Samacheer Kalvi Class 9 Maths Solutions Chapter 8 Statistics Exercise 8.4

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.4

Question 1. Let m be the mid point and b be the upper limit of a class in a continuous frequency distribution. The lower limit of the class is ......
(a) \( 2m - b \)
(b) \( 2m + b \)
(c) \( m - b \)
(d) \( b - 2m \)
Answer: (a) 2m - b
In simple words: The lower limit can be found by taking twice the midpoint and subtracting the upper limit. This formula helps determine the starting point of a class interval when you know its middle and end.

🎯 Exam Tip: Remember the formula for the midpoint of a class: \( m = \frac{\text{lower limit} + \text{upper limit}}{2} \). Rearranging this helps solve for unknown limits.

Question 2. The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is ......
(a) 101
(b) 100
(c) 99
(d) 98
Answer: (c) 99
Total of 8 numbers = \( 81 \times 7 = 567 \)
Total of 7 numbers = \( 78 \times 6 = 468 \)
The number is = \( 567 - 468 \)
\( \implies \) \( = 99 \)
In simple words: First, find the total sum of all numbers. Then, find the sum of the numbers left after one is removed. The difference between these two sums will be the value of the number that was taken out.

🎯 Exam Tip: When a value is added or removed from a data set, always calculate the new total sum before finding the new mean or the value itself.

Question 3. A particular observation which occurs maximum number of times in a given data is called its ......
(a) frequency
(b) range
(c) mode
(d) median
Answer: (c) mode
In simple words: The mode is the number that shows up most often in a list of numbers. It's like finding the most popular item in a group.

🎯 Exam Tip: Clearly distinguish between mean (average), median (middle value), and mode (most frequent value) to avoid confusion in statistics questions.

Question 4. For which set of numbers do the mean, median and mode all have the same values?
(a) 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 2, 1, 5
Answer: (b) 1, 3, 3, 3, 5
In simple words: For the set (1, 3, 3, 3, 5), if you add them up and divide by how many there are, the average (mean) is 3. The middle number (median) is also 3. And the number that appears most often (mode) is also 3. This is a special case where all three central measures are identical.

🎯 Exam Tip: To find the mean, median, and mode, always list the numbers in ascending order first. This makes it easier to spot the middle number and the most frequent number.

Question 5. The algebraic sum of the deviations of a set of n values from their mean is .......
(a) 0
(b) n - 1
(c) n
(d) n + 1
Answer: (a) 0
In simple words: If you take each number in a set and subtract the average (mean) from it, and then add up all those differences, the total will always be zero. This shows how the mean balances the data.

🎯 Exam Tip: This property is fundamental to the definition of the mean; it indicates that the mean is the center of gravity of the data set.

Question 6. The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is ......
(a) 24
(b) 36
(c) 26
(d) 34
Answer: (d) 34
Mean = 28
\( a + b + c + d + e = 28 \times 5 = 140 \) ...... (1)
But \( \frac{a+c+e}{3} = 24 \)
\( \implies a + c + e = 72 \)
Now, substitute \( a+c+e \) into equation (1):
\( (a+c+e) + b + d = 140 \)
\( 72 + b + d = 140 \)
\( \implies b + d = 140 - 72 \)
\( \implies b + d = 68 \)
Mean of b and d = \( \frac{68}{2} \)
\( \implies = 34 \)
In simple words: We know the average of five numbers and the average of three of those numbers. By using these averages, we can find the total sum for each group. Then, we can find the sum of the remaining two numbers and calculate their average.

🎯 Exam Tip: When dealing with means of subsets, always use the total sum of values. This helps isolate the values you need to find the mean for.

Question 7. If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is ......
(a) 9
(b) 11
(c) 13
(d) 15
Answer: (a) 9
Mean = \( \frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5} \)
\( 11 = \frac{5x+20}{5} \)
To find x, multiply both sides by 5:
\( 5x + 20 = 55 \)
\( 5x = 55 - 20 \)
\( 5x = 35 \)
\( \implies x = \frac{35}{5} \)
\( \implies x = 7 \)
Now, find the first three observations: \( x \), \( x+2 \), \( x+4 \). These are 7, 9, 11.
Mean of first 3 observations is = \( \frac{7+9+11}{3} \)
\( \implies = \frac{27}{3} \)
\( \implies = 9 \)
In simple words: First, use the average of all five numbers to find the value of \( x \). Once you have \( x \), you can find the first three numbers in the list. Then, calculate the average of just those first three numbers.

🎯 Exam Tip: Always make sure to answer the specific question asked; in this case, finding \( x \) is an intermediate step, not the final answer.

Question 8. The mean of 5, 9, x, 17 and 21 is 13, then find the value of x.
(a) 9
(b) 13
(c) 17
(d) 21
Answer: (b) 13
Mean = \( \frac{5+9+x+17+21}{5} \)
\( 13 = \frac{52+x}{5} \)
Multiply both sides by 5:
\( 65 = 52 + x \)
\( x = 65 - 52 \)
\( \implies x = 13 \)
In simple words: The average of five numbers is 13. This means if you add them all up, the total should be \( 13 \times 5 = 65 \). By adding the known numbers and subtracting from 65, you can find the missing number \( x \).

🎯 Exam Tip: Remember that the sum of observations equals the mean multiplied by the number of observations. This is key for finding missing values.

Question 9. The mean of the square of first 11 natural numbers is
(a) 26
(b) 46
(c) 48
(d) 52
Answer: (b) 46
The sum of the squares of the first n natural numbers is given by the formula \( \frac{n(n+1)(2n+1)}{6} \).
For n = 11, Sum = \( \frac{11(11+1)(2 \times 11+1)}{6} \)
\( = \frac{11 \times 12 \times 23}{6} \)
\( = 11 \times 2 \times 23 \)
\( = 506 \)
Mean = \( \frac{\text{Sum of squares}}{\text{Number of terms}} \)
Mean = \( \frac{506}{11} \)
\( \implies = 46 \)
In simple words: First, you square each of the first 11 whole numbers (1, 2, 3... up to 11). Then, you add all these squared numbers together. Finally, you divide this total sum by 11 to get the average. There's a quick formula to find the sum of squares, which makes this calculation faster.

🎯 Exam Tip: Knowing the formulas for the sum of the first n natural numbers and the sum of their squares can save a lot of time in competitive exams.

Question 10. The mean of a set of numbers is \( \bar { X } \). If each number is multiplied by z, the mean is ......
(a) \( \bar { X } +z \)
(b) \( \bar { X } -z \)
(c) \( z\bar { X } \)
(d) \( \bar { X } \)
Answer: (c) \( z\bar { X } \)
If each observation is multiplied by a constant k (where \( k \neq 0 \)), then the arithmetic mean is also multiplied by the same constant. In this case, \( z \) is the constant.
In simple words: If you take all the numbers in a group and multiply each one by the same amount, the new average will also be the old average multiplied by that same amount. It's a direct change to the mean.

🎯 Exam Tip: Remember how the mean is affected by transformations: adding/subtracting a constant shifts the mean, while multiplying/dividing by a constant scales the mean.