Question

The variance of the first 10 natural numbers which are multiples of 3 is:

• A. \( 53 \)
• B. \( 73 \)
• C. \( 52.5 \)
• D. \( 74.25 \)

Hint:

For arithmetic sequences, use the sum formula to quickly compute the mean. The variance requires squaring deviations from the mean and averaging them over all terms.

Answer 1

The Correct Option is D

Step 1: Identify the data set 
The first 10 natural numbers that are multiples of 3 are: \[ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 \] This forms an arithmetic sequence where: - First term (\( a \)) = 3 - Common difference (\( d \)) = 3 - Number of terms (\( n \)) = 10 

Step 2: Compute the mean (\( \mu \)) 
The mean of a sequence is given by: \[ \mu = \frac{\sum x_i}{n} \] Using the sum formula for an arithmetic sequence: \[ S_n = \frac{n}{2} (2a + (n-1)d) \] \[ S_{10} = \frac{10}{2} (2(3) + (10-1)3) \] \[ = 5(6 + 27) = 5(33) = 165 \] \[ \mu = \frac{165}{10} = 16.5 \] 

Step 3: Compute the variance (\( \sigma^2 \)) 
Variance is given by: \[ \sigma^2 = \frac{1}{n} \sum (x_i - \mu)^2 \] Computing the squared differences: \[ (3 - 16.5)^2 = (-13.5)^2 = 182.25 \] \[ (6 - 16.5)^2 = (-10.5)^2 = 110.25 \] \[ (9 - 16.5)^2 = (-7.5)^2 = 56.25 \] \[ (12 - 16.5)^2 = (-4.5)^2 = 20.25 \] \[ (15 - 16.5)^2 = (-1.5)^2 = 2.25 \] \[ (18 - 16.5)^2 = (1.5)^2 = 2.25 \] \[ (21 - 16.5)^2 = (4.5)^2 = 20.25 \] \[ (24 - 16.5)^2 = (7.5)^2 = 56.25 \] \[ (27 - 16.5)^2 = (10.5)^2 = 110.25 \] \[ (30 - 16.5)^2 = (13.5)^2 = 182.25 \] Summing these: \[ 182.25 + 110.25 + 56.25 + 20.25 + 2.25 + 2.25 + 20.25 + 56.25 + 110.25 + 182.25 = 742.5 \] \[ \sigma^2 = \frac{742.5}{10} = 74.25 \] Thus, the correct answer is: \[ \mathbf{74.25} \]