The set of numbers k - 2, k, k + 1, and k + 5 is given, where k is any real constant. What… — JAMB UTME

The set of numbers $k - 2$ , $k$ , $k + 1$ , and $k + 5$ is given, where $k$ is any real constant. What is the variance of this set of numbers?

Answer: 6.5

The variance of the given set of numbers is 6.5.

The mean of the set is

\bar{x} = k + 1

. Subtracting the mean from each data point gives deviations of

-3

-1

0

, and

4

. The squares of these deviations are

9

1

0

, and

16

, which sum to

26

. Dividing this sum by

4

gives a variance of

6.5

. A key statistical property illustrated here is that adding or subtracting a constant

k

from every value in a dataset shifts the mean by

k

but leaves measures of dispersion (such as variance and standard deviation) unchanged.

Find the mean (\bar{x}) of the given set {k - 2, k, k + 1, k + 5}.

\bar{x} = \frac{(k - 2) + k + (k + 1) + (k + 5)}{4} = \frac{4k + 4}{4} = k + 1

The mean is calculated by summing all values and dividing by the total count of numbers.

Determine the deviation of each value from the mean, (x_i - \bar{x}).

(k - 2) - (k + 1) = -3, k - (k + 1) = -1, (k + 1) - (k + 1) = 0, (k + 5) - (k + 1) = 4

Deviations measure how far each data value lies from the mean.

Square each individual deviation and sum the results.

(-3)^2 + (-1)^2 + 0^2 + 4^2 = 9 + 1 + 0 + 16 = 26

Squaring converts all deviations into non-negative values.

Divide the sum of squared deviations by the total number of observations (N = 4) to find the variance.

\text{Variance} = \frac{26}{4} = 6.5

Variance is defined as the arithmetic mean of the squared deviations from the mean.

Variance and Invariance under Constant Translation

Question