Question

The variance of the numbers 8, 21, 34, 47, \dots, 320, is:

Hint:

For an arithmetic progression, the variance can be calculated using the formula \( \frac{n}{12} d^2 \), where \( n \) is the number of terms and \( d \) is the common difference.

Answer 1

Correct Answer: 8788

The variance of the arithmetic sequence 8, 21, 34, 47, ..., 320 is calculated. The first term is \(a=8\), the common difference is \(d=13\), and the last term is \(l=320\).

Step 1: Determine the Number of Terms (n)

Using the nth term formula \( a_n = a + (n-1)d \), with \( a_n = 320 \):

\(320 = 8 + (n-1) \times 13\)

\(312 = (n-1) \times 13\)

\(n-1 = \frac{312}{13} = 24\)

\(n = 25\)

Step 2: Calculate the Mean (\(\bar{x}\))

The sum of the arithmetic sequence is \(S_n = \frac{n}{2}(a + l)\).

\(S_{25} = \frac{25}{2}(8 + 320) = \frac{25}{2} \times 328 = 4100\)

The mean is \(\bar{x} = \frac{S_n}{n}\).

\(\bar{x} = \frac{4100}{25} = 164\)

Step 3: Calculate the Variance (\(\sigma^2\))

The variance for an arithmetic sequence is given by \(\sigma^2 = \frac{1}{12}(n^2-1)d^2 \).

\(\sigma^2 = \frac{1}{12}(25^2-1)\times 13^2\)

\(\sigma^2 = \frac{1}{12}(624)\times 169\)

\(\sigma^2 = \frac{105456}{12}\)

\(\sigma^2 = 8788\)

The variance is 8788.