CBSE Class 9 Mathematics Measures of Central Tendency Assignment Set 05

Question. Find the mean of the marks of 10 students, having marks as, 75, 66, 69, 76, 110, 85, 52, 96, 67, 82.
Answer: Number of observations \( = 10 \)
Sum of observations \( = 75 + 66 + 69 + 76 + 110 + 85 + 52 + 96 + 67 + 82 = 778 \)
\( \text{Mean} = \frac{\text{Sum of observations}}{\text{Total number of observations}} = \frac{778}{10} = 77.8 \)

Question. Find the mean of first 15 natural numbers.
Answer: First 15 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
Sum of first \( n \) natural numbers \( = \frac{n(n + 1)}{2} \)
Sum of first 15 natural numbers \( = \frac{15 \times 16}{2} = 120 \)
\( \text{Mean} = \frac{120}{15} = 8 \)

Question. Find the mean of first 10 prime numbers.
Answer: First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Sum of first 10 prime numbers \( = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129 \)
\( \text{Mean} = \frac{129}{10} = 12.9 \)

Question. Form a discrete series of the following data, then calculate the mean.
21, 23, 22, 25, 18, 24, 20, 20, 19, 21, 17, 24, 20, 18, 25, 17, 23, 22, 22, 21.
Answer:
The discrete frequency distribution table is as follows:

\( \text{Mean} = \frac{\sum fx}{\sum f} = \frac{422}{20} = 21.1 \)

Question. Find median of 25, 34, 31, 23, 22, 26, 35, 29, 20, 32.
Answer: Arrange the data in ascending order:
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Number of observations, \( n = 10 \) (even)
\( \text{Median} = \frac{\left(\frac{n}{2}\right)\text{th term} + \left(\frac{n}{2} + 1\right)\text{th term}}{2} \)
\( \text{Median} = \frac{\text{5th term} + \text{6th term}}{2} = \frac{26 + 29}{2} = \frac{55}{2} = 27.5 \)

Question. The following data have been arranged in ascending order of magnitudes:
59, 62, 65, \( x \), \( x+2 \), 72, 85, 94. If the median of the data is 69. Find the value of \( x \).
Answer: Number of observations, \( n = 8 \) (even)
The data is already in ascending order:
59, 62, 65, \( x \), \( x+2 \), 72, 85, 94
\( \text{Median} = \frac{\left(\frac{n}{2}\right)\text{th term} + \left(\frac{n}{2} + 1\right)\text{th term}}{2} \)
\( \text{Median} = \frac{\text{4th term} + \text{5th term}}{2} \)
\( 69 = \frac{x + (x + 2)}{2} \)
\( 69 = \frac{2x + 2}{2} \)
\( 69 = x + 1 \)
\( x = 68 \)

Question. The mean of 10, 12, 16, 20, \( p \) and 26 is 17. Find the value of \( p \).
Answer: Number of observations \( = 6 \)
\( \text{Mean} = \frac{10 + 12 + 16 + 20 + p + 26}{6} = 17 \)
\( \frac{84 + p}{6} = 17 \)
\( 84 + p = 102 \)
\( p = 18 \)

Question. Find the mode of the following data, in each case
(i) 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
(ii) 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
Answer:
(i) In the data: 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
The number 14 occurs 4 times, which is the highest frequency.
Therefore, the Mode is 14.

(ii) In the data: 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
The number 7 occurs 5 times, which is the highest frequency.
Therefore, the Mode is 7.

Question. The demand of different shirt sizes are given below. Find the modal shirt sizes

Answer: The modal shirt size is the size with the maximum number of persons (highest frequency). Here, the maximum frequency is 39, which corresponds to the shirt size 39. Therefore, the modal shirt size is 39.

Question. Prove \( \sum_{i=1}^{n} (x_i - \bar{x}) = 0 \), \( \bar{x} \) is the mean of \( n \) observations, \( x_1, x_2, x_3, \dots, x_n \)
Answer:
We know that the mean \( \bar{x} \) of \( n \) observations \( x_1, x_2, x_3, \dots, x_n \) is given by:
\( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \)
Multiplying both sides by \( n \), we get:
\( \sum_{i=1}^{n} x_i = n\bar{x} \)

Now, consider the LHS of the equation to be proved:
\( \sum_{i=1}^{n} (x_i - \bar{x}) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} \bar{x} \)

Since \( \bar{x} \) is a constant value representing the mean, summing it \( n \) times gives:
\( \sum_{i=1}^{n} \bar{x} = n\bar{x} \)

Substituting these values back into the equation:
\( \sum_{i=1}^{n} (x_i - \bar{x}) = n\bar{x} - n\bar{x} = 0 \)
Hence Proved.