1}{2} = 7 \). The 7th value in the sorted list is 32. So, the median time spent searching for food is 32 seconds. Understanding median is crucial for analyzing experiment results.
In simple words: First, sort all the times from smallest to largest. There are 13 times (an odd number). The median is the middle number, which is the 7th one. The 7th time in the sorted list is 32 seconds.

🎯 Exam Tip: Even for word problems, the first step to finding the median is always to arrange the given numbers in ascending order.

 

Question 5. The following are the marks scored by the students in the Summative Assessment exam. Calculate the median.
Answer: To calculate the median for grouped data, we first find \( \frac{N}{2} \), where \( N \) is the total number of students. Here, \( N = 50 \), so \( \frac{N}{2} = \frac{50}{2} = 25 \). Next, we locate the median class by finding the cumulative frequency that is just greater than or equal to 25. This is 34, which corresponds to the class interval 30-40. So, the median class is 30-40. Now, we use the formula for the median of grouped data: \( Median = l + \frac{(\frac{N}{2}-m)}{f} \times c \). Here are the values from the table:

From the median class (30-40):
\( l \) (lower limit of median class) = 30
\( f \) (frequency of median class) = 10
\( m \) (cumulative frequency of the class preceding the median class) = 24
\( c \) (class width) = 10
Plugging these values into the formula:
\( Median = 30 + \frac{(25-24)}{10} \times 10 \)
\( \implies \) \( Median = 30 + \frac{1}{10} \times 10 \)
\( \implies \) \( Median = 30 + 1 \)
\( \implies \) \( Median = 31 \). So, the median score is 31. Understanding median for grouped data is important in larger datasets.
In simple words: First, find half of the total number of students, which is 25. Then, find the class interval where the cumulative frequency first reaches or goes past 25 (this is 30-40). Use the median formula for grouped data: lower limit + \( (\frac{N}{2} - \text{cumulative frequency before median class}) / \text{frequency of median class} \times \text{class width} \). Plugging in the numbers gives \( 30 + \frac{(25-24)}{10} \times 10 \), which simplifies to \( 30 + 1 = 31 \).

🎯 Exam Tip: Always create a cumulative frequency column when calculating the median for grouped data. Clearly identify all variables for the formula before substituting.

 

Question 6. The mean of five positive integers is twice their median. If four of the integers are 3, 4, 6, 9 and median is 6, then find the fifth integer.
Answer: Let the five positive integers be 3, 4, 6, 9, and \( x \). We are given that the median of these five integers is 6. Since there are five numbers (an odd quantity), the median is the middle number when arranged in order. With 3, 4, 6, 9, and \( x \), for 6 to be the median, \( x \) must be 6 or greater, so the sorted list could be 3, 4, 6, 9, \( x \). The mean of these five integers is calculated as:
\( \text{Mean} = \frac{3+4+6+9+x}{5} = \frac{22+x}{5} \)
The problem states that the mean is twice the median. So, we can set up the equation:
\( \frac{22+x}{5} = 2 \times 6 \)
\( \implies \) \( \frac{22+x}{5} = 12 \)
\( \implies \) \( 22+x = 12 \times 5 \)
\( \implies \) \( 22+x = 60 \)
\( \implies \) \( x = 60 - 22 \)
\( \implies \) \( x = 38 \). This problem effectively combines concepts of mean and median.
In simple words: Let the fifth number be \( x \). The five numbers are 3, 4, 6, 9, \( x \). We know the median is 6. The mean is found by adding all numbers and dividing by 5, which is \( \frac{22+x}{5} \). The problem says the mean is twice the median, so \( \frac{22+x}{5} = 2 \times 6 \). Solving this equation gives \( x = 38 \).

🎯 Exam Tip: When given the median for an odd number of values, that value is directly the middle term. Ensure all values are considered when calculating the mean.